Physics भौतिकशास्त्र

Quick summary

A bipolar junction transistor (BJT) has three regions — emitter, base and collector — forming two p-n junctions. A small base current controls a much larger collector current, so a transistor can amplify signals or act as a switch. This topic covers transistor action, configurations, characteristics, α and β, biasing, load line and the CE amplifier.

Key points

• Types: n-p-n and p-n-p. The emitter is heavily doped, the base is thin and lightly doped, and the collector is moderately doped and largest.
• For normal (active) operation: emitter–base junction forward biased, collector–base junction reverse biased.
• Transistor action (n-p-n): electrons from the emitter enter the thin base; very few recombine; most are swept into the collector. So I_C is almost equal to I_E.
• Current relation: I_E = I_B + I_C.
• Three configurations: common base (CB), common emitter (CE) and common collector (CC).
• CE is the most used amplifier: high current, voltage and power gain; output is 180° out of phase with the input.
• CB: low input and high output resistance; current gain slightly less than 1. CC (emitter follower): voltage gain about 1; high input and low output resistance; used for impedance matching.
• Regions of operation: active (amplifier), saturation (switch ON), cut-off (switch OFF).
• Biasing fixes a stable operating point (Q-point). Fixed bias is simple but unstable with temperature; voltage-divider bias gives good stability.
• DC load line: straight line on output characteristics joining (V_CC, 0) and (0, V_CC/R_C). The Q-point should lie near its middle for undistorted amplification.
• h-parameters model a transistor for small signals: h_ie (input resistance), h_fe (current gain), h_re, h_oe.

Important terms

• α (current gain in CB) = I_C/I_E, slightly less than 1.
• β (current gain in CE) = I_C/I_B, typically 20–300.
• Q-point (कार्यबिंदू) – DC operating values (I_C, V_CE).
• Thermal runaway – rise in temperature increases I_C, which further heats the transistor.

Formulas

• I_E = I_B + I_C
• α = I_C/I_E; β = I_C/I_B; β = α/(1 − α); α = β/(1 + β)
• Fixed bias: I_B = (V_CC − V_BE)/R_B; I_C = βI_B; V_CE = V_CC − I_C R_C
• Load line: V_CE = V_CC − I_C R_C
• CE voltage gain (approx.): A_v = −β R_L/r_in (minus sign = phase reversal)

5 important questions

Q1. Why is the base of a transistor made thin and lightly doped?

So that very few carriers coming from the emitter recombine in the base. Almost all of them reach the collector, giving a large collector current and high current gain.

Q2. If α = 0.98, find β. If I_B = 50 μA, find I_C and I_E for β = 100.

β = α/(1 − α) = 0.98/0.02 = 49. For β = 100: I_C = βI_B = 100 × 50 μA = 5 mA; I_E = I_B + I_C = 5.05 mA.

Q3. Derive the relation between α and β.

1. I_E = I_B + I_C. Divide by I_C: I_E/I_C = I_B/I_C + 1.
2. Since α = I_C/I_E and β = I_C/I_B: 1/α = 1/β + 1.
3. 1/β = 1/α − 1 = (1 − α)/α.
4. So β = α/(1 − α), and α = β/(1 + β).
5. As α → 1, β becomes large (α = 0.99 gives β = 99).

Q4. In a fixed-bias CE circuit, V_CC = 12 V, R_B = 300 kΩ, R_C = 2 kΩ, β = 100, V_BE = 0.7 V. Find I_B, I_C and V_CE.

1. I_B = (V_CC − V_BE)/R_B = (12 − 0.7)/300000 = 11.3/300000 ≈ 37.7 μA.
2. I_C = βI_B = 100 × 37.7 μA ≈ 3.77 mA.
3. V_CE = V_CC − I_C R_C = 12 − 3.77 × 10⁻³ × 2000 = 12 − 7.53 ≈ 4.47 V.
4. V_CE is well above saturation (~0.2 V), so the transistor is in the active region.

Q5. Explain the working of a CE amplifier.

1. Voltage-divider bias sets the Q-point; coupling capacitors pass the AC signal; emitter bypass capacitor keeps gain high.
2. Input signal adds to the base-emitter bias, changing I_B slightly.
3. I_C changes β times more, so the drop across R_C changes a lot.
4. Output V_CE = V_CC − I_C R_C: when input rises, I_C rises and output falls — 180° phase reversal.
5. Gives high voltage and power gain; frequency response curve shows mid-band gain with fall at low and high frequencies.

Exam tip

Draw input and output characteristics for CE with the active, saturation and cut-off regions marked. In biasing numericals, always check V_CE at the end to confirm the active region.

Quick summary

An operational amplifier (op-amp) is a high-gain, direct-coupled differential amplifier made as an integrated circuit (the 741 is the classic example). With a few external resistors and capacitors it can amplify, add, subtract, integrate and differentiate signals. This topic covers the ideal op-amp, the concepts of virtual ground and CMRR, the inverting and non-inverting amplifiers, the voltage follower, the summing amplifier, the integrator and the differentiator.

