NCERT Solutions Class 12 Physics - All 14 Chapters

Chapter 1: Electric Charges and Fields

Key Formulas:
• Coulomb's Law: F = kq₁q₂/r², k = 9 × 10⁹ Nm²/C²
• Electric Field: E = F/q = kQ/r²
• Gauss's Law: ∮ E · dA = q_enc/ε₀
• Force on charge in E-field: F = qE

Q1. Two point charges q₁ = 2 × 10⁻⁶ C and q₂ = −2 × 10⁻⁶ C are placed 15 cm apart. Find the electric force between them. Is it attractive or repulsive?

Solution

F = k|q₁||q₂|/r²
= (9 × 10⁹)(2 × 10⁻⁶)(2 × 10⁻⁶)/(0.15)²
= (9 × 10⁹)(4 × 10⁻¹²)/(2.25 × 10⁻²)
= 36 × 10⁻³/(2.25 × 10⁻²)
= 1.6 N

Since charges are opposite, force is attractive.

Q2. An electric dipole of dipole moment p is placed in a uniform electric field E. Write the expression for the torque experienced by the dipole. Identify two orientations for which torque is (i) maximum (ii) zero.

Solution

Torque: τ = p × E = pE sinθ

(i) Maximum when sinθ = 1, i.e., θ = 90° (dipole ⊥ to E)
(ii) Zero when sinθ = 0, i.e., θ = 0° or 180° (dipole || to E)

Q3. Using Gauss's law, find the electric field due to an infinitely long straight uniformly charged wire with linear charge density λ.

Solution

Consider a Gaussian cylinder of radius r and length l.
By symmetry, E is radial and constant on the curved surface.

∮ E · dA = E × 2πrl = q_enc/ε₀ = λl/ε₀

E = λ/(2πε₀r)

Q4. Calculate the electric field at a point on the equatorial line of an electric dipole of length 2a with dipole moment p, at a distance r from the centre (r >> a).

Solution

For equatorial line (r >> a):
E = −kp/r³ or E = kp/(4πε₀r³)

Direction is opposite to the dipole moment (antiparallel to p).

Q5. A point charge +q is placed at the centre of a cube of side a. What is the electric flux through one face of the cube?

Solution

Total flux through cube = q/ε₀ (by Gauss's law)
By symmetry, flux through each face is equal.

Flux through one face = q/(6ε₀)

Q6. Two point charges −q and +q are placed at distance d apart. Find the electric field at a point on the perpendicular bisector at distance d/2 from the midpoint.

Solution

Distance from each charge to the point:
r = √((d/2)² + (d/2)²) = d/√2

E₁ = E₂ = kq/r² = 2kq/d²

Horizontal components cancel; vertical components add:
E = 2E₁ cosθ = 2(2kq/d²)(d/2)/(d/√2)
= 2(2kq/d²)(1/√2)
E = 2√2 kq/d² (along the perpendicular bisector)

Chapter 2: Electrostatic Potential and Capacitance

Key Formulas:
• V = kQ/r
• U = kq₁q₂/r
• C = Q/V, C = ε₀A/d (parallel plate)
• C_series = 1/C₁ + 1/C₂, C_parallel = C₁ + C₂
• Energy: U = Q²/2C = CV²/2 = QV/2

Q1. A charge of 2 μC is placed at the origin. Find the work done in bringing a charge of 4 μC from infinity to point (2m, 0, 0).

Solution

W = q₂V₁ = q₂(kq₁/r)
= (4 × 10⁻⁶)(9 × 10⁹)(2 × 10⁻⁶)/2
= (4 × 10⁻⁶)(9 × 10³)
= 36 × 10⁻³ J = 36 mJ

Q2. A parallel plate capacitor has capacitance 8 μF. What will be its capacitance if the separation between the plates is halved and a dielectric of constant 4 is introduced?

Solution

C₁ = ε₀A/d = 8 μF

C₂ = Kε₀A/(d/2) = 2K × (ε₀A/d) = 2K × C₁
= 2 × 4 × 8 = 64 μF

Q3. Two capacitors of 3 μF and 6 μF are connected in series. Find the equivalent capacitance and the charge on each if connected to a 9V battery.

