Wednesday, February 20, 2008

Study Guide and Formula Revision Posts

In this blog, I am posting two types of material.

One study guides on each chapter of the book H C Verma. I am giving the JEE syllabus relevant to the chapter, contents of the chapter, a brief on the concepts covered in the chapter and JEE questions from the chapter.

In the formula revision posts, the formulas given in each chapter are being collected for a quick reference and revision.

You can access both types of posts by clicking the chapter name in labels.

I am myself now reading first, then spending time on doing some problems and then do the posting. Currently I am putting a target of 31st May 2008 for completing all the chapters.

Friday, February 8, 2008

IIT JEE Physics Formula Revision 3. Rest and motion :kinematics

1. vav = s/(t2 - t1)
where vav = average speed

2. v = limΔt→0 Δs/Δt = ds/dt
where v = instantaneous speed

3. s = ∫vdt from t1 to t2
where s = distance travelled during time t1 to t2

4. vav = (r2 - r1)/(t2-t1)
Where
vav = average velocity
r2, r1 = position vectors of a particle
t2, t1 = time instants

5. v = lim Δt→0 Δr/Δt = dr/dt

6. v = |dr|/dt = ds/dt
Where v = magnitude of velocity
s = displacement, for small time intervals magnitude of displacement will be equal to distance.

7. aav = (v2 - v1)/(t2-t1)

8. a = lim Δt→0 Δv/Δt = dv/dt

For Motion in straight line

9. Velocity is v = dx/dt
10. a = dv/dt
11. acceleration is a = dv/dt = d²x/dt²

For constant acceleration
12. v = u+at
13. x = distance moved in time t = ut+½at²
14. v² = u²+2ax

For motion in plane

22. Time of flight of the projectile = (2u sin θ)/g

23. OB = (u²sin 2θ)/g

24. t = (u sin θ)/g
At t vertical component of velocity is zero.

25. Maximum height reached by the projectile (in t = (u sin θ)/g) = (u² sin²θ)/2g

27. V(B,S) = V(B,S')+V(S',S)

Where
V(B,S) = velocity of body wrt to S)
V(B,S') = velocity of body wrt to S')
V(S',S) = velocity of S' wrt to S)

we can rewrite above equation as

28. V(B,S') = V(B,S)- V(S',S)

IIT JEE Physics Formula Revision 5. Newtons law of motion

1. Acceleration vectoS a(P,S) = Acceleration vector a(P,S') if S is an inertial frame and S' is a frame moving uniformly with respect to S.

IIT JEE Physics Formula Revision 6. Friction

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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IIT JEE Physics Formula Revision 7. Circular motion

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

Linear variables of circular motion

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

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

at = dv/dt = rdω/dt = r α

Unit Vectors along the Radius and Tangent on a point on the circle on which the particle is moving.

If x-axis is horizontal and y axis is vertical (normal representation)

er = unit vector along the radius at a point on the circle

er = i cos θ+j sin θ

et = unit vector along the tangent at a point on the circle

et = - i sin θ+ j cos θ

Position vector of the point P

If the angular position is θ, x coordinate is r cos θ, and y coordinate is r sin θ.

Hence position vector is r = r(i cos θ + j sin θ)

Differentiating position vector with respect to time we get velocity

v = r ω(-i sin θ + j cos θ)

Differentiating v with respect to time

a = dv/dt = - ω²r er + dv/dt et

Uniform Circular Motion

If the particle moves in the circle with a uniform speed (v = constant), it is uniform
circular motion.

In this case dv/dt = 0
Hence
a = dv/dt = - ω²r er

Acceleration of the particle is in the direction of - er , i.e, towards the centre of the circle. The magnitude of the acceleration is

a = ω²r = v²/r (as v = rω)

Nonuniform circular motion

In nonuniform circular motion, speed is not constant and the acceleration of the particle has both radial and tangential components.

a = - ω²r er + dv/dt et

the radial component is ar = - ω²r = -v²/r
the tangential component is at = dv/dt

The magnitude of the acceleration

a = √ (ar² + at²)

= √[( ω²r) ² + (dv/dt) ²]

The direction of this acceleration makes an angle α with the radius connecting the centre of the circle with the point at that instant.

Tan α = (dv/dt)/ (ω²r)

Forces in Circular Motion

If a particle of mass m is moving along a circle with uniform speed, the force acting on it has to be

F = ma = mv²/r = m ω²r

This force is called centripetal force.

Centrifugal Force

It is equal to the centripetal force

F = mv²/r = m ω²r

Circular Turnings on the Road

When a turn is made, the speed has to be less so that friction can make the vehicle turn in stead of skid

For a safe turn we need to have

v²/r = fs/M

Or M v²/r ≤ fs

As fs≤ μsMg

M v²/r ≤ μsMg
That means v²/gr ≤ μs
v≤√( μsgr)

Instead of relying on friction, road are given an angle to provide centripetal force.

Tan θ = v²/gr

θ is the angle to be given to the road based v. Road are designed for assumed normal speed.

Effect of Earth’s Rotation on Apparent Weight

Axis of rotation of earth = The line joining the north and south pole = SN
Every point P on earth moves in circle.
Draw a perpendicular PC to SN from P.
Centre of the earth = O.
The distance between a point on the equator and centre of earth O = R radius of earth.

The angle between OP and PC is θ

θ is called the colatitude of the place where P is there.