Key points

• An op-amp has two inputs — inverting (−) and non-inverting (+) — and one output. Output V_o = A(V₂ − V₁), where A is the open-loop gain.
• Ideal op-amp: infinite open-loop gain, infinite input impedance, zero output impedance, infinite bandwidth, infinite CMRR, zero offset.
• Real 741: A ≈ 2 × 10⁵, input impedance about 2 MΩ, output impedance about 75 Ω, slew rate 0.5 V/μs.
• With negative feedback, the two inputs are held at almost the same voltage. If the + input is grounded, the − input is at 0 V without being connected to ground: a virtual ground.
• No current enters the input terminals (infinite input impedance).
• Inverting amplifier: gain = −R_f/R₁ (output is 180° out of phase).
• Non-inverting amplifier: gain = 1 + R_f/R₁ (in phase).
• Voltage follower: gain 1, very high input impedance; used as a buffer.
• Summing amplifier adds input voltages; used in audio mixers and D/A converters.
• Integrator (capacitor in the feedback path) gives output proportional to the integral of the input; a square wave input gives a triangular output.
• Differentiator (capacitor at the input) gives output proportional to the rate of change of the input.

Important terms

• CMRR – common-mode rejection ratio, A_d/A_c; how well the op-amp rejects signals common to both inputs (often in dB).
• Slew rate – maximum rate of change of the output voltage.
• Offset voltage – small input voltage needed to make the output zero.
• Closed-loop gain – gain with feedback connected.
• Virtual ground – a point at ground potential that carries no current to ground.

Formulas

• Inverting: A_v = V_o/V_in = −R_f/R₁
• Non-inverting: A_v = 1 + R_f/R₁
• Summing (inverting): V_o = −R_f(V₁/R₁ + V₂/R₂ + V₃/R₃)
• Difference amplifier (equal resistors): V_o = (R_f/R₁)(V₂ − V₁)
• Integrator: V_o = −(1/RC)∫V_in dt
• Differentiator: V_o = −RC dV_in/dt
• CMRR (dB) = 20 log₁₀(A_d/A_c)

5 important questions

Q1. List the characteristics of an ideal op-amp.

Infinite open-loop voltage gain; infinite input impedance (no input current); zero output impedance; infinite bandwidth; infinite CMRR; zero offset voltage and drift; infinite slew rate.

Q2. An inverting amplifier has R₁ = 10 kΩ and R_f = 100 kΩ. Find the gain and the output for an input of 0.2 V.

A_v = −R_f/R₁ = −100/10 = −10. V_o = −10 × 0.2 = −2 V.

Q3. Derive the gain of an inverting amplifier using the virtual ground concept.

1. The + input is grounded, so the − input is at virtual ground (0 V).
2. Current through R₁: I = (V_in − 0)/R₁.
3. No current enters the op-amp, so the same I flows through R_f: I = (0 − V_o)/R_f.
4. Equating: V_in/R₁ = −V_o/R_f, so V_o/V_in = −R_f/R₁. The gain depends only on the resistors, not on the op-amp's own gain.

Q4. Design a summing amplifier to give V_o = −(2V₁ + 3V₂ + V₃) using R_f = 30 kΩ. Find the output for V₁ = 0.1 V, V₂ = 0.2 V, V₃ = 0.5 V.

From V_o = −R_f(V₁/R₁ + V₂/R₂ + V₃/R₃): R₁ = 30/2 = 15 kΩ, R₂ = 30/3 = 10 kΩ, R₃ = 30/1 = 30 kΩ.
V_o = −(2 × 0.1 + 3 × 0.2 + 0.5) = −(0.2 + 0.6 + 0.5) = −1.3 V.

Q5. Explain the op-amp integrator and find its output for a constant input.

1. A resistor R carries the input to the − input (virtual ground); a capacitor C is in the feedback path.
2. Current I = V_in/R flows through R and charges C: I = −C dV_o/dt.
3. So dV_o/dt = −V_in/(RC), giving V_o = −(1/RC)∫V_in dt.

Example: V_in = +1 V constant, R = 100 kΩ, C = 1 μF (RC = 0.1 s): the output falls at 10 V per second, i.e. a ramp, reaching −5 V after 0.5 s (unless it saturates first). A square-wave input therefore gives a triangular output. A large resistor across C is used in practice to stop drift to saturation.

Exam tip

Always start op-amp derivations with the two golden rules: no current into the inputs, and the inputs are at the same voltage. Draw each circuit neatly — diagrams carry half the marks.

Quick summary

Around 1900, classical physics failed to explain several experiments involving light and atoms. New ideas — energy quanta, photons, matter waves and the uncertainty principle — led to quantum mechanics. This topic covers blackbody radiation (briefly), the photoelectric effect, the Compton effect, de Broglie waves, the Davisson–Germer experiment and Heisenberg's uncertainty principle.