Solution

1/C_eq = 1/3 + 1/6 = 3/6 = 1/2
C_eq = 2 μF

Q = C_eqV = 2 × 9 = 18 μC
In series, charge on each capacitor = 18 μC

Q4. Find the potential at the centre of a uniformly charged spherical shell of radius R and total charge Q.

Solution

Potential is constant everywhere inside a charged shell.
V_centre = V_surface = kQ/R = Q/(4πε₀R)

Q5. A 10 μF capacitor is charged to 50V. The energy stored in the capacitor is:

Solution

U = ½CV² = ½(10 × 10⁻⁶)(50)²
= ½(10⁻⁵)(2500)
= 1.25 × 10⁻² J = 12.5 mJ

Q6. Four identical capacitors are connected (i) all in series, (ii) all in parallel. Find the ratio of equivalent capacitances.

Solution

Let each capacitor = C

(i) Series: 1/C_s = 4/C ⇒ C_s = C/4
(ii) Parallel: C_p = 4C

Ratio C_s/C_p = (C/4)/(4C) = 1:16

Chapter 3: Current Electricity

Key Formulas:
• Ohm's Law: V = IR
• Resistance: R = ρL/A
• Power: P = VI = I²R = V²/R
• Kirchhoff: KVL (ΣV = 0 at junction), KCL (ΣI = 0 at junction)
• Series: R = R₁ + R₂, Parallel: 1/R = 1/R₁ + 1/R₂

Q1. A wire of resistance 12 ohms is bent in the form of a circle. Find the effective resistance between two points on any part of the circumference.

Solution

The two arcs are in parallel.
Each arc has resistance = 12/2 = 6 Ω (assuming equal halves)

1/R_eq = 1/6 + 1/6 = 1/3
R_eq = 3 Ω

Q2. A battery of emf 10V and internal resistance 3Ω is connected to a resistor. If the current in the circuit is 0.5A, find (i) the resistance of the resistor (ii) terminal voltage of the battery.

Solution

(i) I = ε/(R + r)
0.5 = 10/(R + 3)
R + 3 = 20 ⇒ R = 17 Ω

(ii) V = ε − Ir = 10 − 0.5(3) = 8.5V

Q3. A toaster of 800W and a water heater of 2000W are designed to work on 230V supply. (a) Which has greater resistance? (b) What is the resistance of each?

Solution

R = V²/P

(a) Toaster: R = 230²/800 = 52900/800 = 66.125 Ω
Heater: R = 230²/2000 = 52900/2000 = 26.45 Ω

(b) The toaster has greater resistance (lower power ⇒ higher resistance).

Q4. In the given circuit, find the current through each resistor using Kirchhoff's laws.

Solution

Consider a circuit with two loops containing batteries ε₁ = 10V, ε₂ = 5V and resistors R₁ = 2Ω, R₂ = 3Ω, R₃ = 5Ω.

By Kirchhoff's voltage law:
Loop 1: 10 − 2I₁ − 5(I₁ − I₂) = 0
Loop 2: −5 − 3I₂ − 5(I₂ − I₁) = 0

Simplify:
7I₁ − 5I₂ = 10 ... (i)
−5I₁ + 8I₂ = −5 ... (ii)

Solving: I₁ = 55/31 ≈ 1.77A, I₂ = 25/31 ≈ 0.81A
Current through R₃ = I₁ − I₂ = 30/31 ≈ 0.97A

Q5. Find the equivalent resistance between A and B of the network shown (3 resistors each of 6Ω in delta configuration).

Solution

For a delta of three equal resistors R connected between A, B, C:
Equivalent resistance between any two terminals = 2R/3

R_eq = 2(6)/3 = 4 Ω

Q6. The resistivity of a material is 1 × 10⁻⁷ Ωm and its dimensions are: length = 10 cm, cross-sectional area = 2 mm². Find its resistance.

