Transmutation of elements

Question

Was it found by Rutherford and Chadwick that oxygen and carbon cannot be transmuted by bombarding them with alpha particles???


I searched internet and got this one.

Russell was able to demonstrate the transmutation of gases in the Bloomfield, New Jersey research laboratory at the Westinghouse Lamp Company on Sept. 30, 1927. Transmutation of hydrogen and oxygen to nitro- gen, and, nitrogen to oxygen and hydrogen was accomplished.
See reference
http://padrak.com/ine/NEN_4_11_3.html

Early Atomic Models

The idea that all matter is made of very small indivisible particles is very old.

Robert Boyle’s study of compression and expansion of air brings out the idea that air is made of tiny particles with lot of empty space between the particles.

The smallest unit of an element which carries all the properties of the element is called an atom.

Experiments on discharge tube, measurement of e/m by Thomson etc. established the existence of negatively charged electrons in the atoms.

Because atoms are electrically neutral, a search for the positive charge inside the atom was started.

Thomson’s Model of the atom

Thomson (1898) suggested that the atom is a positively charged solid sphere in which electrons are embedded in sufficient number to create a neutral atom. This model of the atom could explain why only negatively charged particles are being emitted when a metal is heated. This model was also useful to explain the formation of ions and ionic compounds.

Lenard's Suggestion on Atomic Structure

Lenard observed that cathode rays are passing through thin material without any deviation. According to him, this was so because, there is a lot of empty space in atoms. Hence the positive charged particles are also tiny like electrons.

Hydrogen Spectra

If hydrogen gas is enclosed in a sealed tube and heat to high temperatures, it emits radiation. If this radiation is passed through a prsim, components are different wavelengths get deviated by different amounts and we get the hydrogen spectra on a screen. In the spectra of hydrogen atom, it is observed that light of wavelength 656.3 nm and then light of wave length 486.1 nm are present. Hydrogen atoms do not emit any radiation between 656.3 nm and 486.1 nm. Similarly radiation is observed at 434.1 nm and 4202.nm.

In the invisible region also, there is radiation emitted by the hydrogen atom at discrete wavelengths.

The wavelengths nicely fit the equation

1/λ = R [1/n² - 1/m²]

where R = 1.09737*10^7 m^-1.
n and m are integers with m>n.

The spectrum in the ultraviolet region is called Lyman series and you get the series by setting n = 1.

The hydrogen spectrum in the visible region is called Balmer series and you get the series by setting n = 2.

The hydrogen spectrum in infrared region is called Paschen series and you get the series by setting n= 3.

Rutherford's Model

Rutherford experimented with alpha rays or particles. When he bombarded gold foils with alpha particles, Many went without deviation, some had some deviation and some were deflected by more than 90 and came back. Hence he made a conclusion that there was a particle with a mass equivalent to alpha particle inside the atom. The mass of an atom is concentrated in this particle.

The size of this particle was also estimated by Rutherford. Its linear size is 10 fermi ( 1 fermi is equal to 1 femtometre = 10^-15 m).

Rutherford proposed that the atom contains a positively charged tiny particle called nucleus. It contains the entire mass of the atom. Outside this nucleus, at some distance, electrons move around. The positive of charge of nucleus is exactly equal to the negative charge of the electrons of the atom.

Because electrons are very very light compared to the nucleus, due to heat only electrons come out.

Movement of electrons is to brought in and the coulomb force between the nucleus and the electron is assumed to provide only centripetal force to make the electron rotate in a circular motion.

Difficulties with Rutherford’s Model

Rutherford’s model assumes that the electron rotates around the nucleus. Maxwell’s equations of a electromagnetism show that accelerated electron must continuously emit electromagnetic radiation. But a hydrogen does not emit radiation at ground level energy or normal energy. It emits radiation only when heated. Also, if it emits radiation it will lose energy and the radius of its circular motion will decreases and finally it will fall into the nucleus. Hence, the atomic model proposed by Rutherford needs modification. Bohr proposed such modifications.

Bohr's Postulates and Model

1. The electrons revolve around the nucleus in circular orbits.
2. the orbit of the electron around the nucleus can be only some special values of radius. In these special radii orbits, the electron does not radiate energy as expected from Maxwell’s laws. These orbits are called stationary orbits.
3. The energy of the atom has a definite value when electrons are in a given stationary orbit. But the if more energy is provided to the atom, the electron can jump from one stationary orbit to another stationary orbit of higher energy. If it jumps from an orbit of higher energy (E2) to an orbit of lower energy (E1), it emits a photon of radiation. The energy of the emitted photon will be E2 – E1.

The wave length of the emitted radiation is given by the Einstein-Planck equation

E2-E1 = hυ = hc/λ

4. In stationary orbits, the angular momentum l of the electron about the nucleus is an in integral multiple of the Planck constant h divided by 2 π.

l = nh/2 π

This assumption is called Bohr’s quantization rule.

Energy of an Hydrogen Atom

Assume that the nucleus has a positive charge Ze ( there are z protons each with positive charge e).

By equating the coulomb force acting between Ze and e to the centripetal acceleration mv²/r, we get r the radius at which the electron revolves.

r = Ze²/4π ε₀v²

From Bohr’s quantization rule,

mvr = nh/2 π
where n is a positive integer

Eliminating v from both the equations we get

r = ε₀h²n²/πmZe²

We get expression for v as

v = Ze²/2 ε₀hn


Hence allowed radii are proportinal to n² and for each value of n = 1,2,3…we allowed orbits.

The smallest radius orbit will have n = 1.



As we have expression for v, we can give an expression for kinetic energy when electron is in nth orbit is

K = ½ mv² = mZ²e⁴/8 ε₀²h²n²

The potential energy of the atom is

V = - Ze²/4π ε₀r = -mZ²e⁴/4ε₀²h²n²

The expression for potential energy is obtained by assuming the potential energy to be zero when the nucleus and the electron are widely separated.

The total energy of the atom is

E = K+V = - mZ²e⁴/8 ε₀²h²n²

When an atom is n^th stationary orbit, it is said to be in the n^th energy state.

In giving an expression for the total energy of the atom, kinetic energy of the electron and potential energy of the electron-nucleus pair are considered. Kinetic energy of the nucleus is assumed to be negligible.

Bohr’s postulates can be used to find the allowed energies of the hydrogen atom when its single electron is in various stationary orbits. The methodology can be used any hydrogen like ions which have only one electron.. Therefore it is valid for He⁺, Li⁺⁺, Be⁺⁺⁺ etc.

Radii of Different Orbit of Hydrogen Like Ion

r the radius at which the electron revolves.

r = Ze²/4π ε₀v²


For hydrogen, z =1 and we get r1 as 53 picometre ( 1pm = 10^-12 m) or 0.053 nm. This length is called the Bohr radius and is a convenient unit for measuring lengths in atomic physics. It is denoted by me as the symbol a₀ (In HC Verma a different symbol is given. I am using this symbol as a convenience).

In terms of Bohr radius the second allowed radius is 4 a₀ and third is 9 a₀ and so on. In general nth orbit of hydrogen atom is n²a₀.

Hydrogen Ion - Ground and Excited States

The state of an atom with the lowest energy is called its ground state.

The states with higher energies are called excited states.
Energy of hydrogen atom in the ground state is -13.6 eV.
Energy of hydrogen atom in the next excited state, that is n = 2 state is -3.4 eV.

Hydrogen Spectra

When heated some atoms in the hydrogen become excited and when electrons jump from higher energy levels to lower energy levels in those excited atoms, photons with specific wavelengths are emitted or radiated.

If an electron jumps from mth orbit to nth orbit (m>n), the energy of the atom gets reduced from Em to En. The wavelength of the emitted radiation will be

1/ λ = (Em – En)/hc = RZ²{1/n² - 1/m²]

where R is the Rydberg constant.
R = 1.0973*10^7 m^-

In terms of the Rydberg constant total energy of the atom in the nth state is E = -RhcZ²/n²

For hydrogen atom, when n =1, E = -Rhc and we know its value is -13.6 eV.
Energy of 1 rydberg means -13.6 eV.
Rhc = 13.6eV

Hydrogen Spectra - Series Structure

Lyman series:
All transitions to n =1 state from higher state give the radiation in Lyman series.

Jumping of the electron from n =2 to n =1 gives
1/ λ = R[1 – 1/2²] = R(1 – ¼) which will give λ = 121.6 nm.

Jumping of the electron from n = ∞ to n = 1 gives

1/ λ = R[1 – 1/∞²] = R(1 -0) which gives λ = 91.2 nm.

Balmer seires
All transitions to n = 2 from higher states given radiations within the range of 656.3 nm and 365.0 nm. These wavelengths fall in the visible region.

Paschen series
The transitions or jumps to n = 3 from higher energy levels give Paschen series in the range 1875 nm to 822 nm.

Ionization Potential

The energy of the hydrogen atom in ground state is -13.6 eV. If we supply more than 13.6 eV to the hydrogen atom, the electron and the nucleus get separated and electron moves with some kinetic energy independently (Remember plasma in nuclear fusion).
The minimum energy needed to ionize an atom is called ionization energy. The potential difference through which an electron should be accelerated to acquire this much energy is called ionization potential.

Thus ionization energy of hydrogen atom in ground state is 13.6 eV and ionization potential is 13.6 V.

Binding Energy

Binding energy of a system is defined as the energy released when its constituents are brought from infinity to form the system. Now we know that binding energy of a hydrogen atom is 13.6 eV. The energy is zero when the electron and nucleus at infinite distance. When the electron is brought into n = 1 orbit, the energy becomes -13.6 eV and hence 13.6 eV is released which is the binding energy.

Excitation potential

The energy needed to take the atom form its ground state to an excited state is called the excitation energy of that excited state.

As the hydrogen atom’s ground state energy is -13.6 eV and its energy when electron is in n =2 orbit is -3.4 eV, we have to supply 10.2 eV to excite a hydrogen atom to its first excited state which is electron in n = 2 orbit.

The potential through which an electron should be accelerated to acquire the excitation energy is the excitation potential.

The excitation potential needed bring hydrogen to its first excited state is 10.2 V.