If heavy particle of mass m is suspended through a string at P, the forces on it will be

1. gravitational attraction mg towards earth along PO
2. Centrifugal force = m ω² r
3. Tension in the string

The resultant of mg and m ω² r =
m√(g²+ ω4R²sin² θ-2g ω²R sin² θ) = mg’

The direction of this resultant force with OP is given by

Tan α = ω²Rsin θ cos θ/(g - ω²Rsin² θ)

A plumbline stays along g’

g’ = √(g²- ω²Rsin² θ(2g- ω²R ))

As 2g> ω²R; g’
At equator θ = 90° and

G’ = g- ω²R

At poles θ = 0° and hence

g’ = g

IIT JEE Physics Formula Revision 8. Work and energy

1. K(v) = 1/2mv^2

2. dK = F.dr (both F and dr are vectors)

3. W = ∫F cosθ dr

IIT JEE Physics Formula Revision 15. Wave motion and waves on a string

1. Equation of a wave travelling in the positive x-direction with a constant speed v.

The displacement of the particle at x at time t i.e., y(x,t) is generally abbreviated as y and the wave equation is written as

y = f(t - x/v) ... (1)

2. Equation of a wave travelling in the negative x-direction with a constant speed v.

The displacement of the particle at x at time t i.e., y(x,t) is generally abbreviated as y and the wave equation is written as

y = f(t + x/v) ... (2)

3. The wave equation in (1) can be written as y = f((vt-x)/v) which can be transformed into y = g(x-vt)...(3)

g is a different function. Function g can have the following meaning. If you put t = 0 in equation (3), you get the displacement of various particles at t = 0;

y(x, t = 0) = g(x)

If displacement at t = 0 of all particles of the string is represented by g(x) then the displacement of the particle at x at time t will be y = g(x-vt).

4. Similarly if the wave is travelling along the negative x-direction and the displacement of different particles at t = 0 is g(x), the displacement of the aprticle at x at time t will be

y = g(x+vt)

Function f in equation 1 and 2 represents the displacement of the point x = 0 as time passes,and g in (3) and (4) represents the diplacement at t = 0 of different particles.

5. sine wave or sinusoidal wave

When a person vibrates the left end of a string x = 0 in a simple harmonic motion, the equation of motion of this end may be written as

f(t) = A sin ωt ... (5)

A represents the amplitude
ω = the angular frequency
Time period of oscillation is T = 2π/ω
Frequency of oscillation = 1/T = ω/2π

This wave is called a sine wave or sinusoidal wave.

If the displacement of the particle at x = 0 is given by f(t) = A sin ωt, the displacement of the particle at x at time t will be given by

y = f(t - x/v) = A sin ω(t - x/v)...(6)

7. Velocity of the particle at x at time t is given by

ðy/ðt = Aω cos ω(t - x/v)...(7)

This velocity is different from the velocity of the wave. The wave moves on the string at a constant velocity v along the x axis, but the particles at various points move up and down with velocity ðy/ðt which changes with x and t according to equation (7).

8. λ = (v/ω)2π = vT .... (8)

9. v = λ/T = νλ ... (9)

where ν = 1/T is the frequency of the wave.

Alternative sine wave equations

y = A sin ω(t - x/v)

y = A sin (ωt - kx)... (10)

y = A sin 2π(t/T - x/λ)... (11)

y = A sin k(vt-x) ... (12)

General equation will be

y = A sin [ω(t - x/v)+ Φ)... (13)

Φ is the phase constant.

This equation will allow us to write equation based on the displacement of the left at t = 0.

The constant Φ will be π/2 is we choose t = 0at an instant when the left end reaches its extreme position y = A. The equation will then be

y = A cos ω(t - x/v)... (14)

If t = 0 is taken at the instant when the left end is crossing the mean position from upward to downward direction, Φ will be π and the equation will be

y = A sin (kx-ωt)... (15)

16. Velocity of a wave on a string

v = √(F/μ)

F = tension in the string
μ = linear mass density of the string ... (16)

17. Power transmitted along the string by a sine wave

Pav = 1/2 [ω² A² F²/v]... (17)

IIT JEE Physics Formula Revision 23. Heat and Temperature

1. conversion formula is F = 32 + 9C/5

2. The temperature of triple point of water is assigned a value of 273.16 K

The temperature of ice point on the ideal gas scale is 273.15 K and of the steam point is T = 373.15 K. The interval between the two is 100 K.

Centigrade scale or Celsius scale is defined to have ice point at 0°C and steam point at 100°C. The interval is 100°C.

Hence if θ represents the Celsius or centigrade temperature

V = T – 273.15 K

3. Ideal gas equation

pV = nRT

n = the amount of gas in moles
R = universal constant = 8.314 J/mol-K

Thermal expansion

Average coefficient of linear expansion

Average (α) = (1/L)*∆L/∆T

Coefficient of linear expansion at temperature T
α = Lim (∆T->0)(1/L)*∆L/∆T = (1/L)dL/dT

Suppose the length of a rod is L0 at 0° C and Lθ at temperature θ measured in Celsius. If α is small and constant over the given temperature interval,

α = Lθ-L0/L0*θ or

Lθ = L0(1+αθ)

The coefficient of volume expansion is defined in a similar way.

γ = (1/V)dV/dT

It is also known as coefficient of cubical expansion.

Vθ = V0(1+γθ)

γ = 3α

IIT JEE Physics Formula Revision 24. Kinetic theory of Gases

1. p = 1/3*ρ*Mean(v²)

p = pressure of gas
ρ = density of gas
Mean(v²) = mean square speed of molecules of gas

2. pV = 1/3*M*Mean(v²)

V = Volume of gas
M = Total mass of the gas taken

3. pV = 1/3*Nm*Mean(v²)

N = total number of molecules in the sample
m = mass of one molecule

4. Mean(v²) = ΣV²/N
Square root of Mean(v²) is also called root-mean square speed vrms

Equation 1 can be written as p = 1/3*ρ*v²rms

So vrms = SQRT(3p/ρ)

6. In thermal equilibrium in a mixture of two gases, the average kinetic energies of all molecules are equal. If v1 an v2 are rms speeds of the molecules of A and B

½m1v1² = ½m2v2²

7. pV = Nkt

k is known as the Boltzmann constant.
Its value is 1.38*10^(-23) J/K

R = NA*k (NA is Avogadro's number = 6.02*10^23
R = 8.314 J/mol-K

8. pV = nRT

9. RMS spped in terms of temperature
v(rms) = SQRT(3kT/m)

10. v(rms) = SQRT(3RT/M0)

11. Average kinetic energy of a molecule
½mv(rms)² = (3/2) kT

12. Total kinetic energy of all molecules
U = (3/2)nNA kT
or
U = 3/2nRT ( R = NA*k)
n = number of moles of substance

13. maxwells speed distribution law

dN = 4 π[m/2 πkT]3/2v²e-(mv²/2kT)dv

14. The speed at which dN/dv is maximum is called the most probable speed.

vp = SQRT(2kT/m)

15. van der Waals equation for real gases

For an ideal gas
pV = nRT

This is called the equation of state for that substance

For a real gas

[p + a/V²][V-b] = nRT
where a and b are small positive constants.
a is related to the average force of all attracion between the molecules.
bis related to the total volume of the molecules.
This equation is given by van der Waals.