Solution

R = ρL/A
L = 0.1 m, A = 2 × 10⁻⁶ m²

R = (10⁻⁷)(0.1)/(2 × 10⁻⁶)
= 10⁻⁸/(2 × 10⁻⁶)
= 0.005 Ω = 5 mΩ

Chapter 4: Moving Charges and Magnetism

Key Formulas:
• Biot-Savart Law: dB = (μ₀/4π)(Idl × r/r³)
• Ampere's Law: ∮ B · dl = μ₀I_enc
• Force on charge: F = qvB sinθ
• Cyclotron radius: r = mv/(qB)
• Cyclotron frequency: f = qB/(2πm)

Q1. An electron moves with a speed of 3 × 10⁷ m/s at an angle of 30° with the direction of a magnetic field of 0.1 T. Find the force on the electron.

Solution

F = qvB sinθ
= (1.6 × 10⁻¹⁹)(3 × 10⁷)(0.1) sin 30°
= (1.6 × 10⁻¹⁹)(3 × 10⁷)(0.1)(0.5)
= 2.4 × 10⁻¹³ N

Q2. Find the magnetic field at the centre of a circular coil of radius 10 cm carrying current 2A, with 50 turns.

Solution

B = μ₀NI/(2r)
= (4π × 10⁻⁷)(50)(2)/(2 × 0.1)
= 4π × 10⁻⁷ × 500
= 2π × 10⁻⁴ T ≈ 6.28 × 10⁻⁴ T

Q3. A proton moves in a circle of radius 10 cm with a speed of 10⁶ m/s. Find the magnetic field. (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C)

Solution

r = mv/(qB)
B = mv/(qr) = (1.67 × 10⁻²⁷)(10⁶)/((1.6 × 10⁻¹⁹)(0.1))
= 1.67 × 10⁻²¹/(1.6 × 10⁻²⁰)
= 0.104 T

Q4. A circular loop of radius R carries a current I. Show that at a point on its axis at distance x from the centre, B = μ₀IR²/2(R² + x²)^3/2.

Solution

Using Biot-Savart law:
dB = (μ₀I dl sin 90°)/(4π)(R² + x²)

Only axial component survives:
dB_axial = dB · R/√(R² + x²)

B = ∮ dB_axial = (μ₀I)/(4π(R²+x²)) · R/√(R²+x²) · 2πR
= μ₀IR²/2(R² + x²)^3/2

Q5. A charged particle of mass m and charge q is released from rest in a uniform electric field E and uniform magnetic field B (perpendicular to each other). Describe the path of the particle.

Solution

When E and B are perpendicular, the particle follows a cycloid path. Starting from rest, the electric field accelerates the particle, while the magnetic field deflects it. The path is a combination of circular motion superimposed on a linear drift, producing a cycloid.

Q6. An electron enters a region of uniform magnetic field B = 0.1 T with velocity v = 2 × 10⁷ m/s perpendicular to the field. Find the time period of its circular motion.

Solution

T = 2πm/(qB)
= 2π(9.1 × 10⁻³¹)/((1.6 × 10⁻¹⁹)(0.1))
= 2π(9.1 × 10⁻³¹)/(1.6 × 10⁻²⁰)
= 3.57 × 10⁻¹⁰ s ≈ 0.36 ns

Chapter 5: Magnetism and Matter

Q1. Define magnetic susceptibility and permeability. How are they related for a paramagnetic material?

Solution

Magnetic susceptibility (χ): Ratio of intensity of magnetization (M) to magnetic field strength (H): χ = M/H

Magnetic permeability (μ): Ratio of B to H: μ = B/H

Relation: μ = μ₀(1 + χ)
For paramagnetic materials, χ is small and positive.

Q2. Classify the following as dia-, para-, or ferromagnetic materials: Aluminium, Sodium, Copper, Iron, Nickel, Cobalt, Water.

Solution

Diamagnetic: Copper, Water
Paramagnetic: Aluminium, Sodium
Ferromagnetic: Iron, Nickel, Cobalt

Q3. What is the earth's magnetic field? What are its components?

Solution

The earth behaves as a magnetic dipole with its south pole near the geographic north. The magnetic field at any point on the earth's surface has three components:

1. Horizontal component (H): Component along the geographic meridian.
2. Declination (θ): Angle between geographic and magnetic meridian.
3. Dip (δ): Angle that the total field makes with the horizontal.

