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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Showing posts with label July-Dec revision. Show all posts
Showing posts with label July-Dec revision. Show all posts
Monday, October 13, 2008
Thursday, October 9, 2008
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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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.
Join orkut community
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Saturday, October 4, 2008
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 nth stationary orbit, it is said to be in the nth 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,
rn = 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 ml.
For each combination of n,l, and ml 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 ml 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 ml. 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 ml = 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 ms and it can take values of +1/2 or -1/2.
Therefore a wave function is described by four characteristics n,l, ml and ms.
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 ml = 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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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 nth stationary orbit, it is said to be in the nth 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,
rn = 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 ml.
For each combination of n,l, and ml 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 ml 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 ml. 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 ml = 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 ms and it can take values of +1/2 or -1/2.
Therefore a wave function is described by four characteristics n,l, ml and ms.
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 ml = 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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Saturday, September 27, 2008
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.
µr = 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 - Tc)
Where
Tc is the Curie point and
c’ = constant
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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.
µr = 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 - Tc)
Where
Tc 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)
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)
Tuesday, September 23, 2008
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 vrms
Avg(v²) = (vrms)²
The equation (1) can be written as
p = (1/3)ρ*(vrms)²
Then
vrms) = √[3p/ρ] = √[3pV/M]
Total translational kinetic energy of all the molecules of the gas is
K = Σ (1/2mv² = (1/2)M(vrms)² ... (4)
The average kinetic energy of a molecule = (1/2)m(vrms)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]
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 vrms
Avg(v²) = (vrms)²
The equation (1) can be written as
p = (1/3)ρ*(vrms)²
Then
vrms) = √[3p/ρ] = √[3pV/M]
Total translational kinetic energy of all the molecules of the gas is
K = Σ (1/2mv² = (1/2)M(vrms)² ... (4)
The average kinetic energy of a molecule = (1/2)m(vrms)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]
Friday, September 12, 2008
Friction - July Dec Revision
1. Normal force = Nf = Mg
Where
M = mass of the object
g = acceleration due to gravity
2. fk = µk Nf
where
fk = magnitude of kinetic friction
µk = coefficient of kinetic friction
3. fmax = µs Nf
where
fmax = maximum static friction
µs = coefficient of static friction
The actual static friction can be less than maximum static friction if the force applied is less than fmax .
4. On an adjustable inclined plane, a block is kept and the angle is gradually increased so that the block begins to move.
Then fmax = mg sin θ
Nf = mg cos θ
There coefficient of static friction µs
µs = fmax / Nf
= 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
µk = coefficient of kinetic friction = tan θ’ = h’/d’
(h is going to decrease and d is going to increase)
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Where
M = mass of the object
g = acceleration due to gravity
2. fk = µk Nf
where
fk = magnitude of kinetic friction
µk = coefficient of kinetic friction
3. fmax = µs Nf
where
fmax = maximum static friction
µs = coefficient of static friction
The actual static friction can be less than maximum static friction if the force applied is less than fmax .
4. On an adjustable inclined plane, a block is kept and the angle is gradually increased so that the block begins to move.
Then fmax = mg sin θ
Nf = mg cos θ
There coefficient of static friction µs
µs = fmax / Nf
= 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
µk = coefficient of kinetic friction = tan θ’ = h’/d’
(h is going to decrease and d is going to increase)
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Tuesday, September 2, 2008
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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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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Thursday, August 28, 2008
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.
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.
Wednesday, August 27, 2008
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
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
Tuesday, August 26, 2008
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
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
Saturday, August 23, 2008
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 miri²α
Σ miri² 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
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 miri²α
Σ miri² 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
Monday, July 28, 2008
July - December 2008 Revision for JEE 2009
I plan to go through each chapter in Chemistry, Physics and Mathematics in revision mode during July-December 2008 apart from reading or studying chapter which I have not read so far.
My thinking is that from 1st January 2009 onwards, the aspirants should focus on memorizing things and a revision during July-Dec at leisurely pace, one chapter per day would help in that. From 1 January 2009, the memorization process should take up three chapters per day.
My thinking is that from 1st January 2009 onwards, the aspirants should focus on memorizing things and a revision during July-Dec at leisurely pace, one chapter per day would help in that. From 1 January 2009, the memorization process should take up three chapters per day.
Chapter 1 Introduction - July-Dec Revision
Physics is the study of nature and its laws.
The nature around us is like a big chess game played by Nature. Various events that happen are like the moves made by a chess players. We are allowed to watch the events that happen, and guess the rules, and then play to derive benefits that we want from nature. We may across new events which do not follow old rules that we have formulated and we need to guess the new rules.
Great scientists or scientists in general guess the rules from the observations available at that time and prove the usefulness of those rules by further experiments or events that happen subsequently. These rules may require modification subsequently if they are not able to explain some events observed that happen subsequently.
The description of nature becomes easy if we have the freedom to use mathematics.
Mathematics is the language of physics.
Units: Fundamental and derived
While there are a large number of physical quantities to be measured, only seven fundamental quantities are found to be sufficient. All other quantities can be measured using these seven fundamental quantities.
The set of fundamental quantities must have the following properties.
a. The fundamental quantities should be independent of each other; and
b. All other quantities may be expressed in terms of the fundamental quantities.
Fundamental quantities are also referred to as base quantities.
Who decides the units?
A body named Conference Generale des Poids or CGPM (General Conference on Weight and Measures in English) has been gibven the authority by international agreement.