16. Relative humidity = RH =

Amount of water vapour present in a given volume of air at a given temperature
--------------------------------------------------------------------------------------------------
Amount of water vapour required to saturate the same volume of air at the same tempeture

RH may also be defined as

Vapour pressure of air/SVP at the same temperature

SVP = Saturation vapour pressure: The pressure exerted by a saturated vapour is called saturation vapour pressure.

The RH may also be defined as

SVP at the dew point/SVP at the air temperature

As the vapour pressure of air at the actual temperature is equal to the SVP at the dew point.

IIT JEE Physics Formula Revision 25. Calorimetry

1. Q = mS∆θ

S = specific heat capacity of a substance

2. Q = mC∆θ
C = molar specific heat capacity of a substance

3. W = JH
W = work measured in joule
H = Heat measured in calories
J = mechanical equivalent to heat
= joule/calorie = 4.186 J/cal

IIT JEE Physics Formula Revision 26. Laws of Thermodynamics

First law of thermodynamics and its applications (only for ideal gases) 26.1;

26.1 ΔU = ΔQ - ΔW or

ΔQ = ΔU + ΔW

26.2 ΔW =∫pΔV from V1 to V2

26.3 Work done in an Isothermal process:
W is given by the expression nRTln[V2/V1]

Work doen in an Isobaric process - pressure is constant

W = p[V2 - V1]

Work done in an isochoric process: volume is constant. No work is done by the system.

4. Efficiency of an engine

η = W/Q1 = 1 - Q2/Q1

Q1 = Amout of heat taken by the engine
Q2 = Heat rejected by the engine to low temperature bodies

5. Entropy

∆S = ∆Q/T

6. Sf - Si = ∫∆Q/T from i to f

IIT JEE Physics Formula Revision 27. Specific Heat of Capacities of Gases

1. Specific heat capacity of a substance s

s = ΔQ/mΔT

2. Specific heat capacity of a substance at constant volume is denoted by symbol sv

sv = [ΔQ/mΔT] heat given at constant volume

3. Specific heat capacity of a substance at constant pressure is denoted by symbol sp

sp = [ΔQ/mΔT] heat given at constant pressure

4. Molar heat capacity of a substance at constant volume is denoted by symbol Cv

Cv = [ΔQ/nΔT] heat given at constant volume

5. Molar heat capacity of a substance at constant pressure is denoted by symbol Cp

Cp = [ΔQ/mΔT] heat given at constant pressure

6. Relation between Cp and Cv for an ideal gas

Cp - Cv = R
or
Cp = Cv + R

7. Cv = i/n (dU/dT)
dU = n(Cv)dT
8. Taking energy to be zero at t = 0
U = n(Cv)T

9. pVγ = constant

γ = Cp/Cv

10. In a reversible adiabatic process

Tγ/pγ-1 = constant

11. In a reversible adiabatic process

TVγ-1 = constant

12. Work done in an adiabatic process

W = (p1V1 - p2V2)/(γ-1)

For monoatomic gas

U = n (3/2) RT

Cv = 3/2R

Cp = 5/2R

γ = 5/3 = 1.67

For diatomic gas without vibration

Cv = 5/2R

Cp = 7/2R

γ = 7/5 = 1.40

If the molecules vibrate

Cv = 7/2R

Cp = 9/2R

γ = 9/7 = 1.29

IIT JEE Physics Formula Revision 28. Heat transfer

1. ∆Q/∆t = KA(T1-T2)/x

2. ∆Q/∆t = -KAdT/dx

3. x/KA is called the thermal resistance R.
∆Q/∆t is termed as heat current i

i = T1-T2/R

4. In the experimental determination of thermal conductivity of a solid

K = [xms(θ3-θ4)]/[A(θ1-θ2)t]

5. Wien's displacement law
λmT = b = constant

6. Stefan-Boltzmann Law

The energy of thermal radiation emitted by per unit time by a black body of surface area A is given by

U = σAT4

Where
σ = Stefan Boltzmann constant = 5.67*10-8 W/m²-K4

The energy of thermal radiation emitted by per unit time by a body (not black) of surface area A is given by

U = eσAT4

e is a constant for the given surface called emissivity of the surface. Its value is between 0 and 1. It is one for black body and 0 for completely reflecting surface.

7. Net loss of thermal energy of body with temperature T kept in a room at temperature T0

u = eσA(T4 - T04)

8. Newton’s law of cooling

dT/dt = -bA(T-T0)

b = a constant depends on the nature of the surface involved
a = surface area exposed of the body
T- T0) = temperature difference between the body and surrounding

IIT JEE Physics Formula Revision 29. Electric Field and Potential

For revision points of the chapter

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

The charge on a proton is
e = 1.60218*10^-19 C

Coulomb's formula for the electric force between two charges.

(1) F = k*q1*q2/r²

q1,q2 charges
r = separation between charges
k = constant
In SI units k is measured to be 8.98755*10^9 N-m²/C²

The constant k is often written as 1/4πε0.
The constant ε0 is called the permittivity of the space and its value is

ε0 = 8.85419*10^-12 C²/N-m²

The intensity of field is defined as Vector E = Vector F/q

(3) E = F/q

Electric field due to a point charge (Q)

(4) E = F/q = Q/4πε0

(5) U(r2) - U(r1) = -W = [q1*q2/4πε0]* [(1/r2) - (1/r1)]

Choosing potential energy at infinite separation as zero

(6) U(r) = U(r) - U(∞) = [q1*q2/4πε0]* (1/r)
= q1*q2/4πε0r

(7) VB - VA = (UB - UA)/q

We can define absolute electric potential at any point by choosing a reference point P and saying that the potential at this point is zero.