Key points

• Failures of classical physics: blackbody spectrum (ultraviolet catastrophe), photoelectric effect, Compton effect, line spectra and stability of atoms, specific heat of solids at low temperature.
• Planck (1900): energy is emitted and absorbed in quanta E = hν.
• Photoelectric effect: light ejects electrons from a metal. Observations: no emission below a threshold frequency; maximum KE depends on frequency, not intensity; number of electrons depends on intensity; emission is instantaneous.
• Einstein (1905) explained it: light consists of photons of energy hν; one photon ejects one electron: hν = φ + KE_max.
• Compton effect (1923): X-rays scattered by electrons have a longer wavelength. Explained by treating the photon as a particle with momentum h/λ colliding elastically with an electron.
• de Broglie hypothesis (1924): every moving particle has a wave of wavelength λ = h/p.
• Davisson–Germer experiment (1927): electrons scattered from a nickel crystal showed diffraction, confirming matter waves (strong peak for 54 eV electrons at a 50° scattering angle).
• Wave–particle duality: light and matter show both wave and particle nature, but not in the same experiment.
• Heisenberg's uncertainty principle (1927): position and momentum cannot both be known exactly at the same time.
• Consequences: an electron cannot exist inside a nucleus; a confined particle has non-zero minimum (zero-point) energy.

Important terms

• Work function (कार्यफलन) φ – minimum energy needed to free an electron from a metal surface.
• Stopping potential – reverse voltage that stops the fastest photoelectrons: eV₀ = KE_max.
• Compton wavelength – h/(m₀c) ≈ 2.43 × 10⁻¹² m (0.0243 Å) for an electron.
• Photon – quantum of light with energy hν and momentum h/λ.

Formulas

• E = hν = hc/λ; photon momentum p = h/λ
• Einstein: hν = φ + ½mv²_max; φ = hν₀; eV₀ = hν − φ
• Compton shift: Δλ = (h/m₀c)(1 − cos θ)
• de Broglie: λ = h/p = h/(mv) = h/√(2mE)
• Electron accelerated through V volts: λ = h/√(2meV) ≈ 12.27/√V Å
• Uncertainty: Δx Δp ≥ ħ/2; ΔE Δt ≥ ħ/2 (ħ = h/2π)
• Useful: hc ≈ 1240 eV nm

5 important questions

Q1. State Einstein's photoelectric equation and the meaning of each term.

hν = φ + KE_max, where hν is the photon energy, φ = hν₀ is the work function of the metal (ν₀ = threshold frequency) and KE_max = ½mv²_max is the maximum kinetic energy of the emitted electrons.

Q2. State Heisenberg's uncertainty principle.

It is impossible to measure the position and momentum of a particle at the same time with unlimited accuracy. The product of their uncertainties is at least ħ/2: Δx Δp ≥ ħ/2.

Q3. Light of wavelength 300 nm falls on a metal with work function 2.3 eV. Find the maximum KE of the photoelectrons and the stopping potential.

1. Photon energy E = hc/λ = 1240 eV nm/300 nm ≈ 4.13 eV.
2. KE_max = E − φ = 4.13 − 2.3 = 1.83 eV.
3. Stopping potential V₀ = KE_max/e = 1.83 V.
4. Threshold wavelength λ₀ = 1240/2.3 ≈ 539 nm; light longer than this ejects no electrons.

Q4. Find the de Broglie wavelength of an electron accelerated through 100 V, and the Compton shift for X-rays scattered at 90°.

1. λ = 12.27/√V Å = 12.27/√100 = 1.227 Å.
2. (This is of the order of atomic spacings, so crystals diffract electrons.)
3. Compton shift Δλ = (h/m₀c)(1 − cos 90°) = 0.0243 Å × (1 − 0).
4. Δλ = 0.0243 Å = 2.43 × 10⁻¹² m, independent of the X-ray wavelength.

Q5. Describe the Davisson–Germer experiment and its result.

1. An electron gun accelerates electrons through a known voltage V.
2. The beam falls normally on a nickel crystal; a movable detector measures the scattered electrons at different angles.
3. The intensity is plotted against the scattering angle for different voltages.
4. A sharp peak appears at 54 V and a scattering angle of 50°.
5. Using the known spacing of nickel atoms, the diffraction peak gives λ ≈ 1.65 Å, very close to de Broglie's value h/p ≈ 1.67 Å for 54 eV electrons.
6. Conclusion: electrons behave as waves — matter waves are real.

Exam tip

Keep hc = 1240 eV nm and λ = 12.27/√V Å ready; they turn long numericals into one line. For Compton and photoelectric effects, always mention why classical wave theory fails.

Quick summary

The Schrödinger equation is the basic equation of quantum mechanics. It tells how the wave function of a particle behaves. This topic covers the meaning of the wave function, its conditions, operators, the time-dependent and time-independent Schrödinger equations, expectation values, and the particle in a one-dimensional box.