Limitations of Bohr Model

Maxwell’s theory of electromagnetism is not replaced or refuted in Bohr's model but it is arbitrarily assumed that in certain orbits, electrons get the licence to disobey the laws of electromagnetism and are allowed not to radiate energy.

Quantum Mechanics of the Hydrogen Atom

Electron has a wave character as well as a particle character. The wave function of the electron ψ(r,t ) is obtained by solving Schrodinger’s wave equation. The probability of finding an electron is high where | ψ(r,t )|² is greater. Not only the information about the electron’s position but information about all the properties including energy etc. that we calculated using the Bohr’s postulates are contained in the wave function of ψ(r,t).

Quantum Mechanics of the Hydrogen Atom

The wave function of the electron ψ(r,t) is obtained from the Schrodinger’s equation

-(h²/8π²m) [∂²ψ /∂x² + ∂²ψ /∂y² + ∂²ψ/∂z²] - Ze²ψ/4πε₀r = E ψ


where
(x.y,z ) refers to a point with the nucleus as the origin and r is the distance of this point from the nucleus.
E refers to the energy.
Z is the number of protons.

There are infinite number of functions ψ(r,t) which satisfy the equations.

These functions may be characterized by three parameters n,l, and m_l.

For each combination of n,l, and m_l there is an associated unique value of E of the atom of the ion.

The energy of the wave function of characterized by n,l, and m_l depends only on n and may be written as


En = - mZ²e⁴/8 ε₀²h²n²

Production of X-rays

When highly energetic electrons are made to strike a metal target, electromagnetic radiation comes out. A large part of this radiation has wavelength of the order 0.1 nm (appx 1 A) and is known as X-ray.

Continuous and Characteristic X-rays

If the X-rasy coming from a coolidge tube are examined for wavelengths present, and the intensity of different wavelengths are measurea and plotted, we can observe that there is a minimum wavelength below which no X-ray is emitted.

The wavelength below which no X-rays are emitted is called the cut-off wavelength or the threshold wavelength.

From the plot it can also be observed that at certain sharply defined wavelengths, the intensity of X-rays is very large. These X-rays are called characteristic X-rays.

At other wavelengths the intensity varies gradually and these X-rays are called continuous X-rays.

K X-Rays

X-rays emitted due to electronic transition from a higher energy state to a vacancy created in the K shell are called K X-rays.

Soft and Hard X-rays

The X-rays of low wave length are called hard X rays and X rays of large wave length are called soft X rays.

In terms of energy, harder means more energy in is each photon.

Moseley's law

Square root of frequency of X rays = a(Z-b)

√(v) = a(Z-b)

where Z = position number of element.
a and b are constants

Bragg’s law

2d sin θ = n λ

d = interplanar spacing of the crystal on which X-rays are incident
θ = is the incident angle at which X-rays are strongly reflected.
n = 1,2,3 …
λ = wave length of X-rays

Application of Bragg’s law:By using a monochromatic X-ray beam (having a single wave length) and noting the angles of strong reflection, the interplanar spacing d and several information about the structure of the solid can be obtained.


(Bragg Bday 2 July 1862)

Updated 23 Nov 2015, 9 Nov 2008

Properties of X rays

X rays are electromagnetic waves of short wave lengths.
So they have many properties common with light.

1. They travel in straight lines in vacuum at a speed equal to that of light.
2. They are diffracted by crystals according to Bragg's law.
3. x-rays do not contain charged particles. hence they are not deflected by electric or magnetic field.
4. They effect a photographic plate. The effect is stronger than light.

Properties which are different than light

1. When incident on certain materials barium platinocyanide, X rays cause fluorescence.
2. When passed through a gas, X rays ionize the molecules of the gas.

Atomic nucleus

Properties of Nucleus

A nucleus is made of protons and neutrons.

Studies have shown that average nucleus R of a nucleus may be written as

R = R0A^(1/3) .. (46.1)

where R0 = 1.1*10^-15 m ≈ 1.1 fm and A is the mass number

Density within a nucleus is independent of A.

Nuclear Forces

When nucleons are kept at a separation of the order of femtometre (10^-15 m), a new kind of force, called nuclear force starts acting.

Binding Energy

If the constituents of a hydrogen atom (a proton and an electron) are brought from infinity to form the atom, 13.6 3V of energy is released. Thus, the binding energy of a hydrogen atom in ground state is 13.6 eV. Also 13.6 eV energy must be supplied to the hydrogen atom in ground state to separate the constituents to large distances.

Similarly, the nucleons are bound together in a nucleus and energy must be supplied to the nucleus to separate the constituent nucleons to large distances. The amount of energy needed to do this is called the binding energy of the nucleus. If nucleons are brought together to form the nucleus from large separation this much energy is released.

It is evident from the above discussion that the rest mass energy of a nucleus is smaller than the rest mass energy of its constituents.

Radioactive decay

Two main processes by which an unstable nucleus decays are alpha decay and beta decay.

Alpha decay

In alpha decay, the unstable nucleus emits an alpha particle reducing its proton number Z as well as its neutron number N by 2. As the proton number is changed, the element itself is changed and hence the chemical symbol of the residual nucleus is different from that of the original nucleus (Parent nucleus is original nucleus and the resulting nucleus due to decay is called daughter nucleus).

Alpha decay occurs in all nuclei with mass number A>210.

Beta Decay

Beta decay is a process in which either a neutron is converted into a proton or a proton is converted into a neutron.
When a neutron is converted into a proton, an electron and a new particle named antineutrino are created and emitted from the nucleus. The electron emitted from the nucleus is called a beta particle and is denoted by the symbol β-.


If the unstable nucleus has excess protons than needed for stability, a proton converts itself into a neutron. In the process, a positron and a neutrino are created and emitted from the nucleus.

Positron is represented by e+. The neutrino is represented by ν.

When a positron and electron collide, both the particles are destroyed and energy is made available.

The decay which gives beta rays consisting of positrons is called beta plus decay


Electron capture

A nucleus captures one of the atomic electrons, most likely an electron from the K shell, and a proton in the nucleus combines with the electron and converts itself into a neutron. A neutrino is created in process and emitted from the nucleus. So a combination of proton and electron results in neutron and neutrino.

Gamma Decay

When a daughter nucleus is formed due to alpha or beta decay, the nucleus may be at higher energy level compared to its ground or normal state. The electromagnetic radiation emitted in nuclear transitions from higher energy or excited state to ground state is called gamma ray.

Law of Radioactive decay;

N = N0e- λ t

where
N = number of active nuclei at time t

N0 = number of active nuclei at t = 0.

λ = decay constant


-dN/dt = λN

-dN/dt gives the number of decays per unit time and is called the activity (A) of the sample

A = λN

A = A0e- λ t

Decay constant;

Law of radioactive decay

N = N0e- λ t

where
N = number of active nuclei at time t

N0 = number of active nuclei at t = 0.

λ = decay constant

Half-life and Mean life;

Half life:

The time elapased before half the active nuclei decay is called half-life and is denoted by t1/2.

t1/2. = 0.693/ λ

where
λ = decay constant.

Average life of the nuclei of a material

tav. = t1/2/0.693

Properties and Uses of Nuclear Radiation

Properties and Uses of Nuclear Radiation

Alpha Ray
1. Each particle contains two protons and two neutrons. It is a helium nucleus.
2. It is made of positive particles and hence deflected by electric field as well as magnetic field.
3. Its penetrating power is low. Few cm in air also.
4. They travel at large speeds of the order of 10^6 m/s.
5. All particles from a source and decay scheme have the same energy.
6. Alpha rays produce scintillation (flashes of light) when they strike certain fluorescent materials such as barium platinocynide.
7. It causes ionization in gases.




Beta ray

1. It is a stream of electrons. Electrons are created during nuclear transformation.
2. They are negative particles and hence deflected by electric as well as magnetic fields.
3. Penetrating power greater than alpha rays. They can travel several meters in air before its intensity drops down to small values.
4. Ionizing power is less than alpha rays.
5. beta rays also produced scintillation but it is weak.
6. The energy of particles is not uniform as they share energy with antineutrinos. Energy of beta particles varies from zero to a maximum


Beta plus ray

It has all the properties of beta rays or beta negative rays, except that it is made of positively charged particles.

Gamma Ray

1. Gamma ray is an electromagnetic radiation of short wavelength. Its wavelength is shorted than X-rays.
2. Many properties are similar to X-rays.
3. As there is no charge no deflection in electric or magnetic fields.
4. All the photons coming from a particular gamma decay scheme has the same energy.
5. As it is electromagnetic wave, gamma ray travels with the velocity ‘c’ in vacuum.

Binding energy and its calculation

Energy from the Nucleus

Nuclear energy may be obtained either by breaking a heavy nucleus into two nuclei of middle weight (fission) or by combining two light nuclei to form a middle weight nucleus (fusion).

Reason: The middle weight nuclei are more tightly bound than heavy weight nuclei. When the nucleons of a heavy nucleus regroup in two middle weight nuclei called fragments the total binding energy increases and the rest mass energy decreases. The difference in energy appears as the kinetic energy of the fragments or in some other form.

In the case of fusion, the light weight nuclei are less tightly bound than the middle weight nuclei. Therefore, if two light weight nuclei combine, the binding energy increases and the rest mass decreases. Energy is released in the form of kinetic energy or in some other external form.

Nuclear Fission

The rest mass energy of the heavy nucleus represented by E1 is greater than the rest mass energy of the fragments represented by E3. But the energy level of the heavy nucleus is to be increased to E2 to get the fission process started according to classical physics.

But according to quantum mechanics, fission can take place even if no external energy is given. Such a fission process is termed as barrier penetration. The amount of energy created and the time for which it is created through a barrier penetration process are related through Heisenberg uncertainty relation.

∆E. ∆t ≈ h/2 π

Where h is the Planck constant

Barrier penetration is possible but is not easy.

Uranium Fission Reactor

Breeder Reactors

92238U can capture neutrons and become
92239U. On B radiation it becomes
93239Np. On beta radiation it becomes
94239Pu.


94239Pu when hit by neutron becomes 94240Pu, which is a fissionable material. Thus if out of the 2.47 neutrons produced on average in fission reaction, one neutron is absorbed by 238U we produce fuel equivalent to what is consumed. Such a reactor is called breeder reactor.