(8) VA = VA - VP = (UA - UP)/q

The potential due to a point charge Q (placed at A) at a point P with distance AP = r is

(9) VP = Q/4πε0r
The electric potential at a point due to more than one charge in the system is obtained by finding potential due to individual charges and then adding them.

10. Relation between electric field and potential

dV = -E.dr

Integrating between limits r1 and r2

V2 – V1 = - ∫ E.dr

If we choose r1 as infinity (reference point)

V(r) = - ∫ E.dr (from infinity to r)

Hence we can calculate potential V if we know E and r.

If we know V we can find E through the relation

Ex = - ∂V/∂x
Ey = -∂V/∂y
Ez = -∂V/∂z

We can find the x,y and z components of E.

E can be written as Ex i +Ey j + Ez k

We can write dV as dV = -E dr cos θ where θ is the angle between the field E and the small displacement dr.

-dV/dr = E cos θ

15. Electric Dipole

Electric dipole moment
It is defined as a vector p = q*distance vector d
where distance vector is the vector joining the negative charge to the positive charge.
The line along the direction of the dipole moment is called the axis of the dipole.

16. Electric potential due to a dipole at a point P

Point is at distance r from the centre of the diploe (d/2) and theline joining the point P to the centre of the dipole make an angle θ with the direction of dipole movement (from –q to +q)

Potential at P due to charge –q = - [1/4πε0][q/(r + (dcos θ)/2)]
Potential at P due to charge q = [1/4πε0][q/(r - (dcos θ)/2)]
Net potential due to q and –q = [1/4πε0](qd cos θ)/r²

17. Electric field due to a dipole

Er = [1/4πε0](2p cos θ)/r³

Eθ = [1/4πε0](p sin θ)/r³

Resultant electric field at P = E= √ (Er²+ Eθ²)

= [1/4πε0](p/r³)√(3 cos²θ + 1)

19. Torque on an electric dipole placed in an electric field.

If the dipole axis makes an angle θ with the electric field magnitude of the torque = | Γ| = pE sin θ

In vector notation Γ = p × E

20. Potential energy of a dipole placed in a uniform electric field

dipole axis makes an angle θ with the electric field magnitude of the torque

Change in potential energy = U(θ) – U(90°) = -pE cos θ = -p.E

If we choose the potential energy of the dipole to be zero when θ = 90° , above equation becomes
U(θ) = -pE cos θ = -p.E

IIT JEE Physics Formula Revision 30. Gauss's Law

1. ∆Φ = E ∆s cos θ

∆Φ is the flux of the electric field (E) through the surface ∆s. θ is the angle between positive normal to the surface and electric field.

Using the techniques of integration flux over a surface is:

Φ = ∫ E.∆s
Flux over a closed surface

Φ = ∮ E.∆s (* Intregration over a closed surface)

2. Solid angle

Ω = S/r²

Ω = solid angle (dimensionless figure)

S = the area of the part of sphere intercepted by the cone
r = radius of the sphere assumed on which we are assuming the cone

A complete circle subtends an angle 2 π

Any closed surface subtends a solid angle 4 π at the centre.

How much is the angle subtended by a closed plane curve at an external point? Zero.

3. Gauss's law

The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by ε0.

In symbols

E.∆s (* Intregration over a closed surface) =
qin0.

Where
qin = charge enclosed by the closed surface
ε0 = emittivity of the free space

4. Electric field due to a uniformly charged sphere
A total charge Q is uniformly distributed in a spherical volume of radius R. what is the electric field at a distance r from the centre of the charge distribution outside the sphere?
E = Q/4π ε0

5. Field at an internal point

At centre E = O

At any other point inside (rless than R) radius of the sphere

E = Qr/4π ε0

6. c. Electric field due to a linear charge distribution

The linear charge density (charge per unit length) is λ.

Electric field at a distance r from the linear charge distribution

E = λ/2π ε0r

7. d. Electric field due to a plane sheet of charge

Plane sheet with charge density (charge per unit area) σ.

Field at distance d from the sheet = E = σ/2ε0

We see that the field is uniform and does not depend on the distance from the charge sheet. This is true as long as the sheet is large as compared to its distance from P.

8. Electric field due to a charged conducting surface

To find the field at a point near this surface but outside the surface having charge density σ.

E = σ/ ε0

Updated 12 October 2008

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IIT JEE Physics Formula Revision - Ch. 31. Capacitors

1. For a given capacitor, the charge Q on the capacitor is proportional to the potential difference V between the two plates

So Q α V
or Q = CV

C is called the capacitance of the capacitor.
SI unit of capacitance is coulomb/volt which is written as farad. The symbol F is used for it.

2. For parallel plate capacitor

C = ε0A/d

A = area of the flat plates (each used in the capacitor)
d = distance between the plate

3. Spherical capacitor

It consists of a solid or hollow spherical conductor surrounded by another concentric hollow spherical conductor.

If inner sphere radius is R1 and Outer sphere radius is R2

Inner sphere is given positive charge and outer sphere negative charge.

C = 4πε0R1R2/[R2-R1]

If the capacitor is an isolated sphere (outer sphere is assumed to be at infinity, hence R2 is infinity and

C = 4πε0R1

V becomes Q/C = Q/4πε0R1

V = potential

Parallel limit: if both R1 and R2 are made large but R2-R1 = d is kept fixed

we can write
4πR1R2 = 4πR² = A; where R is approximately the radius of each sphere, and A is the surface area of the sphere.

C = ε0A/d; where A = 4πR1R2 = 4πR²

4. Cylindrical Capacitor

If inner cylinder radius is R1 and Outer cylinder radius is R2 and length is l,
Inner cylinder is given positive charge and outer cylinder negative charge

C = 2πε0l/ln(R2/R1)

5. Combination of capacitors

Series combination

1/C = 1/C1 + 1/C2 + 1/C3 ...

Parallel combination

C = C1 + C2 + C3

6. Force between plates of a capacitor

Plates on a parallel capacitor attract each other with a force

F = Q²/2Aε0

7. Energy stored in a capacitor

Capacitor of capacitance C has a stored energy

U = Q²/2C = CV²/2 = QV/2

Where Q is the charge given to it.