Definitions of base units
Length metre (m)
Mass kilogram (kg)
Time second (s)
Electric current ampere (A)
Thermodynamic temperature kelvin (K)
Amount of substance mole (mol)
Luminous intensity candela (cd)
Metre
The distance travelled by light in vacuum in 1/[299,792,458] second is called 1 m.
Kilogram
the mass of cylinder made of planitum-iridium alloy kept at International Bureau of Weights and Measures is defined as 1 kg.
Second
The time duration in 9,192,631,770 time periods of the selected transition of radiation of Cesium-133 atom is defined as 1 s. (To understand this idea see Bohr model chapter)
Ampere
To under stand the definition of this unit, you have to read the chapter on electric field.
Two long straight wires with negligible cross section are to be placed parallel to each other at a separation of 1 m and electric current in the same amout is sent through them in the same direction. If the attractive force between the two wires adjusted, the current in each wire (one of the wires) that gives a force between them of 2*10^-7 newton per metre of the wires, is defined as 1 A of current.
Kelvin
The fraction 1/9273.16) of the thermodynamic temperature of triple point of water is called 1 K.
Mole
The amount of a substance that contain as many elementary entities (molecules or atoms if the substance is monatomic) as there are number of atoms in 0.012 kg of carbon-12 is called a mole.
Candela
It is the luminous intensity of a blackbody of surface area 1/[600,000] m² placed at the temperature of freezing platinum and at a pressure of 101,325 N/m², in the direction perpendicular to its surface.
Dimensions of physical quantities and Dimensional Formula
When a physical quantity is expressed in terms of the base quantities, it is written as a product of different powers of the base quantities.
[Force] = MLT-2
M,L, and T are base quantities – mass, length and time
The exponent of a base quantity that enters into the expression (MLT-2) is called the dimension of the quantity in that base.
The dimensions or force are 1 in mass, 1 in length and -2 in time. The dimensions of all other base quantities are zero.
The base quantities are denoted as follows in writing expressions. The symbols used are M for mass, L of length, T for time, I for current, K for temperature, mol for mole and cd for candela.
The physical quantity that is expressed in terms of the base quantities is enclosed in square brackets to inform that it is expressed in dimensions of base quantities. Such expression of a physical quantity in terms of dimensions of base quantities is called the dimensional formula.
Order of magnitude
The structure of the world
The nature around us is like a big chess game played by Nature. Various events that happen are like the moves made by a chess players. We are allowed to watch the events that happen, and guess the rules, and then play to derive benefits that we want from nature. We may across new events which do not follow old rules that we have formulated and we need to guess the new rules.
Great scientists or scientists in general guess the rules from the observations available at that time and prove the usefulness of those rules by further experiments or events that happen subsequently. These rules may require modification subsequently if they are not able to explain some events observed that happen subsequently.
The description of nature becomes easy if we have the freedom to use mathematics.
Mathematics is the language of physics.
Units: Fundamental and derived
While there are a large number of physical quantities to be measured, only seven fundamental quantities are found to be sufficient. All other quantities can be measured using these seven fundamental quantities.
The set of fundamental quantities must have the following properties.
a. The fundamental quantities should be independent of each other; and
b. All other quantities may be expressed in terms of the fundamental quantities.
Fundamental quantities are also referred to as base quantities.
Who decides the units?
A body named Conference Generale des Poids or CGPM (General Conference on Weight and Measures in English) has been gibven the authority by international agreement.
Definitions of base units
Length metre (m)
Mass kilogram (kg)
Time second (s)
Electric current ampere (A)
Thermodynamic temperature kelvin (K)
Amount of substance mole (mol)
Luminous intensity candela (cd)
Metre
The distance travelled by light in vacuum in 1/[299,792,458] second is called 1 m.
Kilogram
the mass of cylinder made of planitum-iridium alloy kept at International Bureau of Weights and Measures is defined as 1 kg.
Second
The time duration in 9,192,631,770 time periods of the selected transition of radiation of Cesium-133 atom is defined as 1 s. (To understand this idea see Bohr model chapter)
Ampere
To under stand the definition of this unit, you have to read the chapter on electric field.
Two long straight wires with negligible cross section are to be placed parallel to each other at a separation of 1 m and electric current in the same amout is sent through them in the same direction. If the attractive force between the two wires adjusted, the current in each wire (one of the wires) that gives a force between them of 2*10^-7 newton per metre of the wires, is defined as 1 A of current.
Kelvin
The fraction 1/9273.16) of the thermodynamic temperature of triple point of water is called 1 K.
Mole
The amount of a substance that contain as many elementary entities (molecules or atoms if the substance is monatomic) as there are number of atoms in 0.012 kg of carbon-12 is called a mole.
Candela
It is the luminous intensity of a blackbody of surface area 1/[600,000] m² placed at the temperature of freezing platinum and at a pressure of 101,325 N/m², in the direction perpendicular to its surface.
Dimensions of physical quantities and Dimensional Formula
When a physical quantity is expressed in terms of the base quantities, it is written as a product of different powers of the base quantities.
[Force] = MLT-2
M,L, and T are base quantities – mass, length and time
The exponent of a base quantity that enters into the expression (MLT-2) is called the dimension of the quantity in that base.
The dimensions or force are 1 in mass, 1 in length and -2 in time. The dimensions of all other base quantities are zero.
The base quantities are denoted as follows in writing expressions. The symbols used are M for mass, L of length, T for time, I for current, K for temperature, mol for mole and cd for candela.
The physical quantity that is expressed in terms of the base quantities is enclosed in square brackets to inform that it is expressed in dimensions of base quantities. Such expression of a physical quantity in terms of dimensions of base quantities is called the dimensional formula.
Order of magnitude
The structure of the world
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