Q4. A bar magnet of magnetic moment 4 A·m² is placed in a uniform magnetic field of 0.2 T at an angle of 30°. Find the torque on it.

Solution

τ = MB sinθ = (4)(0.2) sin 30°
= 0.8 × 0.5 = 0.4 N·m

Q5. Derive the expression for the energy stored in a solenoid.

Solution

Energy = Work done in building up current from 0 to I
U = ½LI²

where L = μ₀n²Al is the self-inductance
n = number of turns per unit length
A = cross-sectional area, l = length

U = ½(μ₀n²Al)I² = ½μ₀n²AlI²

Chapter 6: Electromagnetic Induction

Key Formulas:
• Faraday's Law: ε = −dΦ/dt
• Induced EMF in a rod: ε = Blv
• Self-inductance: ε = −L(dI/dt)
• Mutual inductance: ε₁ = −M(dI₂/dt)
• Energy in inductor: U = ½LI²

Q1. A circular coil of area 100 cm² and 50 turns is placed perpendicular to a magnetic field of 0.2 T. The field changes at 0.01 T/s. Find the induced emf.

Solution

ε = NA(dB/dt)
= 50 × 100 × 10⁻⁴ × 0.01
= 50 × 10⁻² × 0.01
= 5 × 10⁻³ V = 5 mV

Q2. A conducting rod of length 20 cm rotates about one end with angular velocity 10 rad/s in a uniform magnetic field of 0.3 T perpendicular to the plane of rotation. Find the induced emf.

Solution

ε = ½Bωl²
= ½(0.3)(10)(0.2)²
= ½(0.3)(10)(0.04)
= 0.06 V = 60 mV

Q3. State and explain Lenz's law. Give one application.

Solution

Lenz's Law: The direction of induced current is such that it opposes the change in magnetic flux that produced it.

Application: Electromagnetic braking in trains. When a conducting disc rotates in a magnetic field, eddy currents are induced. By Lenz's law, these currents oppose the motion, causing the disc to slow down. This principle is used in braking systems.

Q4. An inductor of 2H carries a steady current of 5A. How can a 200V emf be induced in it?

Solution

ε = L(dI/dt)
200 = 2 × (dI/dt)
dI/dt = 100 A/s

The current must change at 100 A/s to induce 200V.

Q5. Two coils have self-inductances 10 mH and 20 mH. If the mutual inductance between them is 5 mH, find the coupling coefficient.

Solution

k = M/√(L₁L₂)
= 5/√(10 × 20)
= 5/√200
= 5/(10√2)
= 1/(2√2) = 0.354

Q6. A coil of 200 turns and area 10 cm² is suddenly removed from a magnetic field of 0.1 T in 0.1s. Find the average induced emf.

Solution

ΔΦ = NBA = 200 × 0.1 × 10 × 10⁻⁴ = 0.02 Wb

ε = ΔΦ/Δt = 0.02/0.1 = 0.2 V

Chapter 7: Alternating Current

Key Formulas:
• V = V₀ sin ωt, I = I₀ sin(ωt − φ)
• Impedance: Z = √(R² + (X_L − X_C)²)
• X_L = ωL, X_C = 1/(ωC)
• Resonance: X_L = X_C, f = 1/(2π√LC)
• Power: P = VI cosφ

Q1. An AC source has ω = 300 rad/s. Find the value of L and C for series resonance at 500 Hz.

Solution

At resonance: ω₀ = 1/√(LC)
300 = 1/√(LC)

If we assume f = 500 Hz, then ω = 2π(500) = 3141.6 rad/s
For ω = 300: LC = 1/300² = 1/90000

For a given L, C = 1/(L × 90000)
If L = 0.1 H: C = 1/(9000) ≈ 111 μF

Q2. In an LCR circuit with R = 20Ω, L = 0.5H, C = 100μF, connected to 220V, 50Hz supply, find (i) impedance (ii) current (iii) phase angle.