8. Change in capacitance of a capacitor with dielectric in it.

Polarization P (which is dipole moment induced per unit volume - where is the dipole? in the diectric slab as the two sides have opposite charges)

If σp is the magnitude of the induced charge per unit area on the faces.

The dipole moment (q*Vr(d)) of the slab is then charge*l (distance between faces)
= σpAl.
where

A is area of cross section of the dielectric slab

As polarization is defined as dipole moment induced per unit volume,
P = σpAl/Al (Al = volume of slab)

= σp

The induced surface charge density is equal in magnitude to the polarization P.

9. Capacitance of a parallel plate capacitor with dielectric

C = KC0
where C0 is capacitance of a similar capacitor without dielectric.

Because K>1, the capacitance of a capacitor is increased by a factor of K when the space between the parallel plates is filled with a dielectric.
Magnitude of induced charge in term of K

QP = Q[1 - (1/K)]

QP = induced charge in the dielectric
Q = Applied charge
K = dielectric constant

10. Gauss's law when dielectric materials are involved

∮KE.dS = Qfree0

Where integration is over the surface, E and dS are vectors, Qfree is the free charge given (charge due to polarisation is not considered) and K is dielectric constant.

The law can also be written as

D.dS = Q(free)
where D = Eε0 + P; E and P are vectors
E = electric field and P is polarisation

11. Electric field due to a point charge placed inside a dielectric

E = q/4πε0Kr²

Energy in the electric field in a dielectric

u = ½Kε0

IIT JEE Physics Formula Revision 32. Electric Current in Conductors

1. Average current i(bar) = ∆Q/∆t

The current at time t = I = lim ∆t→0 ( ∆Q/∆t) = dQ/dt

Q is charge, t is time

2. Current density

Average current density j (bar) = Δi/ΔA

i is current and A is area of the conductor

The current density at a point P is
j = lim ∆t→0 (Δi/ΔA) = di/dS

If current i is uniformly distributed over an area S and is perpendicular to it

j = i/S

For a finite area i = ∫j.dS

Where
j = density of current (vector)
dS = area (vector)

3. If τ be the average time between successive collisions, the distance drifted during this period is

l = ½ a(τ) = ½ (eE/m)( τ) ²

The drift speed is

vd = l/ τ = ½ (eE/m)τ = kE

τ the average time between successive collisions, is constant for a given material at a given temperature.

Relation between current density and drift speed

j = i/A = nevd

4. Ohm's law

j = σE

E is field and σ is electrical conductivity of the material.

Resistivity of a material ρ = 1/σ

Another form of Ohm's law

V = voltage difference between the ends of a conductor = El (l = length of the conductor)

V = Ri

R = resistance of the conductor = ρ*l/A

1/R is called conductance

5. As temperature of a resistor increases its resistance increases. The relation can be expressed as

R(T) = R(T0)[1 + α(T - T0)]

α is called temperature coefficient of resistivity.

6. The work done by the battery force per unit charge is

Є = W/q = Fb*d/q

This Є is called the emf of the battery. Please note that emf is work done/charge.

If nothing is externally connected
Fb = qE or
Fb*d = qEd = qV (because V =Ed)

V = potential difference between the terminals
As Є = W/q = Fb*d/q = qV/q = V

Therefore Є = V

7. Thermal energy produced in a register

U = i²Rt

Power developed = P = U/t = i²R = Vi

8. Combination of resistors in series

Equivalent resistance = R1 + R2 +R3+...

Combination of resistors in parallel

1/Equivalent resistance = 1/R1 + 1/R2 +1/R3+...

Division of current in resistors joined in parallel

i1/i2 = R2/R1

i1 = iR2/(R1 + R2)

9. Batteries connected in series

i = (Є 1 + Є 2)/(R + r0)
Where R = external resistance
r0 = r1 + r2 r1, r2 are internal resistances of two batteries

Batteries connected in parallel

Equivalent emf = Є 0 = [Є 1r2 + Є 2r1]/(r1+r2)
where Є 1, Є 2 are emfs of of batteries , and r1, r2 are internal resistances.
equivalent internal restance = r0 = r1r2/(r1 + r2)

So i in the circuit = ε0/(R + r0)

10. Wheatstone Bridge ; R1 and R2 are two resistances connected in series. R3 and R4 are the other two resistance connected series. If the there is no deflection in the galvanometer

R4 = R3R2/R1

as R1/R2 = R3/R4

11. Charging of the capacitor

q = Є (1 - e^-t/CR)
q is charge on the capacitor, t is time, Є = emf of the battery, C = capacitance, R is resistance of battery and connecting wires,
CR has units of time and is termed time constant. In one time constant τ (=CR) the charge accumulated becomes 0.63 Є C.

12. Discharging of the capacitor

q = Qe^(-t/CR)
where q is charge remaining on the capacitor
Q is the initial charge
In one time constant 0.63% is discharged.

IIT JEE Physics Formula Revision 33. Thermal and Chemical Effects of Electric Current

1. Joule's laws of heating

Work done by the electric field on the free electrons in time t is

W = potential difference * charge
= V(it)
= (iR)it = i²Rt

2. If θc, θn, θi are temperature of the cold junction, neutral temperature and inversion temperature respectively
c

θn – θc = θi - θn

3. Thermo emf depends on temperature with the relation

ЄAB = aABθ + ½ bsub>ABθ²

Where aAB and bAB are constants for a pair of metals A and B.

This gives dЄAB/dθ = aAB + bsub>ABθ

The quantity dЄAB/dθ is called thermoelectric power at temperature θ.

5. Law of intermediate Metal

Suppose ЄAB, ЄAC, ЄBC are emfs from thermocoupes AB,AC, and BC.

If hot junctions and cold junctions of the three thermocouples are at the same temperature,
Then,

ЄAB = ЄAC - ЄBC

6. Law of intermediate temperature

Let Є θ1, θ2, represent the thermo-emf of a given thermocouple when the temperatures of junctions are maintained at θ1and θ2. Then

Є θ1, θ2 = Є θ1, θ3 + Є θ3, θ2

7. The actual emf produced in a thermocouple loop is the algebraic sum of the net Peltier effect and the net Thomspon effect.