Nuclear Fusion

For he light nuclei to come together with in a distance of 1 fm (femtometer), we need a temperature of the order of 10^9 K. At that temperature electrons are completely detached from atoms and only nuclei remain. It is called plasma. In Sun, the temperature is 1.5*10^7 K and fusion is taking place. So fusion can take place due to barrier penetration process at 10^7 K.

Fusion in Laboratory

The major problem on earth for fusion reaction is holding plasma at high temperature for extended period of time.

Lawson criterion for fusion reactor

In order to get an energy output greater than the energy input, a fusion reactor should achieve

n τ >10^14 s/cm³

where
n = the density of the interacting particles

τ = confinement time

The quantity n τ in s/cm³ is called Lawson number

Tokamak Design

In this design, the deuterium plasma is contained in a toroidal region by specially designed magnetic field. The directions and magnitudes of the magnetic field are so managed in the toroidal space that whenever a charge plasma particle attempts to go out qv×B force tends to push it back into the toroidal volume.

With such designs, confinement of the plasma has been achieved for short duration of few microseconds.

A large fusion machine known as Joint European Torus (JET) is designed to achieve fusion energy on this principle.

At the Institute of Plasma Research (IPR) Ahmedabad, a small machine named Aditya is functioning on the Tokamak design. This machine is being used to study properties of plasma.

Inertial Confinement

It is an alternate method of confinement of plasma. A small solid pellet is made that contains deuterium and tritium. Intense laser beams are directed on the pellet from many directions on all the sides. The laser vaporizes the pellet converting it into plasma and then compresses it. The density increases by 10^3 to 10^4 time the initial density and temperature raises to high values. Fusion occurs in these conditions. In this method also so far, confinement for very small duration is only achieved.

The source of fuel for fusion is water only and water is abundant in oceans. Also these reactions do not result in radioactive emissions like that of fission.

46. Nucleus - Revision Facilitator

Recollect points under the topic

If required right click on the topic if link is provided, read the material, close it and come back.


46.1 Atomic nucleus;

46.2 Nuclear Forces

46.3 Binding Energy

46.4 Radioactive decay

46.5 Law of Radioactive decay;

Decay constant

Half-life and Mean life

46.6 Properties and Uses of Nuclear Radiation

46.7 Energy from the Nucleus

46.8 Nuclear Fission

46.9 Uranium Fission Reactor

46.10 Nuclear Fusion

46.11 Fusion in Laboratory



http://iit-jee-physics.blogspot.com/2008/03/concept-review-ch46-nucleus.html

1. Introduction to Physics - Revision Facilitator

Sections

1. Introduction to physics

1.1 What is Physics
1.2 Physics and Mathematics
1.3 Units
1.4 Definitions of base units
1.5 Dimension
1.6 Uses of dimension
1.7 Order of magnitude
1.8 The structure of world

2. Physics and Mathematics - Revision Facilitator

2. Physics and mathematics

Sections

2.1 Vectors and scalars
2.2 Equality of vectors
2.3 Addition of vectors
2.4 Multiplication of a vector by a number
2.5 Subtraction of vectors
2.6 Resolution of vectors
2.7 DCT product or scalar product of two vectors
2.8 Cross product or vector product of two vectors
2.9 Differential calculus: dy/dx as rate measure
2.10 Maxima and Minima
2.11 Integral calculus
2.12 Significant digits
2.13 Significant digits in calculations
2.14 Errors in measurements

3. Rest and Motion: Kinematics - Revision Facilitator

3. Rest and Motion: Kinematics


Sections

3.1 Rest and Motion
3.2 Distance and displacement
3.3 average speed and instantaneous speed
3.4 Average velocity and instantaneous velocity
3.5 Average acceleration and instantaneous acceleration
3.6 Motion in a straight line
3.7 Motion in a plane
3.8 Projectile motion
3.9 Change of frame

Mind Map

Rest - Motion – displacement – Speed – Velocity – Acceleration – Frame of Reference

Displacement – Distance moved

Speed - average speed - instantaneous speed

Velocity - Average velocity - instantaneous velocity – Acceleration

Acceleration - Average acceleration - instantaneous acceleration

Motion - straight line - Motion in a plane - Projectile motion

Frame of Reference – Change in Frame of Reference

4. The forces - Revision Facilitator

4. The forces

Sections

4.1 Introduction
4.2 Gravitational forces
4.3 Electromagnetic (EM) forces
4.4 Nuclear Forces
4.5 Weak forces
4.6 Scope of Classical physics


Mind Map

Forces - Gravitational forces - Electromagnetic (EM) forces - Nuclear Forces - Weak forces - Scope of Classical physics

Gravitational forces - G (universal constant 6.67 *106-11 N-m^2/kg^2) – Acceleration due to gravity g = GM/R^2) - Spherical body treated as a point mass at their centres

Electromagnetic (EM) forces – Coulomb forces – Forces between two surfaces in contact – Tension in a string or rope – Force due to a spring

Nuclear Forces – Exerted when interacting particles are protons or neutrons

Weak forces – Forces responsible for beta decay – antinutrino - positron

Scope of Classical physics – Applicable to bodies of linear sizes greater than 10^-6 m – Subatomic bodies – Quantum physics applicable – If the velocity of bodies are comparable to 3*10^* m/s relativistic mechanics is applicable.

26. Laws of Thermodynamics - Revision Facilitator

Recollect some points for each topic.
If you want to read about the topic, right click on the link to open it in a new window, read it, close it and come back here(If a link is there).


26.1 The first law of thermodynamics
26.2 Work done by a gas
26.3 Heat engines
26.4 The second law of thermodynamics
26.5 Reversible and irrerversible processes
26.6 entropy
27.7 Carnot engine

Links will be created in due course.

29. Electric Field and Potential - Revision Facilitator

Recollect some points for each topic.
If you want to read about the topic, right click on the link to open it in a new window, read it, close it and come back here(If a link is there).


29.1 What is electric charge?
29.2 Coulomb's law
29.3 electric field
29.4 Lines of electric force
29.5 Electric potential energy
29.6 Electric potential
29.7 Electric potential due to a point charge
29.8 Relation between electric field and potential
29.9 Electric dipole
29.10 Torque on an electric dipole placed in an lectric field
29.11 Potential energy of a diple placed in a uniform electric field
29.12 Conductors, insulators nad semiconductors
29.13 The electric field inside a conductor

Links will be created in due course.

Full chaper concept review

http://iit-jee-physics.blogspot.com/2008/03/concept-review-ch-29-electric-field-and.html

30. Gauss's Law - Revision Facilitator

Recollect some points for each topic.
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30.1 Flux of electric field through a surface
30.2 Solid angle
30.3 Gauss’s law and its derivation from Couloms’s law
30.4 Gauss's law's application
30.5.Spherical charge distributions
30.6 Earthing a conductor

Links will be created in due course.

Concept review of the full chapter

http://iit-jee-physics.blogspot.com/2008/03/concept-review-ch30-gausss-law.html

31. Capacitors - Revision Facilitator

Recollect some points for each topic and right click on the link to open it in a new window, read it, close it and come back here.

31.1 Capacitance;
31.2 Calculation of capacitance
31.3 Capacitors in series and parallel;
31.4 Parallel plate capacitor
31.5 Energy stored in a capacitor.
31.6 Dielectrics
31.7 Parallel plate capacitor with dielectrics;
31.8 An alternative form of Gauss's law
31.9 Electric field due to a point charge q placed in a an infinite dielectric
31.10 Energy in the Electric field in a dielectric
31.11 Corona discharge
31.12 High voltage generator


Links will be created in due course.

32. Electric Current in Conductors - Revision Facilitator

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32.1 Electric current and current density
32.2 Drift speed
32.3 Ohm's law
32.4 Temperature dependence of resistivity
32.5 Battery and EMF
32.6 Energy transfer in an electric current
32.7 Kirchhoff's Laws
32.8 Combination of resistors in series and parallel
32.9 Grouping of batteries
32.10 Wheatstone bridge
32.11 Ammeter and Voltmeter
32.12 Stretched wire potentiometer
32.13 Chargin and discharging of capactiros
32.14 Atmospheric electricity

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33. Thermal and Chemical Effects of Electric Current - Revision Facilitator

Recollect some points for each topic and right click on the link to open it in a new window, read it, close it and come back here.

33.1 Joule's law of heating
33.2 Verification of Joule's Laws
33.3 Seebeck effect
33.4 Peltier effect
33.5 Thomson effect
33.6 Explanation Seebeck, Peltier, Thomson effects
33.7 Electrolysis
33.8 Faraday's Laws of electrolysis
33.9 Voltameter or Coulomb meter
33.10 Primary and Secondary cells
33.11 Primary cells
33.12 Secondry Cell: Lead accumulator

Links will be created in due course.

34. Magnetic Field - Revision Facilitator

Recollect some points for each topic and right click on the link to open it in a new window, read it, close it and come back here.

1. Introduction
2. Definition of Magnetic field
3. Relation between electric and magentic fields
4. Motion of a charged aprticle in a uniform magentic field
5. Magnetic force on a current carrying wire
6. Torque on a current loop.

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35. Magnetic Field Due to a Current - Revision Facilitator

Recollect some points for each topic and right click on the link to open it in a new window, read it, close it and come back here.

35.1 Bio Savart Law
35.2 Magnetic field due to current in a straight wire
35.3 Force between parallel currents
35.4 Field due to a circular current
35.5 Ampere's law
35.6 Magnetic field at a point due to a long straight current
35.7 Solenoid

Links will be created in due course.

Ch. 36 PERMANENT MAGNETS - Revision Facilitator

38. Electro Magnetic Induction - Revision Facilitator

Try to recollect and if required right click on the link, open in separate window and read. Close it and come back to this page.

38.1 Faraday's law,
38.2 Lenz's law;
38.3 The origin of induced emf
38.4 Eddy Current
38.5 Self and mutual inductance;
38.6 RC, LR and LC circuits with d.c. and a.c. sources.
38.7 Energy stored in an inductor
38.8 mutual inductance;
38.9 Induction coil

Links to various topic will be provided for facilitating revision if required

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Ch.43 Bohr's Model and Physics of the Atom

43.1 Early atomic models
43.2 Hydrogen Spectra
43.3 Difficulties in Rutherford Model
43.4 Bohr's model
43.5 Limitations of Bohr's Model
43.6 The wave Function of an Electron
43.7 Quantum Mechanics of the Hydrogen Atom
43.8 Nomencluature in Atomic Physics
43.9 Laser

44. X-Rays - Revision Facilitator

Try to recollect relevant points on the topic and if required right click on the topic if link is given and open in a new window to read the relevant material. Close the window and come back.