Solution

ω = 2π(50) = 314 rad/s
X_L = ωL = 314 × 0.5 = 157 Ω
X_C = 1/(ωC) = 1/(314 × 10⁻⁴) = 31.85 Ω

(i) Z = √(20² + (157 − 31.85)²) = √(400 + 15663) = √16063 = 126.7 Ω
(ii) I = V/Z = 220/126.7 = 1.74 A
(iii) tanφ = (X_L − X_C)/R = 125.15/20 = 6.26
φ = 80.9° (inductive)

Q3. A transformer steps down 220V to 22V. If the primary coil has 1000 turns, find the number of turns in the secondary.

Solution

V₁/V₂ = N₁/N₂
220/22 = 1000/N₂
N₂ = 1000/10 = 100 turns

Q4. Find the power dissipated in a pure inductor of inductance 0.5 H connected to an AC source of 200V, 50Hz.

Solution

In a pure inductor, phase angle φ = 90°
Power P = VI cos 90° = 0 W

The average power dissipated in a pure inductor is zero because it only stores and releases energy.

Q5. Why is the power factor important in AC circuits? What is its ideal value?

Solution

Power factor = cosφ = R/Z

Power factor determines the ratio of actual power (dissipated) to apparent power.

Ideal value = 1 (purely resistive circuit)

When cosφ = 1, maximum power is delivered to the load. Low power factor means more current is needed for the same power, causing higher losses in transmission lines.

Chapter 8: Electromagnetic Waves

Q1. Arrange the following in order of increasing wavelength: X-rays, infrared rays, microwaves, UV rays, visible light, radio waves, gamma rays.

Solution

Increasing wavelength:
Gamma rays < X-rays < UV rays < Visible light < Infrared rays < Microwaves < Radio waves

Q2. Name the electromagnetic waves used (i) in radar (ii) for cooking (iii) in radiotherapy (iv) for sterilizing surgical instruments.

Solution

(i) Radar: Microwaves
(ii) Cooking: Microwaves (2.45 GHz)
(iii) Radiotherapy: Gamma rays
(iv) Sterilizing: UV rays

Q3. Prove that electromagnetic waves are transverse in nature.

Solution

Maxwell's equations show that E and B fields are always perpendicular to the direction of propagation. Since the fields oscillate perpendicular to the direction of travel (and to each other), EM waves are transverse in nature.

Q4. The magnetic field in a plane electromagnetic wave is given by B_y = 2 × 10⁻⁷ sin(0.5 × 10³x + 1.5 × 10¹¹t) T. Find (i) wavelength (ii) frequency.

Solution

Comparing with B = B₀ sin(kx + ωt):
k = 0.5 × 10³ rad/m, ω = 1.5 × 10¹¹ rad/s

(i) λ = 2π/k = 2π/(500) = 1.26 × 10⁻² m = 1.26 cm
(ii) f = ω/(2π) = (1.5 × 10¹¹)/(2π) = 2.39 × 10¹⁰ Hz = 23.9 GHz

Q5. What is the frequency of electromagnetic waves produced by an oscillating charge of frequency 10 MHz?

Solution

The frequency of the EM waves is equal to the frequency of the oscillating charge.
f = 10 MHz

Chapter 9: Ray Optics and Optical Instruments

Key Formulas:
• Mirror: 1/v + 1/u = 1/f
• Lens: 1/v − 1/u = 1/f (thin lens formula)
• Snell's Law: n₁ sinθ₁ = n₂ sinθ₂
• Prism: n = sin[(A + δ_m)/2] / sin(A/2)

Q1. An object 5 cm tall is placed 30 cm from a convex mirror of focal length 15 cm. Find the position and size of the image.

Solution

1/v + 1/u = 1/f
1/v + 1/(−30) = 1/15 (convex mirror, f = +15)
1/v = 1/15 + 1/30 = 3/30 = 1/10
v = 10 cm (behind the mirror)

m = −v/u = −10/(−30) = 1/3
Image height = 5 × 1/3 = 5/3 cm ≈ 1.67 cm (virtual, erect)

Q2. A candle is placed 20 cm from a convex lens of focal length 10 cm. At what distance from the lens should a screen be placed to get a sharp image? What is the nature of the image?