So emf in a thermocouple loop = ЄAB =
AB)T - (Π AB ) T0 +(T – T0) (σA - σA)

(More explanation is needed for the formula.)

IIT JEE Physics Formula Revision 34. Magnetic Field

1. Magnetic field can be defined mathematically as

F = qv × B

Equation uniquely determines the direction of magnetic field B from the rules of the vector product.

F = force (vector)
q = charge
v = velocity (vector)
B = magnetic field (vector)

2. Motion of a Charged particle in a uniform magnetic field

qvB = mv²/r

r = mv/qB

The time taken to complete the circle is
T = 2πr/v = 2πm/qB

Frequency of revolutions is

ν = 1/T = qB/2πm
This frequency is called cyclotron frequency.

6. If a straight wire of length l carry8ng a current i is placed in a uniform magnetic field B, the force on it is

F = il×B

Where
i = current in the conductor
l = vector of length of the conductor
B = magnetic field

7. Formula for Torque on a current loop

If there is a rectangular loop carrying current i in a uniform magnetic field B
then net torque acting on the loop is

Г = iABsin θ

Where i = current in the loop
A = area (magnitude)
B = magnetic field (magnitude)
θ = the angle of inclination of the loop with the plane perpendicular to the plane of magnetic field.

We can also define in terms of vector

Г = iA× B

iA can be termed as μ the magnetic dipole moment or simply magnetic moment of the current loop.

If there are n turns in the loop, each turn experiences a torque.

The net torque is
Г = niA× B

μ = niA

IIT JEE Physics Formula Revision 35. Magnetic field due to a Current

Note: vectors are shown in bold letters

1. Biot Savart Law

dB = [1/4 π ε0c²]*(i) (dl*r/r³) ... (1)
where

dB = magnetic field at point P, due to current element dl
c = speed of light
i = current
r = vector joining the current element to the point P.

[1/ ε0c²] is written as µ0 and is called the permeability of vacuum.
Its value is 4 π*10-7 T-m/A

In terms of µ0 equation (1) becomes

dB = (µ0 /4 π)*(i) (dl*r/r³) ... (2)

The magnitude of the field

dB = (µ0 /4 π)*(idl sin θ/r²)

where θ is the angle between dl and r.
the direction of the field is perpendicular to the palne containing the current element and the point P according to the rules of the cross product.

4. Magnetic field due to current in a straight wire

B = (µ0i /4 πd) (cos θ1 – cos θ2) … (4)

θ1 and θ2 and the value of θ corresponding to the lower end and the upper end respectively.

If the point P is on the perpendicular bisector of the straight wire

B = (µ0i /4 πd) [2a/√(a² +4d²)] … (5)

a = length of the wire
d = distance between the wire and point P (perpendicular distance)

If the wire is very long such that θ1 = 0 and θ2 = π.
The equation is B = (µ0i /2 πd) … (6)

7. Force between two parallel currents

dF/dl = (µ0i1i2 )/2 πd

dF/dl = force per unit length of the wire W2 due to wire W1
i1,i2 = current in wire W1 and W2 respectively
d = distance between wires kept parallel to each other.

8. Magnetic field in an axial point

B = (µ0ia²)/[2(a²+d²)3/2] .. (8)

If the point is far away from the centre d is very large compared to a

B = (µ0ia²)/2d³

If the magnetic dipole moment of due to the circular conductor with area πa² and current flowing through it is ‘i’, the µ = i πa² and B is

B = (µ0i /4π)(2µ/d³) .. (9)

10. Ampere’s law

The circulation of ∫B.dl of the resultant magnetic field along a closed plane curve is equal to µ0 times the total current crossing the area bounded by the closed curve provided the electric field inside the loop remains constant.

The circulation ∫B.dl over closed curve = µ0*I … (10)

11. Magnetic field inside a solenoid

B = µ0ni

n = number of turns per unit length along the length of the solenoid.
i = current passing through the solenoid

12. toroid

B = µ0Ni/2πr
Where
N = total number of turns
i = current in the toroid
r = distance of the point at which field is being calculated from centre of the toroid.

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IIT JEE Physics Formula Revision 36. Permanent Magnets

1. force

A magnetic charge m placed in a magnetic field B experiences a force.

F = mB .. (1)

2. Magnetic field due to magnetic charge

B = (µ0/4 π)(m/r²)

3. Pole strength due to current

m = IA ...(3)

Where
m = pole strength
I = surface current per unit length of the magnet
A = cross sectional of the magnet

4. Magnetic moment of a bar magnet

M = 2ml … (4)

Where
M = Magnetic moment of a bar magnet
M = pole strength
2l = magnetic length of the bar magnet

5. Potential energy at an angle θ

U(θ) = -MB cos θ = -M.B …(5)

6. Magnetic field due to a bar magnet

End on position. A position on the magnetic axis of a bar magnet is called an end on position

B = (µ0/4 π)[(2Md)/(d² – l²)²] …(6)

If d is very large compared to l, then

B = (µ0/4 π)(2M/d³) … (7)

B = (µ0/4 π)[m 2l/(d² + l²)3/2 ] …(8)
= (µ0/4 π)[M/(d² + l²)3/2 ]

If d is very large compared to l,

B = (µ0/4 π)[M/(d³)

10. Magnetic scalar potential

V(r2) – V(r1) = -r1r2 B.dr … (10)

11. The component of the magnetic field in any direction is given by

Bl = -dV/dl ... (11)

12.For a pole of pole strength m, the field at a distance r is

B = (µ0/4 π)[m/r²)

So the potential at a distance r is

V( r )= - r 0/4 π)[m/r²)dr

= (µ0/4 π)[m/r) ... (12)

13. Magnetic scalar potential due to a magnetic dipole

Magnetic scalar potential at a point P which is at a distance r from the mid point of the magnetic dipole, and the angle between the dipole axis and the line joining the mid point of the dipole to the point P is θ

V = (µ0/4 π)[Mcos θ /r²)
Where
M = 2ml =magnetic moment of the dipole

14. Magnetic field due to dipole

Magnetic field at P =

0/4 π)[M /r²)√(1 +3 cos² θ)] .. (14)

17. current in galvanometer

i = K tan θ .. (17)

where K = 2rBH0 for the given galvanometer at a given place.