Key points

• The state of a particle is described by a wave function ψ(x, t), which can be complex.
• Born interpretation: |ψ|² = ψ*ψ is the probability density; |ψ|² dx is the probability of finding the particle between x and x + dx.
• Normalisation: ∫|ψ|² dx over all space = 1 (the particle is somewhere).
• An acceptable (well-behaved) ψ must be single-valued, finite, continuous, and have a continuous first derivative (where the potential is finite).
• Observables are represented by operators: momentum p → −iħ ∂/∂x; energy E → iħ ∂/∂t.
• The time-dependent Schrödinger equation describes how ψ changes in time.
• When the potential does not depend on time, separation of variables gives the time-independent (steady-state) equation, an eigenvalue equation Hψ = Eψ.
• Allowed energies (eigenvalues) are often discrete — quantisation comes out naturally from boundary conditions.
• Expectation value ⟨A⟩ is the average result of many measurements.
• Particle in a 1-D box (infinite walls at x = 0 and L): ψ = 0 outside; inside, energies are Eₙ = n²h²/(8mL²), n = 1, 2, 3 …
• The lowest energy E₁ is not zero (zero-point energy), consistent with the uncertainty principle. Energy levels are spaced in the ratio 1 : 4 : 9 …
• Classical physics is recovered for large n or large mass (correspondence principle).

Important terms

• Wave function (तरंग फलन) – mathematical function describing the quantum state.
• Eigenvalue / eigenfunction – if Âψ = aψ, a is the eigenvalue and ψ the eigenfunction.
• Hamiltonian operator H – −(ħ²/2m)∇² + V, the energy operator.
• Stationary state – state with definite energy whose |ψ|² does not change with time.

Formulas

• Time-dependent: iħ ∂ψ/∂t = −(ħ²/2m) ∂²ψ/∂x² + Vψ
• Time-independent: d²ψ/dx² + (2m/ħ²)(E − V)ψ = 0
• Free particle: ψ = Ae^(i(kx − ωt)), E = ħ²k²/(2m)
• Normalisation: ∫|ψ|² dx = 1
• Expectation: ⟨x⟩ = ∫ψ* x ψ dx; ⟨p⟩ = ∫ψ*(−iħ ∂/∂x)ψ dx
• Box: Eₙ = n²h²/(8mL²) = n²π²ħ²/(2mL²); ψₙ = √(2/L) sin(nπx/L)

5 important questions

Q1. Give the physical meaning of the wave function.

ψ itself has no direct physical meaning, but |ψ|² gives the probability density. |ψ(x)|² dx is the probability of finding the particle between x and x + dx at that time.

Q2. What conditions must an acceptable wave function satisfy?

It must be single-valued, finite everywhere, continuous, have a continuous first derivative (for finite potentials), and be normalisable (∫|ψ|² dx finite).

Q3. Derive the energy eigenvalues of a particle in a one-dimensional box.

1. V = 0 for 0 < x < L, V = ∞ outside, so ψ = 0 outside and at the walls.
2. Inside: d²ψ/dx² + k²ψ = 0, with k² = 2mE/ħ².
3. General solution: ψ = A sin kx + B cos kx.
4. ψ(0) = 0 gives B = 0. ψ(L) = 0 gives sin kL = 0, so kL = nπ (n = 1, 2, 3 …; n = 0 gives ψ = 0 everywhere).
5. k = nπ/L, so E = ħ²k²/(2m) = n²π²ħ²/(2mL²) = n²h²/(8mL²).
6. Normalising: ∫₀ᴸ A² sin²(nπx/L) dx = A²L/2 = 1 → A = √(2/L).

Q4. Find the lowest energy of an electron in a box of width 1 Å (h = 6.63 × 10⁻³⁴ J s, m = 9.11 × 10⁻³¹ kg).

1. E₁ = h²/(8mL²), L = 10⁻¹⁰ m.
2. h² = (6.63 × 10⁻³⁴)² = 4.396 × 10⁻⁶⁷.
3. 8mL² = 8 × 9.11 × 10⁻³¹ × 10⁻²⁰ = 7.288 × 10⁻⁵⁰.
4. E₁ = 4.396 × 10⁻⁶⁷/7.288 × 10⁻⁵⁰ ≈ 6.03 × 10⁻¹⁸ J.
5. In eV: 6.03 × 10⁻¹⁸/1.6 × 10⁻¹⁹ ≈ 37.7 eV. Then E₂ = 4E₁ ≈ 151 eV.

Q5. Derive the time-independent Schrödinger equation from the time-dependent one.

1. Time-dependent: iħ ∂ψ/∂t = −(ħ²/2m) ∂²ψ/∂x² + V(x)ψ, with V independent of t.
2. Try ψ(x, t) = φ(x) f(t).
3. Substitute and divide by φf: iħ (1/f)(df/dt) = −(ħ²/2m)(1/φ)(d²φ/dx²) + V.
4. Left side depends only on t, right only on x, so both equal a constant E.
5. Time part: f = e^(−iEt/ħ). Space part: −(ħ²/2m) d²φ/dx² + Vφ = Eφ.
6. Rearranged: d²φ/dx² + (2m/ħ²)(E − V)φ = 0.

Exam tip

The particle-in-a-box derivation is the most frequently asked long question — practise applying the two boundary conditions and normalisation step by step. Remember Eₙ ∝ n² and ∝ 1/L².