Production of X-rays

Continuous and Characteristic X-rays

Soft and Hard X-rays

Moseley's law

Bragg’s law


Properties of X rays


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JEE Physics Revision Facilitator - All Chapters Combined

Have a quick review to recollect the material for each section.

1. Introduction to physics

1.1 What is Physics
1.2 Physics and Mathematics
1.3 Units
1.4 Definitions of base units
1.5 Dimension
1.6 Uses of dimension
1.7 Order of magnitude
1.8 The structure of world

2. Physics and mathematics

2.1 Vectors and scalars
2.2 Equality of vectors
2.3 Addition of vectors
2.4 Multiplication of a vector by a number
2.5 Subtraction of vectors
2.6 Resolution of vectors
2.7 DCT product or scalar product of two vectors
2.8 Cross product or vector product of two vectors
2.9 Differential calculus: dy/dx as rate measure
2.10 Maxima and Minima
2.11 Integral calculus
2.12 Significant digits
2.13 Significant digits in calculations
2.14 Errors in measurements


3. Rest and Motion: Kinematics

3.1 Rest and Motion
3.2 Distance and displacement
3.3 average speed and instantaneous speed
3.4 Average velocity and instantaneous velocity
3.5 Average accleration and instantaneous aceleration
3.6 Motion in a straight line
3.7 Motion in a plane
3.8 Projectile motion
3.9 Change of frame

4. The forces

4.1 Introduction
4.2 Gravitational forces
4.3 Electromagnetic (EM) forces
4.4 Nuclear Forces
4.5 Weak forces
4.6 Scope of Classical physics


5. Newtons laws of motion

5.1 Newton's First law
5.2 Newton's second law
5.3 Working with Newton's laws
5.4 Newton's third law of motion
5.5 Pseudo forces
5.6 The Horse and the cart
5.7 Inertia


6. Friction

6.1 Friction as the component of contact force
6.2 Kinetic friction
6.3 Static friction
6.4 Laws of friction
6.5 Understanding friction at atomic level
6.6 A Laboratory method to measure


7. Circular motion

7.1 Angular variables
7.2 Unit vectors along the radius and the tangent
7.3 Acceleration in circular motion
7.4 Dynamics of circular motion
7.5 Circular turnings and banking of roads
7.6 Centrifugal force
7.7 Effect of earth's rotation on apprarent weight

8. Work and energy

1. Kinetic enery
2. Work and work energy theorem
3. Calculation of work done
4. Work energy theorem for a system of particles
5. Potential energy
6. Conservative and nonconservative forces
7. Definition of Potential energy and conservation of mechanical energy
8. Change in the potential energy in a rigid body motion
9. Gravitational potential energy
10. Potential energy oif a compressed or extended spring
11. Different forms of energy: Mass energy equivalence

9. Centre of mass,linear momentum,collision

1. Centre of mass
2. Centre of mass of continuous bodies
3. Motion of the Centre of mass
4. Linear momentum and its conservation principle
5. Rocket propulsion
6. Collision
7. Elastic collision in one dimension
8. Perfectly inelastic collision in one dimension
9. Coefficient of restitution
10. Elastic collision in two dimensions
11. Impulse and impulsive force

10. Rotational mechanics

1. Rotation of a rigid body
2. Kinematics
3. Rotational dynamics
4. Torque of a force about the axis of rotation
5. Г = Iα
6. Bodies in equilibrium
7. Bending of a cyclist on a horizontal turn
8. Angular momentum
9. L = Iα
10. Conservation of angular momentum
11. Angular impulse
12. Kinetic energy of a rigid body rotating about a given axis
13. Power delivered and work done by a torque
14. Calculation of moment of inertia
15. Two important theorems on moment of inertia
16. Combined rotation and translation
17. Rolling
18. Kinetic energy of a body in combined rotation and translation
19. Angular momentum of a body in combined rotation and translation
20. Why does a rolling sphere slow down

Solutions of Problems in H C Verma's Books

Today I bought the two books "Solutions of Concepts of Physics" containing solutions of problems in given in H C Verma's Books.

I am yet to use them but I recommend those books for purchase by JEE aspirants.

They are prepared by Govind Verma and published by Raj Publications, Darya Ganj, New Delhi 110002

One should try to do problems in the book and after spending some set time, should see the solution, if not able to do the problem.

Zeroth Law of Thermodynamics

Zeroth law of thermodynamics

If two bodies A and B are in thermal equilibrium and A and C are also in thermal equilibrium then B and C are in also in thermal equilibrium

All bodies in thermal equilibrium are assigned equal temperature.

Heat flows from the body at higher temperature to the body at lower temperature

Measurment Procedures - Ch. 27 Specific Heat Capacities of Gases

Determination of C_p of a gas

Regnault’s apparatus is used to measure C_p of a gas.

It has an arrangement to send gas through a pipe where in two manometers measure the pressure at two point and in between there is an adjusting screw to change the rate of flow. As the gas is flowing through this arrangement, adjusting screw is changed to maintain a constant difference of pressure between two manometers and this ensures that gas is at constant pressure when it is flowing through the calorimeter. This gas flows through an oil bath tank wherein it is given heat and then flows through the calorimeter wherein it gives out heat.

Observations

W = the water equivalent of the calorimeter with the coil
M = mass of water in the calorimeter
θ1 = temperature of the oil bath which gives heat to the gas.
θ2 = initial temperature of water in calorimeter
θ3 = Final temperature of water in calorimeter
n = amount of the gas (in moles) passed through the water
s = specific heat capacity of water

C_p of a gas. = (W + m) s (θ3 - θ2)/[n (θ1 – (θ2+ θ3)/2]

Determination of n = amount of the gas (in moles) passed through the water

It depends on the levels of mercury in the manometer attached to gas tank. If the difference in levels of the manometer is h and the atmospheric pressure (separately measured) is equal to a height H of mercury. The difference h is noted at the start of the experiment and at the end of the experiment. The pressure of the gas varies between p₁ = H+h at the beginning to p₂ = H+h at the end. Under the assumption of ideal gas

p₁V = n1RT and p₂V = n2RT

n is equal to n1 – n2 = (p₁ - p₂)V/RT


Determination of C_v of a gas

Joly’s differential steam calorimeter is used to measure C_v of a gas. The arrangement has two hollow copper spheres attached to two pans of a sensitive balance. In one of the spheres the gas for which C_v is to be measured is filled. At the start the temperature of the steam chamber without any steam is noted. It is the temperature of the gas at the beginning (θ1). Steam is sent through the steam chamber in which these two hollow spheres are there. Steam condenses on the hollow spheres and it collected in the pans attached to the hollow spheres. More steam condenses on the sphere having gas in it. After steady state conditions are reached temperature measurement is taken. This the final temperature of the gas.

Observations

m1 = the mass of gas taken or filled in the hollow sphere
m2 = the mass of extra steam condensed on the pan of the sphere having gas
θ1 = the temperature of the gas at the beginning
θ2 = the temperature of the gas at the end
L = Specific latent heat of vaporization of water.
M = molecular weight of the gas

C_v = Mm2L/[m1(θ1 – θ2)]

ch. 25 Calorimetry - Laws and Theories

Principle of Calorimetry

Neglecting any heat exchange with the surroundings, the principle of calorimetry states that the total heat given by the hot objects equals the total heat received by the cold objects.


Mechanical equivalent of heat

W = JH

Saturated vapour pressure - July Dec Revision

When we place an open flask with a liquid in a closed jar, after sufficient time, volume of the liquid becomes constant. We know that liquid evaporates, but at this point in time, when volume of liquid is constant, we have to interpret that rate of transformation from liquid to vapour equals the rate of transformation from vapour to liquid.

If some vapour from outside is injected into the space above the liquid in the jar, we observe that the volume of liquid will increase. That means more vapour is getting transformed into liquid.

When a space actually contains the maximum possible amount of vapour, the vapour is said to be saturated or it is called saturated vapour. If the amount of vapour in a space is less than the maximum possible, the vapour is called unsaturated vapour.

Saturated vapour increases with temperature. At higher temperature, the space contains more vapour. This is because more liquid molecules escape from liquid surface at higher temperatures.

The pressure exerted by a saturated vapour is called saturated vapour pressure. As a higher temperature more amount of vapour is there of a liquid, saturated vapour pressure is also higher at higher temperatures for a substance. Even into an empty jar, as vapour increases more and more, and pressure increases above the SVP, vapour starts condensing and liquid forms. This is as per definition of vapour. A vapour can be transformed into a liquid at a constant temperature by increasing pressure.

In the atmosphere around us air which is mixture of nitrogen and oxygen and water vapour are mixed with each other. If a given volume air contains maximum amount of vapour possible, the air called saturated with water vapour.


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Declination

A plane passing through the geographical poles(that is axis of rotation of the earth) and a given point P on the earth’s surface is called the geographical meridian at the point P. The plane passing through the geomagnetic poles (dipole axis of the earth) and the point P is called the magnetic meridian at the point P.

The angle between the magnetic meridian and the geographical meridian at a point is called the declination at that point.

Navigators use a magnetic compass needle to locate direction. The needle stays in equilibrium when it is in magnetic meridian. For finding the true north, navigators have to use declination and arrive at the direction of true north.

Chapter permanent magnets

Presently updating revision points and formula revision sheets of all chapters. Planning to complete all chapters by month end.

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Bohr’s Model and Physics of the Atom - July-Dec Revision

Revision - Points of the chapter

Early Atomic Models

The idea that all matter is made of very small indivisible particles is very old.

Robert Boyle’s study of compression and expansion of air brings out the idea that air is made of tiny particles with lot of empty space between the particles.

The smallest unit of an element which carries all the properties of the element is called an atom.

Experiments on discharge tube, measurement of e/m by Thomson etc. established the existence of negatively charged electrons in the atoms.