Solution

1/v − 1/u = 1/f
1/v − 1/(−20) = 1/10
1/v = 1/10 − 1/20 = 1/20
v = 20 cm (on the other side)

m = v/u = 20/(−20) = −1
Image is real, inverted, same size

Q3. A ray of light passes from glass (n = 1.5) to air. Find the critical angle.

Solution

sinθ_c = 1/n = 1/1.5 = 2/3
θ_c = sin⁻¹(2/3) = 41.8°

Q4. Two thin lenses of focal lengths 20 cm and −30 cm are placed in contact. Find the equivalent focal length.

Solution

1/F = 1/f₁ + 1/f₂ = 1/20 + 1/(−30)
= 1/20 − 1/30 = (3−2)/60 = 1/60
F = 60 cm (converging combination)

Q5. A prism has angle A = 60° and minimum deviation δ_m = 37°. Find the refractive index of the material of the prism.

Solution

n = sin[(A + δ_m)/2] / sin(A/2)
= sin[(60 + 37)/2] / sin(30)
= sin(48.5°) / 0.5
= 0.749/0.5
= 1.5

Q6. Draw a ray diagram for a simple microscope when the object is at the near point. Derive the expression for magnifying power.

Solution

When object is at near point (D = 25 cm):
1/v − 1/u = 1/f
1/v = 1/f + 1/(−D) = 1/f − 1/D

Angular magnification: m = D/|u|

From lens formula: 1/|u| = 1/f − 1/D = (D−f)/(fD)
|u| = fD/(D−f)

m = D/|u| = (D−f)/f = D/f − 1 = 1 + D/f
(using D = 25 cm: m = 1 + 25/f)

Chapter 10: Wave Optics

Key Formulas:
• Double slit: d sinθ = nλ (maxima)
• d sinθ = (n + ½)λ (minima)
• Fringe width: β = λD/d
• Single slit minima: a sinθ = nλ

Q1. In a Young's double slit experiment, the slit separation is 0.5 mm and the screen is 1 m away. If the wavelength of light used is 600 nm, find the fringe width.

Solution

β = λD/d
= (600 × 10⁻⁹)(1)/(0.5 × 10⁻³)
= 6 × 10⁻⁴/5 × 10⁻⁴
= 1.2 mm

Q2. In Young's double slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in water of refractive index 4/3, find the new fringe width.

Solution

β' = β/n = 0.4/(4/3) = 0.4 × 3/4
= 0.3 mm

Q3. What is the condition for destructive interference of light waves?

Solution

Path difference Δx = (2n + 1)λ/2
or Phase difference φ = (2n + 1)π

Q4. A single slit of width 0.2 mm is illuminated by light of wavelength 500 nm. Find the angular position of the first minimum.

Solution

a sinθ = nλ
For n = 1: sinθ = λ/a
= (500 × 10⁻⁹)/(0.2 × 10⁻³)
= 2.5 × 10⁻³
θ ≈ 2.5 × 10⁻³ rad = 0.14°

Q5. State Huygens' principle. How does it explain the rectilinear propagation of light?

Solution

Huygens' Principle: Every point on a wavefront acts as a source of secondary wavelets. The new wavefront is the envelope of these secondary wavelets.

For rectilinear propagation: Secondary wavelets from a point source spread in all directions, but the tangent to these wavelets at any instant forms a spherical wavefront, explaining propagation in straight lines in a homogeneous medium.

Q6. What is a diffraction grating? Show that the condition for maxima is d sinθ = nλ.

Solution

A diffraction grating has a large number of equally spaced parallel slits.

For constructive interference, path difference between light from adjacent slits must be an integral multiple of λ:

d sinθ = nλ (n = 0, 1, 2, ...)

where d is the grating element (slit separation).
Higher order maxima (n > 0) are brighter than in Young's experiment because more slits contribute.

Chapter 11: Dual Nature of Radiation and Matter

Key Formulas:
• Photoelectric: E = hν = hc/λ
• K_max = hν − W₀
• de Broglie: λ = h/(mv) = h/p
• Stopping potential: eV₀ = K_max

Q1. The work function of a metal is 2 eV. Find the threshold wavelength and the maximum kinetic energy of photoelectrons when light of wavelength 4000 Å falls on it.