Quick summary

When a particle meets a sudden change in potential energy — a step or a barrier — quantum mechanics gives results that classical physics cannot: a particle can be reflected even when it has enough energy to pass, and it can pass through a barrier even when it does not have enough energy. This topic covers the potential step, the rectangular barrier, reflection and transmission coefficients, quantum tunnelling and its applications such as alpha decay, the tunnel diode and the scanning tunnelling microscope.

Key points

• Solve the time-independent Schrödinger equation −(ħ²/2m)ψ″ + Vψ = Eψ in each region and join the solutions using the boundary conditions: ψ and dψ/dx must be continuous.
• E greater than V₀: classically all particles go on; quantum mechanically some are reflected. R = [(k₁ − k₂)/(k₁ + k₂)]², T = 4k₁k₂/(k₁ + k₂)², R + T = 1.
• E less than V₀: total reflection (R = 1), but the wave function enters the step and decays as e^(−αx) — a finite chance of finding the particle in the classically forbidden region.
• Potential step (V = 0 for x less than 0, V = V₀ for x greater than 0):
• Rectangular barrier of height V₀ and width a, with E less than V₀: the wave decays inside the barrier and a small wave emerges on the far side. The particle "tunnels" through.
• For a wide or high barrier (αa much greater than 1), T ≈ 16(E/V₀)(1 − E/V₀) e^(−2αa), with α = √(2m(V₀ − E))/ħ.
• Tunnelling probability falls exponentially with barrier width and with √(mass × (V₀ − E)), so it matters only for light particles and very thin barriers.
• Applications: alpha decay (Gamow, 1928), tunnel diode, scanning tunnelling microscope (STM), nuclear fusion in the Sun, flash memory, field emission.

Important terms

• Reflection coefficient (R) – ratio of reflected to incident probability current.
• Transmission coefficient (T) – ratio of transmitted to incident probability current.
• Classically forbidden region – region where E is less than V, so classical kinetic energy would be negative.
• Penetration depth – distance 1/α over which ψ falls by a factor e inside the step.
• Probability current density – J = (ħk/m)|A|² for a plane wave.

Formulas

• k₁ = √(2mE)/ħ (outside), k₂ = √(2m(E − V₀))/ħ (above the step)
• α = √(2m(V₀ − E))/ħ (inside the barrier, E less than V₀)
• Step, E greater than V₀: R = [(k₁ − k₂)/(k₁ + k₂)]², T = 4k₁k₂/(k₁ + k₂)²
• Barrier, thick: T ≈ 16(E/V₀)(1 − E/V₀) e^(−2αa)
• Penetration depth δ = 1/α

5 important questions

Q1. What is quantum tunnelling? Give two examples.

It is the passage of a particle through a potential barrier whose height is greater than the particle's energy. Classically this is impossible, but the wave function does not become zero inside the barrier; it decays and leaves a small amplitude on the other side. Examples: alpha particles escaping from nuclei; electrons crossing the thin junction of a tunnel diode; the STM current between tip and surface.

Q2. For a potential step with E = 2V₀, find R and T.

k₂/k₁ = √((E − V₀)/E) = √(1/2) = 0.707.
R = [(1 − 0.707)/(1 + 0.707)]² = (0.293/1.707)² = (0.1716)² ≈ 0.029.
T = 4 × 0.707/(1.707)² = 2.828/2.914 ≈ 0.971. Check: R + T = 1 ✔. About 3% of particles are reflected even though they have twice the step energy.

Q3. Derive R and T for a potential step with E greater than V₀.

1. Region I (x less than 0): ψ₁ = Ae^(ik₁x) + Be^(−ik₁x) (incident + reflected).
2. Region II (x greater than 0): ψ₂ = Ce^(ik₂x) (transmitted only).
3. At x = 0: ψ continuous → A + B = C; ψ′ continuous → k₁(A − B) = k₂C.
4. Solving: B/A = (k₁ − k₂)/(k₁ + k₂), C/A = 2k₁/(k₁ + k₂).
5. R = |B/A|² = [(k₁ − k₂)/(k₁ + k₂)]². T = (k₂/k₁)|C/A|² = 4k₁k₂/(k₁ + k₂)². (T has the factor k₂/k₁ because the particle speed changes.) R + T = 1.

Q4. An electron of energy 1 eV meets a barrier 2 eV high and 0.5 nm wide. Estimate the transmission probability.

1. V₀ − E = 1 eV = 1.6 × 10⁻¹⁹ J.
2. α = √(2 × 9.11 × 10⁻³¹ × 1.6 × 10⁻¹⁹)/(1.055 × 10⁻³⁴) = √(2.915 × 10⁻⁴⁹)/(1.055 × 10⁻³⁴) = 5.40 × 10⁻²⁵/1.055 × 10⁻³⁴ ≈ 5.12 × 10⁹ m⁻¹.
3. 2αa = 2 × 5.12 × 10⁹ × 0.5 × 10⁻⁹ = 5.12; e^(−5.12) ≈ 0.0060.
4. Prefactor 16 × 0.5 × 0.5 = 4. T ≈ 4 × 0.0060 ≈ 0.024 (about 2.4%).