Because atoms are electrically neutral, a search for the positive charge inside the atom was started.

Thomson’s Model of the atom

Thomson (1898) suggested that the atom is a positively charged solid sphere in which electrons are embedded in sufficient number to create a neutral atom. This model of the atom could explain why only negatively charged particles are being emitted when a metal is heated. This model was also useful to explain the formation of ions and ionic compounds.

Lenard’s Suggestion

Lenard observed that cathode rays are pssing through thin material without any deviation. According to him, this was so because, there is a lot of empty space in atoms. Hence the positive charged particles are also tiny like electrons.

Rutherford’s Model of the Atom

Rutherford experimented with alpha rays or particles. When he bombarded gold foils with alpha particles, Many went without deviation, some had some deviation and some were deflected by more than 90 and came back. Hence he made a conclusion that there was a particle with a mass equivalent to alpha particle inside the atom. The mass of an atom is concentrated in this particle.

The size of this particle was also estimated by Rutherford. Its linear size is 10 fermi ( 1 fermi is equal to 1 femtometre = 10^-15 m).

Rutherford proposed that the atom contains a positively charged tiny particle called nucleus. It contains the entire mass of the atom. Outside this nucleus, at some distance, electrons move around. The positive of charge of nucleus is exactly equal to the negative charge of the electrons of the atom.

Because electrons are very very light compared to the nucleus, due to heat only electrons come out.

Movement of electrons is to brought in and the coulomb force between the nucleus and the electron is assumed to provide only centripetal force to make the electron rotate in a circular motion.

Hydrogen Spectra

If hydrogen gas is enclosed in a sealed tube and heat to high temperatures, it emits radiation. If this radiation is passed through a prsim, components are different wavelengths get deviated by different amounts and we get the hydrogen spectra on a screen. In the spectra of hydrogen atom, it is observed that light of wavelength 656.3 nm and then light of wave length 486.1 nm are present. Hydrogen atoms do not emit any radiation between 656.3 nm and 486.1 nm. Similarly radiation is observed at 434.1 nm and 4202.nm.

In the invisible region also, there is radiation emitted by the hydrogen atom at discrete wavelengths.

The wavelengths nicely fit the equation

1/λ = R [1/n² - 1/m²]

where R = 1.09737*10^7 m^-1.
n and m are integers with m>n.

The spectrum in the ultraviolet region is called Lyman series and you get the series by setting n = 1.

The hydrogen spectrum in the visible region is called Balmer series and you get the series by setting n = 2.

The hydrogen spectrum in infrared region is called Paschen series and you get the series by setting n= 3.

Difficulties with Rutherford’s Model

Rutherford’s model assumes that the electron rotates around the nucleus. Maxwell’s equations of a electromagnetism show that accelerated electron must continuously emit electromagnetic radiation. But a hydrogen does not emit radiation at ground level energy or normal energy. It emits radiation only when heated. Also, if it emits radiation it will lose energy and the radius of its circular motion will decreases and finally it will fall into the nucleus. Hence, the atomic model proposed by Rutherford needs modification. Bohr proposed such modifications.

Bohr’s Model

Bohr’s postulates

1. The electrons revolve around the nucleus in circular orbits.
2. the orbit of the electron around the nucleus can be only some special values of radius. In these special radii orbits, the electron does not radiate energy as expected from Maxwell’s laws. These orbits are called stationary orbits.
3. The energy of the atom has a definite value when electrons are in a given stationary orbit. But the if more energy is provided to the atom, the electron can jump from one stationary orbit to another stationary orbit of higher energy. If it jumps from an orbit of higher energy (E2) to an orbit of lower energy (E1), it emits a photon of radiation. The energy of the emitted photon will be E2 – E1.

The wave length of the emitted radiation is given by the Einstein-Planck equation

E2-E1 = hυ = hc/λ

4. In stationary orbits, the angular momentum l of the electron about the nucleus is an in integral multiple of the Planck constant h divided by 2 π.

l = nh/2 π

This assumption is called Bohr’s quantization rule.

Energy of an Hydrogen Atom

Bohr’s postulated can be used to find the allowed energies of the hydrogen atom when its single electron is in various stationary orbits. The methodology can be used any hydrogen like ions which have only one electron.. Therefore it is valid for He⁺, Li⁺⁺, Be⁺⁺⁺ etc.

Assume that the nucleus has a positive charge Ze ( there are z protons each with positive charge e).

By equating the coulomb force acting between Ze and e to the centripetal acceleratin mv²/r, we get r the radius at which the electron revolves.

r = Ze²/4π ε₀v²

From Bohr’s quantization rule,

mvr = nh/2 π
where n is a positive integer

Eliminating v from both the equations we get

r = ε₀h²n²/πmZe²

We get expression for v as

v = Ze²/2 ε₀hn


Hence allowed radii are proportinal to n² and for each value of n = 1,2,3…we allowed orbits.

The smallest radius orbit will have n = 1.



As we have expression for v, we can give an expression for kinetic energy when electron is in nth orbit is

K = ½ mv² = mZ²e⁴/8 ε₀²h²n²

The potential energy of the atom is

V = - Ze²/4π ε₀r = -mZ²e⁴/4ε₀²h²n²

The expression for potential energy is obtained by assuming the potential energy to be zero when the nucleus and the electron are widely separated.

The total energy of the atom is

E = K+V = - mZ²e⁴/8 ε₀²h²n²

When an atom is n^th stationary orbit, it is said to be in the n^th energy state.

In giving an expression for the total energy of the atom, kinetic energy of the electron and potential energy of theelectron-nucleus pair are considered. Kinetic energy of the nucleus is assumed to be negligible.

Radii or different orbits.

For hydrogen, z =1 and we get r1 as 53 picometre ( 1pm = 10^-12 m) or 0.053 nm. This length is called the Bohr radius and is a convenient unit for measuring lengths in atomic physics. It is denoted by me as the symbol a₀ (In HC Verma a different symbol is given. I am using this symbol as a convenience).

In terms of Bohr radius the second allowed radius is 4 a₀ and third is 9 a₀ and so on. In general nth orbit of hydrogen atom is n²a₀.

For a hydrogen like ion with Z protons in the nucleus,

r_n = radius of ‘n’ th orbit = n²a₀/Z

Energy at Ground and Excited states

From the total energy equation or expression we can find for hydrogen total energy when electron is in the state n = 1 (which is the ground state) as E1 = -13.6 eV. The radius corresponding to this energy is 53 picometre or 0.053 nm.

As in the energy equation only n changes with orbits, En is proportional to n².

There En = E1/n² = -13.6eV/n²

Note that the energy is expressed in negative units, so that larger magnitude means lower energy.


The state of an atom with the lowest energy is called its ground state.

The states with higher energies are called excited states.
Energy of hydrogen atom in the ground state is -13.6 eV.
Energy of hydrogen atom in the next excited state, that is n = 2 state is -3.4 eV.


Emission by Hydrogen Atom and Hydrogen spectra

When heated some atoms in the hydrogen become excited and when electrons jump from higher energy levels to lower energy levels in those excited atoms, photons with specific wavelengths are emitted or radiated.

If an electron jumps from mth orbit to nth orbit (m>n), the energy of the atom gets reduced from Em to En. The wavelength of the emitted radiation will be

1/ λ = (Em – En)/hc = RZ²{1/n² - 1/m²]

where R is the Rydberg constant.
R = 1.0973*10^7 m^-

In terms of the Rydberg constant total energy of the atom in the nth state is E = -RhcZ²/n²

For hydrogen atom, when n =1, E = -Rhc and we know its value is -13.6 eV.
Energy of 1 rydberg means -13.6 eV.
Rhc = 13.6eV

Series structures

Lyman series:
All transitions to n =1 state from higher state give the radiation in Lyman series.

Jumping of the electron from n =2 to n =1 gives
1/ λ = R[1 – 1/2²] = R(1 – ¼) which will give λ = 121.6 nm.

Jumping of the electron from n = ∞ to n = 1 gives

1/ λ = R[1 – 1/∞²] = R(1 -0) which gives λ = 91.2 nm.

Balmer seires
All transitions to n = 2 from higher states given radiations within the range of 656.3 nm and 365.0 nm. These wavelengths fall in the visible region.

Paschen series
The transitions or jumps to n = 3 from higher energy levels give Paschen series in the range 1875 nm to 822 nm.


Ionization potential

The energy of the hydrogen atom in ground state is -13.6 eV. If we supply more than 13.6 eV to the hydrogen atom, the electron and the nucleus get separated and electron moves with some kinetic energy independently (Remember plasma in nuclear fusion).
The minimum energy needed to ionize an atom is called ionization energy. The potential difference through which an electron should be accelerated to acquire this much energy is called ionization potential.

Thus ionization energy of hydrogen atom in ground state is 13.6 eV and ionization potential is 13.6 V.

Binding energy

Binding energy of a system is defined as the energy released when its constituents are brought from infinity to form the system. Now we know that binding energy of a hydrogen atom is 13.6 eV. The energy is zero when the electron and nucleus at infinite distance. When the electron is brought into n = 1 orbit, the energy becomes -13.6 eV and hence 13.6 eV is released which is the binding energy.

Excitation potential

The energy needed to take the atom form its ground state to an excited state is called the excitation energy of that excited state.

As the hydrogen atom’s ground state energy is -13.6 eV and its energy when electron is in n =2 orbit is -3.4 eV, we have to supply 10.2 eV to excite a hydrogen atom to its first excited state which is electron in n = 2 orbit.

The potential through which an electron should be accelerated to acquire the excitation energy is the excitation potential.
The excitation potential needed bring hydrogen to its first excited state is 10.2 V.


Limitations of Bohr’s Model

Maxwell’s theory of electromagnetism is not replaced or refuted but it is arbitrarily assumed that in certain orbits, electrons get the licence to disobey the laws of electromagnetism and are allowed not to radiate energy.

The Wave Function of an Electron

Quantum mechanics describes the spectra in a much better way than Bohr’s model.

Electron has a wave character as well as a particle character. The wave function of the electron ψ(r,t ) is obtained by solving Schrodinger’s wave equation. The probability of finding an electron is high where | ψ(r,t )|² is greater. Not only the information about the electron’s position but information about all the properties including energy etc. that we calculated using the Bohr’s postulates are contained in the wave function of ψ(r,t).