18. Current in moving coil galvanometer

i = (k/nAB) θ … (18)

the constant (k/nAB) is called the galvanometer constant and may be found by passing a known current, measuring the deflection θ and putting these values in equation (18).

19 Shunt
Shunt is a small resistance.
Current in galvanometer

ig = [Rs/(Rs+g)]*i

i = main current
ig = current that goes through galvanometer

20. tangent law of perpendicular fields

B = BHtan θ .. (20)

21. Deflection magnetometer

Tan A position

M/ BH = (4 π/µ0)[(d² – l²)²/2d] tan θ

22. Deflection magnetometer Tan B position

M/ BH = (4 π/µ0)[(d² – l²)3/2] tan θ.. (22)

23. Oscillation Magnetometer

Time period T = 2 π/ ω = 2 π√ (I/M BH) .. (23)
From equation (23)

M BH) = 4 π²I/T² … (24)

25. Gauss’s law for magnetism: the flux of the magnetic field through any closed surface is zero

integral over the closed surface ∫B.ds = 0 .. (25)

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IIT JEE Physics Formula Revision 37. Magnetic Properties of Matter

Ch. 37 Magnetic Properties of Matter – Formulae

1. Magnetization vector = Magnetic moment per unit volume

I = M/V

2. Magnetic intensity

H = B0 - 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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IIT JEE Physics Formula Revision 38. Electro Magentic Induction

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)

IIT JEE Physics Formula Revision 39. Alternating current

Ch. 39 Alternating Current

1. Alternating current

i = i0 sin (ωt + φ) .. (1)

where
i = current at time t
i0 = peak current
current repeats after each time interval T = 2 π/ω

2. Emf produced in a AC generator

Є = NBA ω sin ωt = Є0 sin ωt … (2)

Where
N = number of turns on the armature
B = Magnetic field
A = area of the coil on the armature
ω = angular velocity of the armature

3. RMS (rms) current in a AC circuit

irms = i0/√2 … (3)
The alternating current and voltage are generally measured and expressed in termsof their rms values. When the household current is expressed as 220 V AC, it means that the rms value of voltage is 220 V. The peak voltage value is (220 V) √2 = 311 V.

4. current in an AC circuit containing only resistor

i = i0 sin ωt …… (4)
where
i0 = Є0/R ... (5)

6. Ac circuit containing only a capacitor

i = i0 cos ωt … (6)

where
i0 =C Є0ω
= Є0/(1/ωC) … (7)

8. AC circuit containing only an inductor

i = (Є0/ωL) sin (ωt – π/2) …(8)

or i = i0 sin (ωt – π/2)
where i0 = Є0/ωL

10. Impedance

The peak current and the peak emf in Ac circuits may be written as

i0 = Є0/Z … (10)

where Z = R for a purely resistive circuit
Z = 1/ ωC for a purely capacitive circuit
Z = ωL for a purely inductive circuit

11. Vector method to find the current in an AC circuit

i = (Є0/Z) sin (ωt + φ)

where
Z = √(R² + X²)
R = resistance in the circuit
X = reactance in the circuit (X due to capacitance is 1/ ωC and X due to inductance is ωL)

φ is to be determined from tan φ = X/R

15. Power in Ac Circuits

P = Єrms irmscos φ

16. Voltages in primary and secondary of transformer

Є2 = - (N2 / N1) Є1 .... (16)

Power transfer

Є1i1 = Є2 i2 …(17)

i2 = - (N1 / N2) i1

‘-‘ sign shows that i2 is 180° out of phase with i1 .

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IIT JEE Physics Formula Revision 40. Electromagentic Waves

Electromagnetic waves formula revision

1. The wave equation for light propagating in x-direction in vacuum may be written as

E = E0 sin ω(t-x/c)

Where E is the sinusoidally varying electric field at the position x at time t.
c is the speed of light in vacuum.
The electric field is in the Y-Z plane. It is perpendicular to the direction of propagation of the wave.

There is a sinusoidally varying magnetic field associated with the electric field when light is propagating. This magnetic field is perpendicular to the direction of wave propagation and the electric field E.

B = B0 sin ω(t-x/c)

Such a combination of mutually perpendicular electric and magnetic fields constitute an electromagnetic wave in vacuum.

2. Maxwell generalised Ampere’s law to

∫B.dl = µ0(i + id)

id = ε0*(d ΦE/dt)

Where

ΦE/ = the flux of the electric field through the area bounded by the closed curve along which the circulation of B is calculated.

3. Maxwell’s Equations

Gauss’s laws for electricity and magnetism, Faraday’s law and Ampere’s are collectively known as Maxwell’s equations

Gauss’s law of electricity

∮E.ds = q/ ε0

Gauss’s law for magnetism

∮B.ds = 0

∮E.dl = -dΦB/dt

Ampere’s law

∮B.dl = µ0(i + id)

id = ε0*(d ΦE/dt)

These equations are satisfied by a plane electromagnetic wave given by
Ey = E = E0 sin ω(t-x/c)
Bz = B = B0 sin ω(t-x/c)

4. c = i/√(µ0ε0)
the value calculated from this expression comes out to be 2.99793*10^8 m/s which was same as the experimentally measured value of speed of light in vacuum. This also provides a confirmatory proof that light is an electromagnetic wave.

5. Total energy of the electromagnetic wave is

U = ½ ε0E²dV + B²dV/2µ0

When we substitute the values of E and B in the above equation and take an average over a longer period of time

uav = ½ ε0E0² = B0²/2µ0

6. intensity of electromagnetic wave is per unit area per unit time I = U/AΔt = uavc.

In terms of maximum electric field (substituting the value of uavc)

I = ½ ε0E0²c

7. The electromagnetic wave also carries linear momentum with it. The linear momentum carried by the portion of wave having energy U is given by

p = U/c

where U = energy contained in the portion of the wave.

Thursday, February 7, 2008

IIT JEE Physics Formula Revision 41. Electric Current through Gases

1. Paschen's law
Sparking potential of a gas in a discharge tube is a function of the product of the pressure of the gas and the separation between the electrodes.