If the width doubles to 1 nm, T falls to about 4 × e^(−10.24) ≈ 1.4 × 10⁻⁴: tunnelling is very sensitive to width.

Q5. Explain alpha decay using tunnelling.

Inside the nucleus an alpha particle is held by the strong nuclear force; outside, the Coulomb repulsion forms a barrier about 25–30 MeV high. Alpha particles have only 4–9 MeV, so classically they could never escape. Quantum mechanically, they tunnel through the barrier with a tiny probability per collision with the wall. Since they hit the wall about 10²¹ times per second, escape eventually happens. A small increase in energy makes the barrier thinner and the decay much faster: this explains the Geiger–Nuttall law, linking half-life (seconds to billions of years) with alpha energy.

Exam tip

State the boundary conditions clearly before solving any step or barrier problem. In tunnelling numericals, compute α first and remember that the exponent 2αa dominates the answer.

Quick summary

Lagrangian mechanics reformulates Newton's mechanics using energies instead of forces. With generalised coordinates and the Lagrangian L = T − V, the equations of motion of complicated constrained systems can be found easily. This topic covers constraints, degrees of freedom, D'Alembert's principle, Lagrange's equations, cyclic coordinates and an introduction to the Hamiltonian.

Key points

• Constraints restrict motion. Holonomic: expressible as equations among coordinates and time (bead on a wire, rigid body). Non-holonomic: not expressible so (gas in a container; rolling with inequalities).
• Scleronomic constraints do not depend on time; rheonomic constraints do (bead on a moving wire).
• Degrees of freedom = number of independent coordinates needed = 3N − k (N particles, k holonomic constraints).
• Generalised coordinates q₁, q₂, … are any independent quantities (lengths, angles) that fix the configuration.
• Constraint forces (like tension, normal reaction) do no virtual work in ideal cases, so they drop out.
• D'Alembert's principle: Σ(Fᵢ − ṗᵢ)·δrᵢ = 0; Lagrange's equations follow from it.
• Lagrangian L = T − V (kinetic minus potential energy).
• Lagrange's equation for each coordinate: d/dt(∂L/∂q̇) − ∂L/∂q = 0.
• Generalised momentum: p = ∂L/∂q̇.
• Cyclic (ignorable) coordinate: one that does not appear in L; its generalised momentum is conserved.
• Hamiltonian H = Σpq̇ − L; for scleronomic conservative systems H = T + V (total energy).
• Hamilton's equations: q̇ = ∂H/∂p, ṗ = −∂H/∂q.
• Advantages: works with scalars (energies), handles constraints easily, valid in any coordinates.

Important terms

• Virtual displacement – imaginary instantaneous small displacement consistent with constraints (आभासी विस्थापन).
• Configuration space – space of generalised coordinates.
• Generalised force – Qⱼ = ΣFᵢ·∂rᵢ/∂qⱼ.

Formulas

• Degrees of freedom f = 3N − k
• L = T − V
• d/dt(∂L/∂q̇ⱼ) − ∂L/∂qⱼ = 0
• pⱼ = ∂L/∂q̇ⱼ; ∂L/∂qⱼ = 0 ⇒ pⱼ = constant
• H = Σpⱼq̇ⱼ − L; q̇ⱼ = ∂H/∂pⱼ, ṗⱼ = −∂H/∂qⱼ

5 important questions

Q1. What are holonomic and non-holonomic constraints? Give an example of each.

Holonomic constraints can be written as equations f(r₁, r₂, …, t) = 0, e.g. a simple pendulum with a rigid rod (x² + y² = l²). Non-holonomic constraints cannot, e.g. gas molecules inside a container (r ≤ R, an inequality).

Q2. What is a cyclic coordinate? What is its importance?

A coordinate that does not appear in the Lagrangian (only its velocity may appear). Since ∂L/∂q = 0, Lagrange's equation gives d/dt(∂L/∂q̇) = 0, so the corresponding generalised momentum is conserved.

Q3. Using Lagrange's equation, obtain the equation of motion of a simple pendulum.

1. Generalised coordinate: angle θ; bob mass m, length l.
2. T = ½ml²θ̇²; V = −mgl cos θ (taking the suspension point as zero).
3. L = ½ml²θ̇² + mgl cos θ.
4. ∂L/∂θ̇ = ml²θ̇, so d/dt(∂L/∂θ̇) = ml²θ̈; ∂L/∂θ = −mgl sin θ.
5. Equation: ml²θ̈ + mgl sin θ = 0 → θ̈ + (g/l) sin θ = 0.
6. For small θ: θ̈ + (g/l)θ = 0, SHM with T = 2π√(l/g).

Q4. Find the acceleration of Atwood's machine using the Lagrangian method.