Quantum Mechanics of the Hydrogen Atom

The wave function of the electron ψ(r,t) is obtained from the Schrodinger’s equation

-(h²/8π²m) [∂²ψ /∂x² + ∂²ψ /∂y² + ∂²ψ/∂z²] - Ze²ψ/4πε₀r = E ψ


where
(x.y,z ) refers to a point with the nucleus as the origin and r is the distance of this point from the nucleus.
E refers to the energy.
Z is the number of protons.

There are infinite number of functions ψ(r,t) which satisfy the equations.

These functions may be characterized by three parameters n,l, and m_l.

For each combination of n,l, and m_l there is an associated unique value of E of the atom of the ion.

The energy of the wave function of characterized by n,l, and m_l depends only on n and may be written as


En = - mZ²e⁴/8 ε₀²h²n²

These energies are identical with Bohr’s model energies.

The paramer n is called the principal quantum number, l the orbital angular momentum quantum number and m_l. The magnetic quantum number.

When n = 1, the wave function of the hydrogen atom is

ψ(r) = ψ₁₀₀ = √(Z³/ π a₀²) *(e^{-r/ a₀})


ψ₁₀₀ denotes that n =1, l = 0 and m_l = 0

a₀ = Bohr radius

In quantum mechanics, the idea of orbit is invalid. At any instant the wve function is spread over large distances in space, and wherever ψ≠ 0, the presence of electron may be felt.

The probability of finding the electron in a small volume dV is | ψ(r)| ² dV
We can calculate the probability p(r)dr of finding the electron at a distance between r and r+dr from the nucleus.

In the ground state for hydrogen atom it comes out to be

P(r) = (4/ a₀)r²e ^{-2r/ a₀}

The plot of P(r) versus r shows that P(r) is maximum at r = a₀ Which the Bohr’s radius.

But when we put n =2, the maximum probability comes at two radii one near r = a₀ and the other at r = 5.4 a₀. According to Bohr model all electrons should be at r = 4 a₀.

Nomenclature in Atomic Physics

An interesting property of electrons is that each electron has a spin angular momentum. It is characterized by m_s and it can take values of +1/2 or -1/2.

Therefore a wave function is described by four characteristics n,l, m_l and m_s.

Any particular wave function described particular values of the above four characteristics or quantum numbers is termed a quantum state.

For n =1, l = 0 and m_l = 0. Hence there will be two quantum states.

For n = 2 there 8 quantum states.

In general there are 2n² quantum states.

The quantum states corresponding to a particular n are together called a major shell.

n =1 shell is called K shell, n = 2 is called L shell and n = 3 shell is called M shell etc.

Pauli’s exclusion principle says that there cannot be more than one electron in any quantum state.

It is customary to use the symbols s,pd,f etc. to denote the value of the orbital angular momentum quantum number l corresponding to the value of l = 0,1,2,3 etc. respectively. These are called subshell for a given shell.

For an atom having many electrons, the quatum states are gradually filled from lower energy to higher energy to form the ground state of the atom.

Protons and neutrons also obey Pauli principle. They also have quantum numbers even though we do not study them in the current syllabus.

Any particle that obeys Pauli exclusion principle is called a fermion.
Electrons, protons, and neutrons are all fermions.

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IIT JEE Material Updated

Material relating to chapters x-rays and Nucleus updated.

Click on the labels and Xrays and nucleus in labels list

Magnetic Properties of Matter – Formulae - July Dec Revision

Ch. 37 Magnetic Properties of Matter – Formulae

1. Magnetization vector = Magnetic moment per unit volume

I = M/V

2. Magnetic intensity

H = B/µ₀ - I .. (2)

Where

H = magnetic intensity

B - resultant magnetic field

I = intensity of magnetization

Magnetic intensity due to a magnetic pole of pole strength m at a distance r from it is

H = m/(4 πr²) …(5)

6. Magnetic susceptibility

I = χH … (6)

Χ is called the susceptibility of the material.

7. Permeability

B = µH … (7)

µ = µ₀ (1+χ) is a constant and is called the permeability of the material.

µ₀ is the permeability of vacuum.

µ_r = µ/µ₀ = 1+ χ is called the relative permeability of the material.

9. Curie’s law

χ = c/T … (9)
where c = Curie’s constant


For ferromagnetic materials
Χ = c’/(T - T_c)

Where
T_c is the Curie point and
c’ = constant



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Formula revision Electromagnetic induction - July-Dec Revision

1. Faraday’s law of electromagnetic induction
Є = -dФ/dt … (1)

Where

Є = emf produced
Ф = ∫B.dS = the flux of the magnetic field through the area.

2. i = Є/R = -(1/R) dФ/dt …(2)

where i = current in the circuit
R = resistance of the circuit

3. Є = vBl

Where
Є = emf produced
v = velocity of the conductor
B = magnetic field in which the conductor is moving
l = length of the conductor

4. Induced electric field

∫E.dl = -dФ/dt .. (4)

where

E = induced electric filed due to magnetic field B

5. Self induction
Magnetic field through the area bounded by a current-carrying loop is proportional to the current flowing through it.

Ф = Li … (5)

Where

Ф = ∫B.dS = the flux of the magnetic field through the area.
L = is a constant called the self-inductance of the loop.
i = current through the loop.

6. Self induced EMF

Є = -dФ/dt = -Ldi/dt ….(6)

7. Self inductance of a long solenoid

L = µ₀n² πr² l … (7)

8. Growth of current through an LR circuit

i = i₀(1 - e^-tR/L) … (8)
= i₀(1 - e^{-t/ τ} ) … (9)

where

i = current in the circuit at time t

i₀ = Є/R
Є = applied emf
R = resistance of the circuit
L = inductance of the circuit
τ = L/R = time constant of the LR circuit


10. Decay of current in a LR circuit

i = i₀(1 - e^-tR/L) … (10)
= i₀(1 - e^{-t/ τ} ) … (11)

i = current in the circuit at time t

i₀ = current in the circuit at time t = 0
R = resistance of the circuit
L = inductance of the circuit
τ = L/R = time constant of the LR circuit

12. Energy stored in an inductor

U = ½ Li² … (12)

13. Energy density

u = U/V = B²/2µ₀


14. Mutual induction


Ф = Mi … (14)

Where
M = constant called mutual inductance of the given pair of circuits

Є = -Mdi/dt …. (15)

July Dec Revision - Kinetic theory of gases

Any sample of gas is made of molecules.

The observed behaviour of gas results from the behaviour of its large number of molecules.

Kinetic theory of gases attempts to develop a model of the molecular behaviour which should result in the observed behaviour an ideal gas.


Assumptions of kinetic theory of gases

1. All gases are made of molecules moving randomly in all directions
2. The size of molecule is much smaller than the average separation between the molecules.
3. The molecules exert no force on each other or on the walls of the container except during collision (no atraction force or repulsion force).
4. All collisions between two molecules or between a molecule and a wall are perfectly elastic. Also the time spent during a collision is negligibly small.
5. the molecules obey Newton's laws of motion.
6. When a gas is left for sufficient time in a closed container, it comes to a steady state. The density and the distribution of molecules with different velocities are independent of position, direction and time.

The assumptions are close to the real situations at low densities.
The molecular size is roughly 100 times smaller than the average separation between the molecules at 0.1 atm and room temperature.

The real molecules do exert electric forces on each other but these forces can be neglected as the average separation between molecules is large as compared to their size.


Pressure of an ideal gas
p = (1/3)ρ*Avg(v²) ........ (1)

where
ρ = density of gas = mass per unit area
Avg(v²) = average of the speeds of molecules squared

pV = (1/3)M*Avg(v²) ....... (2)

M = Mass of gas in the closed container

pV = (1/3)nm*Avg(v²) ........ (3)
n = number of molecules of gas in the container
m = mass of each molecule


RMS Speed: The square root of mean square speed is called root-mean-square speed or rms speed.
It is denoted by the symbol v_rms

Avg(v²) = (v_rms)²

The equation (1) can be written as

p = (1/3)ρ*(v_rms)²

Then
v_rms) = √[3p/ρ] = √[3pV/M]


Total translational kinetic energy of all the molecules of the gas is

K = Σ (1/2mv² = (1/2)M(v_rms)² ... (4)

The average kinetic energy of a molecule = (1/2)m(v_rms)²

Then from equation (2)
K = (3/2)pV

according to the kinetic theory of gases, the internal energy of an ideal gas is the same as the total translational kinetic energy of its molecules.

For different kinds of gases, it is not the rms speed but average kinetic energy of individual molecules that has a fixed value at a given temperature.

The heavier molecules move with smaller rms speed and the lighter molecules move with larger rms speed.

All gas laws can be deduced from kinetic theory of gases.

Ideal gas equation

pV = nRT

R = universal gas constant = 8.314 J/mol-L

The average speed of molecules is somewhat less than the rms speed.

Average speed = (Σv)/n = √[8kT/πm]

Friction - July Dec Revision

1. Normal force = N_f = Mg
Where
M = mass of the object
g = acceleration due to gravity



2. f_k = µ_k N_f
where

f_k = magnitude of kinetic friction
µ_k = coefficient of kinetic friction

3. f_max = µ_s N_f

where

f_max = maximum static friction
µ_s = coefficient of static friction

The actual static friction can be less than maximum static friction if the force applied is less than f_max .

4. On an adjustable inclined plane, a block is kept and the angle is gradually increased so that the block begins to move.

Then f_max = mg sin θ
N_f = mg cos θ

There coefficient of static friction µ_s

µ_s = f_max / N_f
= tan θ = h/d

where θ = angle of incline when the block starts moving
m = mass of the block placed on incline
h = height of the incline
d= length of the incline

5. T find kinetic friction the angle of the incline is slightly reduced and the block is made to move with uniform velocity

In this case
µ_k = coefficient of kinetic friction = tan θ’ = h’/d’
(h is going to decrease and d is going to increase)


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Earth is not Strictly an Inertial Frame - July Dec Revision

In H C Verma,in Newton's Laws chapter, it was given.

The earth in not strictly an inertial frame. However we can say that the earth in an inertial frame of reference to a good approximation.