Solution

W₀ = hc/λ₀
λ₀ = hc/W₀ = (6.63 × 10⁻³⁴)(3 × 10⁸)/(2 × 1.6 × 10⁻¹⁹)
= 19.89 × 10⁻²⁶/(3.2 × 10⁻¹⁹)
= 6.22 × 10⁻⁷ m = 6220 Å

Energy of incident photon: E = hc/λ = (6.63 × 10⁻³⁴)(3 × 10⁸)/(4 × 10⁻⁷) = 4.97 × 10⁻¹⁹ J = 3.11 eV

K_max = E − W₀ = 3.11 − 2 = 1.11 eV

Q2. Find the de Broglie wavelength of an electron accelerated through a potential difference of 100V.

Solution

λ = 12.27/√V Å (for electrons)
= 12.27/√100 Å
= 12.27/10 Å
= 1.227 Å

Q3. Explain why the photoelectric effect cannot be explained by wave theory of light.

Solution

Wave theory predicts that:

1. Light of any frequency should cause photoelectric effect - but experiments show a threshold frequency exists.

2. Increasing intensity should increase kinetic energy - but KE depends on frequency, not intensity.

3. There should be a time lag - but emission is instantaneous.

These observations contradict wave theory and are explained by Einstein's photon theory where light consists of particles (photons) of energy E = hν.

Q4. The stopping potential for photoelectrons from a metal surface is 3V when light of wavelength 2500 Å is used. Find (i) the work function (ii) threshold wavelength.

Solution

(i) K_max = eV₀ = 3 eV
E = hc/λ = (6.63 × 10⁻³⁴)(3 × 10⁸)/(2.5 × 10⁻⁷) = 4.97 eV

W₀ = E − K_max = 4.97 − 3 = 1.97 eV

(ii) λ₀ = hc/W₀ = 12400/1.97 = 6294 Å

Q5. An alpha particle and a proton have the same kinetic energy. Find the ratio of their de Broglie wavelengths.

Solution

λ = h/p = h/√(2mK)

λ_α/λ_p = √(m_p/m_α) = √(1/4) = 1/2

Chapter 12: Atoms

Bohr's Model:
• E_n = −13.6/n² eV (for hydrogen)
• r_n = 0.529 n² Å
• ν = 13.6(1/n₁² − 1/n₂²) Hz
• Angular momentum: L = nh/(2π)

Q1. Find the radius of the second orbit and the energy of an electron in the second orbit of hydrogen atom.

Solution

r₂ = r₀ × n² = 0.529 × 4 = 2.116 Å

E₂ = −13.6/n² = −13.6/4 = −3.4 eV

Q2. Calculate the wavelength of the spectral line corresponding to the transition from n = 4 to n = 2 in hydrogen atom.

Solution

1/λ = R(1/2² − 1/4²) = R(1/4 − 1/16) = R(3/16)

λ = 16/(3R) = 16/(3 × 1.097 × 10⁷)
= 16/(3.291 × 10⁷)
= 4.86 × 10⁻⁷ m = 486 nm (H-beta line, visible)

Q3. Find the ionization energy of hydrogen atom in the ground state.

Solution

Ionization energy = −E₁ = −(−13.6 eV)
= 13.6 eV

Q4. What are the limitations of Bohr's model of the atom?

Solution

1. It only works for hydrogen-like atoms (single electron systems).

2. It cannot explain the fine structure of spectral lines.

3. It violates Heisenberg's uncertainty principle.

4. It doesn't explain the intensity of spectral lines.

5. It treats electrons as particles in definite orbits, which contradicts the wave nature of electrons.

Q5. The total energy of an electron in the third orbit of hydrogen atom is −1.51 eV. Find its kinetic energy and potential energy.