1. Masses m₁ and m₂ on a light string over a frictionless pulley; x = distance m₁ has descended; m₂ is then at (l − x) below the pulley.
2. T = ½(m₁ + m₂)ẋ².
3. V = −m₁gx − m₂g(l − x).
4. L = ½(m₁ + m₂)ẋ² + m₁gx + m₂g(l − x).
5. Lagrange: (m₁ + m₂)ẍ − (m₁ − m₂)g = 0.
6. ẍ = (m₁ − m₂)g/(m₁ + m₂). The tension never had to be found.

Q5. Write the Hamiltonian of a one-dimensional harmonic oscillator and obtain its equations of motion.

1. L = ½mẋ² − ½kx²; p = ∂L/∂ẋ = mẋ, so ẋ = p/m.
2. H = pẋ − L = p²/m − (p²/2m − ½kx²) = p²/(2m) + ½kx² (= total energy).
3. Hamilton's equations: ẋ = ∂H/∂p = p/m; ṗ = −∂H/∂x = −kx.
4. Combine: mẍ = −kx → ẍ + (k/m)x = 0.
5. SHM with ω = √(k/m).

Exam tip

Follow the same four steps every time: choose generalised coordinates, write T and V, form L, apply Lagrange's equation for each coordinate. State the number of degrees of freedom at the start.

Quick summary

A central force always acts along the line joining a body to a fixed centre, and its size depends only on the distance from that centre. Gravity and the Coulomb force are the main examples. This topic covers the reduction of the two-body problem to a one-body problem, conservation of angular momentum, the effective potential, the differential equation of the orbit, and the derivation of Kepler's laws from the inverse-square law.

Key points

• A central force has the form F = f(r) r̂. Its torque about the centre is zero, so angular momentum L is conserved.
• Because L is constant in direction, the motion stays in one plane.
• Because L is constant in size, areal velocity dA/dt = L/(2m) is constant (Kepler's second law).
• A central force is conservative, so total energy E = ½m(ṙ² + r²θ̇²) + V(r) is conserved.
• Two-body problem: two masses m₁ and m₂ interacting with each other can be replaced by one body of reduced mass μ = m₁m₂/(m₁ + m₂) moving about a fixed centre, plus free motion of the centre of mass.
• Effective potential: V_eff(r) = V(r) + L²/(2μr²). The extra term is the "centrifugal barrier", which keeps the body away from r = 0.
• For the gravitational potential V = −k/r, the energy decides the orbit: E less than 0 → ellipse (or circle at the minimum of V_eff); E = 0 → parabola; E greater than 0 → hyperbola.
• Orbit equation (with u = 1/r): d²u/dθ² + u = −(μ/L²u²) f(1/u). For the inverse-square force this gives a conic section.
• Kepler's third law T² = (4π²/GM) a³ follows from the areal velocity and the area of the ellipse.

Important terms

• Reduced mass (μ) – m₁m₂/(m₁ + m₂); for the Earth–Sun system it is almost the Earth's mass.
• Eccentricity (e) – shape of the conic: e = 0 circle, 0 less than e less than 1 ellipse, e = 1 parabola, e greater than 1 hyperbola.
• Turning points – values of r where ṙ = 0 (perihelion and aphelion).
• Centrifugal barrier – the L²/(2μr²) term in the effective potential.

Formulas

• L = μr²θ̇ = constant; dA/dt = L/(2μ)
• E = ½μṙ² + L²/(2μr²) + V(r)
• μ = m₁m₂/(m₁ + m₂)
• Orbit equation: d²u/dθ² + u = −(μ/(L²u²)) f(1/u), u = 1/r
• Inverse-square force f = −k/r²: 1/r = (μk/L²)(1 + e cos θ)
• e = √(1 + 2EL²/(μk²))
• Kepler III: T² = 4π²a³μ/k = 4π²a³/(G(m₁ + m₂)) (≈ 4π²a³/GM for a planet)

5 important questions

Q1. Show that motion under a central force is confined to a plane.

τ = r × F = r × f(r) r̂ = 0, since r and r̂ are parallel. So dL/dt = 0 and L = r × p is constant. The position vector r is always perpendicular to the fixed vector L, so r always lies in the plane perpendicular to L passing through the centre. The motion is planar.

Q2. Find the reduced mass of the hydrogen atom (m_p = 1836 m_e).

μ = m_e m_p/(m_e + m_p) = m_e × 1836/1837 ≈ 0.99946 m_e. Using μ instead of m_e changes the Rydberg constant slightly (R_H = R_∞ μ/m_e), which explains the small difference between hydrogen and deuterium lines.

Q3. Derive the differential equation of the orbit.

1. In polar form the radial equation is μ(r̈ − rθ̇²) = f(r), with L = μr²θ̇.
2. Put u = 1/r. Then ṙ = −(L/μ) du/dθ and r̈ = −(L²u²/μ²) d²u/dθ².
3. Also rθ̇² = L²u³/μ².
4. Substituting: −(L²u²/μ)(d²u/dθ² + u) = f(1/u).
5. So d²u/dθ² + u = −(μ/L²u²) f(1/u).