Thus for routine affairs, acceleration (a) = 0 if and only if external force (F) = 0 is true in the earth frame of reference.

This fact was identified and formualated by Newton as first law(Newton's first law).

Is Newton's first law a law?

If we restrict the statement to measurements made from earth frame, this becomes a law.

If we try to universalize the statement to different frames, it becomes a definition of inertial frame.


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Newton’s Laws of Motion –Study Plan - Session 2

Day 2 Study Plan

5.2 Newton's second law
5.3 Working with Newton's laws

W.O.E. 3 and 4






5.2 Newton's second law

The acceleration of a particle as measured from an inertial frame is given by the (vector) sum of all the forces acting on the particle divided by its mass.

a = F/m or F = ma

Acceleration and force are measured at the same instant. If force becomes zero at an instant, acceleration also becomes zero at the same instant.


5.3 Working with Newton’s First and Second Law

1. Decide the System

We have to assume that forces are acting on a system and the system is at rest or in motion. In this context, the system may be a single particle, a block, a combination of two blocks one kept over the other or two blocks connected by a string etc. But there is a restriction for treating one as a system. All parts of the system should have identical acceleration.

Step 2. Identify the Forces

Once the system is decided, make a list of the forces acting on the system due to all the objects other than the system. Any force applied by the system should not be included in the list of the forces (material from the chapter on forces should help you in deciding various forces exerted by the system and forces exerted by the objects on the system).

Step 3. Make a Free Body Diagram

Represent the system by a point in a separate diagram and draw vectors representing the forces with this point as the common origin

Step 4: Choose axes and Write Equations.

Example 5.2

The example describes a block being pulled by a man with help of a string.
The system under analysis is the block.
It is accelerating in the horizontal direction. So there is net force in the horizontal direction.
It is not accelerating in the vertical direction. Hence net force is vertical direction is zero.


Worked out examples:
3,4,5,

Obective I

3,5, 6,9,10,

Attempt questions in Objective questions (OBJ II)

4

Exercises

1, 2, 3,

Newton’s Laws of Motion - Study Plan - Session - 1

5.1 First law of motion (From HC Verma Part 1)

Study Plan

Day 1

5.1 First law of motion
-----------------------

Points to be remembered

First law of motion

If the (vector) sum of all the forces acting on a particle is zero then and only then the particle remains unaccelerated (i.e., remains at rest or moves with constant velocity).

We can say in vector notation
a = 0 if and only if resultant force F = 0

A frame of reference in which Newton's first law is valid is called an inertial frame of reference.

A frame of reference in which Newton's first law is not valid is called a noninertial frame of reference. (Example: lamp in an elevator cabin whose cable had broken)

Example of lamp in an elevator whose cable had broken:

In the cabin when one measures with reference to the cabin, the lamp hanging from the ceiling has no acceleration. Hence the forces acting on the lamp, its weight (W) and the tension in the string supporting it are balancing each other. We can infer that W = T.

But for an observer on the ground, lamp is accelerating with acceleration g, when he considers the forces acting on the lamp as W and T once again, W is not equal to T as lamp is acceleating. Both cannot be right at the same time, and it means in one of the frames Newton's first law is not applicable.


Inertial frame: Hence inertial frame is a frame of reference in which Newton's first law is valid.

Is earth an inertial frame of reference?
Strictly it is not. But as a good approximation, earth can be taken as an inertial frame of reference.

All frames moving uniformly with respect to an inertial frame are themselves inertial.

This relation was derived from the relation that converts acceleration with respect to one frame into acceleration with respect to another frame. When the other frame is moving with uniform velocity it acceleration with respect to the frame in reference is zero.

Examples: A train moving with uniform velocity with respect to ground, a plane flying with uniform velocity with respect to a high etc. The sum of forces acting on a suit case kept on the shelves of them with turnout to be zero.

Concepts covered in the session

First law of motion
Frames of reference (concept from Chapter 3)
Inertial frame of reference
noninertial frame of reference
Inertial frames other than earth (frames moving with uniform velocity w.r.t earth)




Example 5.1 deals with forces acting on heavy particle hanging from a string fixed with the roof. the particle is stationery. The forces acting are pull of the earth downward due to gravity and pull of the string vertically upward.



Go through examples in worked out examples

1.
2.


Attempt questions in Objective questions (OBJ I)

1.
5.

Attempt questions in Objective questions (OBJ II)

1.
2.
3.

Angular Momentum - July Dec Revision

Principle of conservation of angular momentum

If the total external torque in a system is zero, its angular momentum remains constant.

Angular momentum of a particle about a point O is defined as

l = r × p

where
p = linear momentum
r = position vector of the particle from the given point O.

The angular momentum of a system of particles is the vector sum of the angular momenta of the particles of the system.

Kepler’s Laws of Planetary Motion - July Dec Revision

Kepler’s Laws of Planetary Motion
1. All planets move in elliptical orbits with the sun at a focus.

2. The radius vector from the sun to the planet sweeps equal area in equal time.

3. The square of the time period of a plant is proportional to the cube of the semimajor axis of the ellipse.

(Chapter: Gravitation)

1. All planets move in elliptical orbits with the sun at a focus.

Circular path is a special case of an ellipse when the major and minor axes are equal. For a circular apth, the planet should have velocity perpendicular to the line joining it with the sun and the magnitude has to be

v = √(GM/a)

If these conditions arenot satisfied, the planet moves in an ellipse.

2. The radius vector from the sun to the planet sweeps equal area in equal time.

For a circular orbit, as the speed of the particle remains constant, it will sweep equal area in equal time.

3. The square of the time period of a plant is proportional to the cube of the semimajor axis of the ellipse.

For circular orbits this law is proved.

Past JEE Questions

According to Kepler’s second law, the radius vector to a plant form the sun sweeps out equal areas in equal intervals of time. This law is a consequence of the conservation of -----------------.

(JEE 1985)

Answer: Angular momentum

Collission - July-December 2008

Collision between two bodies - (For example two balls)

If no external force is acting on the system that is two ball, during whole process of colliosion (before the collision, during the collision, and after the collision) the momentum of the two-body system will remain constant.

If before the collision velocities are v1 and v2, masses m1 and m2

m1v1 + m2v2 = P

If during the collsion (at the instant) both have the same velocity V

(m1 + m2)*V = P

After the collision velocities are say v1' and v2'

m1v1' + m2v2' = P

In the case of energy of the system, assuming that there is no friction, the sum of the kinetic energy before the collision and after the collision will be same in case of elastic collision.

One example of elastic collision given in the book is a spring attached to the ball at the beginning which is getting hit from behind. due to the collision, the spring gets compressed and then expands.

Other example is balls perfectly elastic material. In the case of balls, there will be a deformation due to the collision and the surfaces will be in contact for some interval of time and travel together at the same speed. As the forces develop inside them the balls try to regain their original shape, and in the process push each other. The velocity of the front ball increases while that of the rear ball decreases and the balls separate. After the separation, the balls regain their original shape so that the elastic potential energy is completely converted back into kinetic energy.

Thus during an elastic collision, the initial kinetic energy of the two body system is equal to the final kinetic energy, but it does not remain constant as during the collision some kinetic energy is converted into elastic potential energy.

In the case of inelastic collsion, the two deformed balls have no tendency to regain the shape and tend to remain in contact after the deformation and move with same speed. The kinetic energy of the two body system decreases at the time of deformation and remains constant thereafter.


Collisions – Psat JEE Problems

1. A ball hits the floor and rebounds after an inelastic collision. In this case

a. the momentum of the ball just after the collision is the same as that just before the collision.
b. the mechanical energy of the ball remains the same in the collision.
c. the total momentum of the ball and the earth is conserved.
d. the total energy of the ball and the earth is conserved.

(More than one alternative may be correct)
JEE 1986

Answer ( c ) and (d)

Note: total kinetic energy is not conserved but total energy is conserved.

2. A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments, each of mass m, are found to move with a speed v, each in mutually perpendicular directions. The total energy released in the process of explosion is _________ .


(JEE 1987)

answer (3/2)mv²

Solution;

As the two fragments went in mutual perpendicular directions with velocity v the resultant momentum

2mv cos 45° = 2mv(1/√2) = √2 mv

if the velocity of fragment with 2m mass is V

then 2 mV =/√2mv

V = v/√2

Hence energy released = ½ mv² + ½ mv² + ½ *2 m (v/√2) ²

= (3/2)mv²

Moment of Inertia - July -Dec 2008 Revision

Moment of Inertia

Torque created by external forces in rotating motion

When a particle is rotating it has tangential acceleration and radial acceleration.

Radial acceleration = ω²r

Hence radial force acting on it = m ω²r

Tangential acceleration = dv/dt = rdω/dt = r α
ω = angular velocity
α = angular acceleration
Tangential force = mrα

Torque created by radial force is zero as the force intersects the axis of rotation.
Tangential force and axis are skew (they do not intersect) and they are perpendicular. Thus the resultant torque is Force*radius = mr²α

In the case of a body having ‘n’ particles and rotating, the total torque is equal to torque acting on each of the particles

Total torque on the body = Γ(total) = Σ m_ir_i²α
Σ m_ir_i² is called as moment of inertia.

Moment of inertia can be calculated using the above formula for collection of discrete particles.

If the body is a continuous, the technique of integration needs to be used. We consider a small element of the body with mass dm and having a perpendicular distance from the axis or line about which moment of inertial is to be calculated.

We find ∫ r²dm under proper limits to get the moment of inertia of the body.

r²dm is the moment of inertia of the small element.

Determination of moment of inertia of representative bodies.