Solution

For hydrogen atom:
K.E. = −E = 1.51 eV
P.E. = 2E = 2(−1.51) = −3.02 eV
Total E = K.E. + P.E. = 1.51 − 3.02 = −1.51 eV ✓

Chapter 13: Nuclei

Key Formulas:
• Mass defect: Δm = [Zm_p + (A−Z)m_n] − m_nucleus
• Binding energy: B.E. = Δm × 931.5 MeV
• Radioactivity: N = N₀e^−λt
• Half-life: t_1/2 = 0.693/λ

Q1. The half-life of a radioactive substance is 30 days. What fraction of the original substance remains after 90 days?

Solution

N = N₀(1/2)^t/t_1/2
= N₀(1/2)^90/30
= N₀(1/2)³
= N₀/8

Fraction remaining = 1/8 = 0.125

Q2. Find the binding energy per nucleon of ⁵⁶Fe if the mass of ⁵⁶Fe atom is 55.9349 u. (Mass of proton = 1.00783 u, mass of neutron = 1.00867 u)

Solution

Δm = [26(1.00783) + 30(1.00867)] − 55.9349
= [26.20358 + 30.26010] − 55.9349
= 56.46368 − 55.9349
= 0.52878 u

B.E. = 0.52878 × 931.5 = 492.6 MeV
B.E./nucleon = 492.6/56 = 8.80 MeV/nucleon

Q3. Explain the difference between nuclear fission and fusion. Give one example of each.

Solution

Fission: Heavy nucleus splits into lighter nuclei.
Example: ²³⁵U + n → ²³⁶U* → ¹⁴¹Ba + ⁹²Kr + 3n

Fusion: Light nuclei combine to form a heavier nucleus.
Example: 2²H → ⁴He + energy

Both release energy because the products have higher binding energy per nucleon.

Q4. A radioactive isotope has a half-life of 5 years. How much of a 100 g sample will remain after 20 years?

Solution

N = N₀(1/2)^t/t_1/2
= 100(1/2)^20/5
= 100(1/2)&sup4;
= 100/16 = 6.25 g

Q5. The activity of a radioactive sample drops to 1/16th of its original value in 28 days. Find its half-life.

Solution

N/N₀ = (1/2)ⁿ where n = number of half-lives
1/16 = (1/2)ⁿ
(1/2)&sup4; = (1/2)ⁿ
n = 4 half-lives

t_1/2 = 28/4 = 7 days

Chapter 14: Semiconductor Electronics

Q1. Distinguish between intrinsic and extrinsic semiconductors.

Solution

Intrinsic: Pure semiconductor with equal number of electrons and holes.
Extrinsic: Doped semiconductor with majority carriers (electrons or holes).

Extrinsic types:
n-type: Doped with pentavalent (P, As), majority carriers = electrons
p-type: Doped with trivalent (B, Ga), majority carriers = holes

Q2. Draw the V-I characteristic of a p-n junction diode in forward and reverse bias.

Solution

In forward bias, current increases exponentially after knee voltage (~0.7V for Si, ~0.3V for Ge). In reverse bias, a small saturation current flows until breakdown voltage is reached.

Q3. What is the logic gate equivalent of the Boolean expression Y = A · B + C?

Solution

A · B: AND gate with inputs A, B
(A · B) + C: OR gate with inputs (A · B) and C

Circuit: A, B → AND gate → output with C → OR gate → Y

Q4. Explain the working of an n-p-n transistor as an amplifier in common emitter configuration.

Solution

In CE configuration, the emitter is common, input is applied to base-emitter junction (forward biased), and output is taken from collector-emitter junction (reverse biased).

When a small AC signal is applied at the base:
• Base current changes by ΔI_B
• Collector current changes by ΔI_C = β ΔI_B
• Voltage gain: A_v = βR_C/r_i

The transistor amplifies the weak input signal at the output.

Q5. Draw the truth table for NAND gate and NOR gate. Show that NAND gate is a universal gate.

Solution

NAND Gate: Y = (A · B)'
A B | Y
0 0 | 1
0 1 | 1
1 0 | 1
1 1 | 0

NOR Gate: Y = (A + B)'
A B | Y
0 0 | 1
0 1 | 0
1 0 | 0
1 1 | 0

NAND is universal: NOT = NAND(A,A), AND = NAND(NOT(A),NOT(B)), OR = NAND(NAND(A,A),NAND(B,B))