V = f(pd)

2. Experiment to Determine e/m by Thomson

Cathode rays are subjected to electric and magnetic fields.

If both the electric field and magnetic field are switched on and the values are so chosen that

v = E/B

The magnetic field evB will exactly cancel the electric force eE and the beam will pass undeflected. If the potential difference between the anode and the cathode is V, the speed of the electrons coming out of A is given by

½ mv² = eV

as v = E/B

½ m (e/B) ² = eV
therefore e/m = E²/2B²V

3. If n thermions are ejected per unit time by a metal surface, the thermionic current i = ne. This current is given by the Richardson-Dushman equation.

i = ne = AST²e- φ/kT

where
S = surface area
T = the absolute temperature of the surface.
φ = work function of the metal
K = Boltzmann constant
A = constant which depends only the nature of the metal

IIT JEE Physics Formula Revision 42. Photoelectric Effect and Waveparticle Duality

1. Relation between properties of photon and properties of light waves.

E and p are energy and linear momentum of a photon of light.
ν and λ are the frequency and wavelength of the same light when it is considered (behaves) as a wave.

Then E = hν = hc/λ
p = h/λ = E/c ... (42.1)

wherein h is a universal constant known as the Planck constant and has a value 6.626*10^-34 J-s and is also equal to 4.136*10^-15 eV-s.

C = velocity of light vacuum = 299,792,458 m/s ≈ 3.0*10^8 m/s

2. The maximum kinetic energy of the electron that comes out due to energy E supplied is:

Kmax = E – φ

Some energy from the E – φ is dissipated as the electron may have some collision before escaping from the material.

If monochromatic light of wave length is incident on the metal surface, photons of energy hc/ λ fall on the surface. The maximum kinetic energy of an electrons that comes out due to these photons is:

Kmax = hc/λ - φ = h υ - φ

The above equation is called Einstein's photoelectric equation.

3. Writing work function φ as hυ0 (h multiplied by frequency)

υ λ = c
υ0 = c/ λ0
λ0 = threshold wavelength

Kmax = h(υ - υ0)

4. Relation between Maximum kinetic energy of photoelectrons and stopping potential:

As a photoelectron travels from the cathode to the anode, the potential energy increases by eV0. This is equal to the decrease in the kinetic energy of the photoelectron.

The maximum kinetic energy a photoelectron will have is hc/λ - φ
Hence eV0. = hc/λ – φ

V0 = hc/e(1/ λ) – φ/e

5. A relation for wavelength of electron was proposed by Louis Victor de Broglie.

The proposed expression for wavelength is

λ = h/p

Where p is the momentum of the electron and
h is the Planck constant.

IIT JEE Physics Formula Revision 43. Bohr's Model and Physics of Atom

1. Equation for wavelengths of radiation emitted by the hydrogen atom.
1/λ = R [1/n² - 1/m²]

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

2. Velocity of an electron when it is in a stationary orbit represented by n which is an integer

v = Ze²/2 ε0hn

3. Radius of a stationary orbit based on n which is an integer

r = ε0h²n²/πmZe²

4. Kinetic energy when electron is in nth orbit is

K = ½ mv² = mZ²e4/8 ε0²h²n²

5. 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.

6. The total energy of the atom is

E = K+V = - mZ²e4/8 ε0²h²n²

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

rn = radius of ‘n’ th orbit = n²a0/Z

8. Energy of hydrogen atom when the single electron is in the nth orbit.
En = E1/n² = -13.6/n² eV

-13.6 eV is the energy when the electron is in the n = 1 orbit.

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

9. If an electron jumps from mth orbit to nth orbit (m>n) of a hydrogen like ion, 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]

10. 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.

11. 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

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

ψ(r,) = ψ100 = √(Z³/ π a0²) *(e-r/ a0)

where
ψ100 denotes that n =1, l = 0 and ml = 0

a0 = Bohr radius

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IIT JEE Physics Formula Revision 44. X-Rays

1. K = eV
where K = Kinetic energy of an electron when it hits the target
V = potential difference applied between the target and the filament

2. λ = hc/E

Where
λ = wave length of the X-ray
E = kinteic energy of the electron due to which the X-ray is emitted

3. λmin = hc/eV

λmin = cutoff wavelength below which no X-rays are emitted
V = potential difference applied between the target and the filament

4. Moseley's law

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

√(ν) = a(Z-b)

where
ν = frequency of X-rays
Z = position number of element.
a and b are constants

5. 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

IIT JEE Physics Formula Revision 46. Nucleus

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

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

2. B = [Zm{11H} +Nmn
– m{ Z Z+n}]c²

Where
B = binding energy of the nucleus

m{11H} is the mass of a hydrogen atom

m{ Z Z+n} is the mass of an atom with Z protons and N neutrons

3. Binding energy per nucleon = B/A = a1 - a2/(A1/3) - a3Z(Z-1)/ (A4/3)

4. Mass excess =
(mass of atom – A’)c²

5. Mass excess = 931(m-A)MeV

6. Alpha decay process is represented by

ZAX --> Z-2A-4Y + 24He

7. q value of the process

Q = [m{ZAX} – m{Z-2A-4Y} – m{24He}]c²

8. Beta decay process

N --> p + e + antineutrino

9. Beta decay proces
ZAX --> Z+1AY +e + antineutrino

e is also shown as beta minus.

10. kinetic energy Q available ot the product particles is

Q = [m(ZAX – m{Z+1AY}]c²

11. Proton conversion process – Beta plus decay

P --> n + e+ + v (neutrino)

12.
ZAX --> Z-1AY + e+ + v (neutrino)

13.Q value of the decay

Q = {m(ZAX) – m(Z-1AY) – 2me]c²

IIT JEE Study of Physics

As I have brought my Chemistry study to some stage, I started looking at physics once again. I am trying to write down formulas of each chapter in a notebook first. That way I read the chapter fully. It may take two days for each chapter. I may start the posting the formulas for revision purpose in this blog. Along with it I plan to go through problems/questions of past JEEs and update the study guides with them.