1. Uniform rod about a perpendicular

M = total mass of the body
l = length of the body
I = Ml²/12

I is obtained by taking a small element dx at a distance x from the centre of the rod.
Mass dm of the element = (M/l)*dx (M/l give mass per unit length)

dI = (M/l)*dx*x²
I = ∫(M/l)*dx*x² (limits are from –l/2 to l/2: these limits cover the entire rod).
= M/l[x³/3] from –l/2 to l/2
= M/3l[l³/8 + l³/8] = M/3l(l³/40) = Ml²/12

2. Moment of inertia of uniform rectangular plate about a line parallel to an edge and passing through the centre.

M = total mass of the body
Plate measurements l,b
Axis or line is parallel to b

I = Ml²/12

If the axis of line is parallel to l

I = Mb²/12



3. Circular ring
M = total mass of the body
R = radius

I = MR²

4. Uniform circular plate
M = total mass of the body
R = radius

I = MR²/2

5. Hollow cylinder about its axis

M = total mass of the body
R = radius

I = MR²


6. Uniform solid cylinder about its axis.

M = total mass of the body
R = radius

I = MR²/2

7. Hollow sphere about a diameter

M = total mass of the body
R = radius

I =(2/3) MR²

8. Uniform solid sphere about a diameter

M = total mass of the body
R = radius

I = (2/5) MR²

Parallel axis theorem

Moment of inertia of a body about an axis parallel to its centre of mass is equal to
Moment of the body about the axis passing through its centre of mass plus Mr² (where M is the mass of body and r is the perpendicular distance between two axes).


Past JEE problem

1. A circular disc of radius r/3 is removed from the outer-edge of a bigger circular disc of mass 9M and radius r. the moment of inertia of the remaining portion of the disc about the centre O of the disc perpendicular to the plane of the disc is

a. 4Mr²
b. (37/9)Mr²
c. (40/9)Mr²
d. 9Mr²

JEE 2005

Answer (a)

Solution

Mass of the disc removed

Total mass – M
Total area = π r²
Area of piece removed π (r/3) ²

Hence mass of the disc removed = M[π (r/3) ²]/( π r²) = M

Moment of inertia of removed about its own centre is = ½ (M)*(r/3) ²


According to the parallel axis theorem

Moment of inertia of the removed piece about an axis passing through the centre of the big disc = ½ (M)*(r/3) ² + M(2r/3) ²
=1/2(M) (r²/9) + M(4r²/9) = (½)(M) (r²/9 + 8r²/9) = (½)Mr²

Moment of inertia of bigger disc before cutting the piece

= ½ (9M)(r²)

Hence moment of inertia after removal of the piece
=( ½) (9M)(r²) - (½)Mr² = 4Mr²

Physics Formula Revision Rotational mechanics July - Dec 2008

Formula Sheet – rotational mechanics

Rotational kinematics

Angular variables

θ = angular position of the particle

ω = angular velocity = dθ/dt = lim∆t→0 ∆θ/∆t

α = angular acceleration = dω/dt = d²θ/dt²

If the angular acceleration is constant, formulas similar in form to linear formulas can be used to find the angular variables:

θ = ω₀t + ½ αt²

ω = ω₀ + αt

ω² = ω₀² + 2 α θ

where
ω₀ = angular velocity at the beginning

Relation between the linear motion of a particle of a rigid body and rotation of the rigid body

v = r ω
where
v = linear speed of the particle

a_t = rate of change of speed of the particle in circular motion

a_t = dv/dt = rdω/dt = r α (These relations are from the chapter of circular motion)

Rotational Mechanics - July - Dec Revision 2008

Rotation of a rigid body

If each particle of a rigid body moves in a circle, with centres of all the circles on a straight line and with planes of the circles perpendicular to this line we say that the body is rotating about this line. The straight line is called the axis of rotation. (Particle makes circular motion. Rigid body makes rotation.)

Kinematics of rotation of rigid body

For a rigid body let the axis of rotation be Z-axis.
At time t =0, let particle P be at P₀.
Perpendicular to axis of rotation from P₀ be PQ. (Q is on the axis)
If at time t Particle P moves P₁ and angle P₀Q P₁ = θ.
Hence the particle has rotated through θ.
All particles have rotated through θ.
We can say the whole rigid body has rotated through angle θ.
The angular position of the body at time t is θ.
If P has made a complete revolution its circular path, every particle will do so and hence rigid body has done so. We can say rigid body made a complete revolution and it has rotated through an angle of 2 π radians.
Hence, rotation of a rigid body is measured by the rotation of a line PQ (P is a particle on the rigid body and Q is a point on the axis of rotation and PQ is perpendicular from P to the axis of rotation).

As the rotation of rigid body is defined in terms of the circular motion of a particle on the rigid body, kinematics of circular motion of particle becomes applicable to rotation.

Angular variables

θ = angular position of the particle

ω = angular velocity = d θ/dt = lim∆t→0 ∆θ/∆t

α = angular acceleration = d ω/dt = d²θ/dt²

If the angular acceleration is constant, formulas similar in form to linear formulas can be used to find the angular variables:

θ = ω₀t + ½ αt²

ω = ω₀ + αt

ω² = ω₀² + 2 α θ

Where

ω₀ is velocity at time t = 0.

Given the axis of rotation, the body can rotate in two directions – clockwise or anticlockwise. One of the directions has to be defined as positive direction according to the convenience of the problem.

The SI unit for angular velocity is radian/sec (rad/s).
One revolution/sec = 2 π radian/sec

Similar to the circular motion of the particle, in rotation for a particle P

s = Linear distance traveled by the particle in circular motion

∆s = Linear distance traveled by the particle in circular motion in time ∆t


∆s = r∆θ

Where
r = radius of the circle over which the particle is moving
∆θ = angular displacement in time ∆t

∆s/∆t = r∆θ/∆t

v = r ω
where
v = linear speed of the particle

a_t = rate of change of speed of the particle in circular motion

a_t = dv/dt = rdω/dt = r α




Rotational dynamics

In rotation of a body the resultant force due to external forces is zero, but the resultant of
Torque produced by the external forces is nonzero and this torque produces rotation motion.

Torque of a force about the axis of rotation

First we define torque a force about a point.
For a force F acting on a particle P, to find torque about a point O, define the position vector of P with respect to O. Let this position vector be r.

Torque of force F about O = Γ = F × r.
This is vector product of two vectors hence a vector quantity, as per the rules of vector product, the direction of Γ will be perpendicular to to F and r.

When the torque about an axis of rotation is to be determined, select a point on the axis of the rotation and find the torque of the force acting on a particle about this point. Find the angle between the axis and the line joining the point on axis (about which the torque is calculated) and the particle (one which the force is acting). Let the angle be θ.

Torque about the axis due to a force is the component along the axis, of the torque of the force about a point on the axis.

Magnitude of the torque = | F × r | cos θ

The torque about the axis is same, even if different points are chosen along the axis for determining the torque of the force about those points.

Some special cases of relation between force and the axis of rotation.

1. Force is parallel to the axis of rotation.
Torque along the axis is zero.

2. F and r are collinear. The torque about O is zero and the torque about axis is zero.

3. Force and axis are perpendicular but they do not intersect. In three dimensions, two lines may be perpendicular without intersecting. Example: A vertical line on
a wall and a horizontal line on the opposite wall.

In this case torque about the axis is equal to Force multiplied by the perpendicular to axis from the force direction (line along which the force is acting).


Torque produced by forces on a particle

A particle having circular motion will have two forces acting on it.
One force produces tangential acceleration dv/dt in it. Hence the force named tangential force is 'ma' = mdv/dt = mrα
This force creates a torque of mr²α

the other force creates radial acceleration or centripetal acceleration ω²r. Hence the force named radial force is mω²r
As intersects the axis of rotation, the torque produced by it is zero.

Hence total torque produced by a rigid body consisting of n particles is

Г(total) = Σm_ir_i²α

= αΣm_ir_i² as α is same for all particles

Let I = Σm_ir_i²

Г(total) = Iα

Quantity I is called moment of inertia of the body about the axis of rotation.

I = Σm_ir_i²

where

m_i = mass of the ith particle
r_i = perpendicular distance of ith particle from the axis of rotation.




Angular momentum

Conservation of angular momentum


Work done by a torque

Power delivered by a torque

Moment of inertia

Moment of Inertia

Torque created by external forces in rotating motion

When a particle is rotating it has tangential acceleration and radial acceleration.

Radial acceleration = ω²r

Hence radial force acting on it = m ω²r

Tangential acceleration = dv/dt = rdω/dt = r α
ω = angular velocity
α = angular acceleration
Tangential force = mrα

Torque created by radial force is zero as the force intersects the axis of rotation.
Tangential force and axis are skew (they do not intersect) and they are perpendicular. Thus the resultant torque is Force*radius = mr²α

In the case of a body having ‘n’ particles and rotating, the total torque is equal to torque acting on each of the particles

Total torque on the body = Γ(total) = Σ m_ir_i²α
Σ m_ir_i² is called as moment of inertia.

Moment of inertia can be calculated using the above formula for collection of discrete particles.

If the body is a continuous, the technique of integration needs to be used. We consider a small element of the body with mass dm and having a perpendicular distance from the axis or line about which moment of inertial is to be calculated.

We find ∫ r²dm under proper limits to get the moment of inertia of the body.

r²dm is the moment of inertia of the small element.

Determination of moment of inertia of representative bodies.

1. Uniform rod about a perpendicular

M = total mass of the body
l = length of the body
I = Ml²/12

I is obtained by taking a small element dx at a distance x from the centre of the rod.
Mass dm of the element = (M/l)*dx (M/l give mass per unit length)

dI = (M/l)*dx*x²
I = ∫(M/l)*dx*x² (limits are from –l/2 to l/2: these limits cover the entire rod).
= M/l[x³/3] from –l/2 to l/2
= M/3l[l³/8 + l³/8] = M/3l(l³/40) = Ml²/12

2. Moment of inertia of uniform rectangular plate about a line parallel to an edge and passing through the centre.

M = total mass of the body
Plate measurements l,b
Axis or line is parallel to b

I = Ml²/12

If the axis of line is parallel to l

I = Mb²/12



3. Circular ring
M = total mass of the body
R = radius

I = MR²

4. Uniform circular plate
M = total mass of the body
R = radius

I = MR²/2

5. Hollow cylinder about its axis

M = total mass of the body
R = radius

I = MR²


6. Uniform solid cylinder about its axis.

M = total mass of the body
R = radius

I = MR²/2

7. Hollow sphere about a diameter

M = total mass of the body
R = radius

I =(2/3) MR²

8. Uniform solid sphere about a diameter

M = total mass of the body
R = radius

I = (2/5) MR²


Moment of inertia theorems

Theorem of parallel axes

Thoerem of perpendicular axes

Combined rotation and translation

Rolling