Q4. Using the orbit equation, show that an inverse-square attractive force gives a conic section.

1. For f = −k/r² = −ku²: the right side becomes −(μ/L²u²)(−ku²) = μk/L².
2. d²u/dθ² + u = μk/L², a constant.
3. Solution: u = μk/L² + C cos θ, i.e. 1/r = (μk/L²)(1 + e cos θ) with e = CL²/(μk).
4. This is the polar equation of a conic with the force centre at a focus. Bound orbits (e less than 1) are ellipses — Kepler's first law.

Q5. Using the effective potential, find the radius and speed of a circular orbit for V = −k/r, and solve for a satellite.

V_eff = −k/r + L²/(2μr²). For a circular orbit dV_eff/dr = 0: k/r² − L²/(μr³) = 0 → r₀ = L²/(μk). Equivalently the force balance k/r² = μv²/r gives v = √(k/(μr)).
For a satellite around the Earth, k/μ = GM = 3.99 × 10¹⁴ m³ s⁻². At r = 7.0 × 10⁶ m: v = √(3.99 × 10¹⁴/7.0 × 10⁶) = √(5.70 × 10⁷) ≈ 7.55 km s⁻¹; period T = 2πr/v = 2π × 7.0 × 10⁶/7550 ≈ 5.8 × 10³ s (about 97 min).

Exam tip

State the two conservation laws (angular momentum and energy) at the start of any central-force answer. In the orbit-equation derivation, show the change of variable to u = 1/r clearly; it is where most students lose marks.

Quick summary

Atoms emit and absorb light only at certain wavelengths, giving line spectra. Bohr's model explains the hydrogen spectrum, but the fine structure of lines and the effect of magnetic fields need the vector atom model, which adds electron spin and treats angular momenta as vectors. This topic covers hydrogen series, quantum numbers, spin, coupling schemes, spectral terms, sodium D lines and the Zeeman effect.

Key points

• Bohr model: electrons move in fixed orbits with angular momentum nħ; a photon of energy hν = E₂ − E₁ is emitted when an electron jumps down.
• Hydrogen series: Lyman (to n = 1, ultraviolet), Balmer (to n = 2, visible), Paschen (n = 3), Brackett (n = 4), Pfund (n = 5) — the last three in infrared.
• Bohr's model fails for many-electron atoms, line intensities, fine structure and the Zeeman effect.
• Vector atom model: orbital (l) and spin (s) angular momenta are vectors that combine to give total angular momentum j.
• Quantum numbers: principal n (1, 2, 3 …), orbital l (0 to n − 1), magnetic mₗ (−l to +l), spin s = ½ with mₛ = ±½, total j = l ± ½ (for one electron), and mⱼ.
• Space quantisation: an angular momentum vector can take only certain directions with respect to a field.
• Stern–Gerlach experiment (1922): a beam of silver atoms split into two in a non-uniform magnetic field, showing space quantisation and (later understood) electron spin.
• Spin: proposed by Uhlenbeck and Goudsmit (1925).
• Pauli exclusion principle: no two electrons in an atom can have all four quantum numbers (n, l, mₗ, mₛ) the same.
• L–S (Russell–Saunders) coupling: for lighter atoms, all l's add to L, all s's add to S, then J = L + S. j–j coupling: for heavy atoms, each electron's l and s combine first to j.
• Spectral term: ²ˢ⁺¹L_J, with L = 0, 1, 2, 3 written S, P, D, F.
• Sodium D lines: the 3P level splits into ²P₃/₂ and ²P₁/₂ by spin–orbit interaction; transitions to ²S₁/₂ give D₂ (5890 Å) and D₁ (5896 Å).
• Selection rules: Δl = ±1, Δj = 0, ±1 (j = 0 to j = 0 not allowed).
• Zeeman effect: splitting of spectral lines in a magnetic field. Normal (lines split into three — Lorentz triplet; spin plays no part, S = 0). Anomalous (more components; depends on Landé g factor). In very strong fields: Paschen–Back effect.

Important terms

• Fine structure (सूक्ष्म संरचना) – splitting of a spectral line into close components.
• Landé g factor – factor relating magnetic moment to total angular momentum.
• Multiplicity – 2S + 1 (singlet, doublet, triplet).
• Bohr magneton μ_B – eħ/(2m) ≈ 9.27 × 10⁻²⁴ J/T.

Formulas

• Hydrogen: 1/λ = R(1/n₁² − 1/n₂²), R ≈ 1.097 × 10⁷ m⁻¹
• Energy levels: Eₙ = −13.6/n² eV
• |l| = √(l(l + 1)) ħ; |s| = √(s(s + 1)) ħ; |j| = √(j(j + 1)) ħ
• Landé: g = 1 + [j(j + 1) + s(s + 1) − l(l + 1)]/[2j(j + 1)]
• Normal Zeeman shift: Δν = eB/(4πm)
• Energy shift in a field: ΔE = g mⱼ μ_B B

5 important questions

Q1. State the Pauli exclusion principle.

No two electrons in an atom can have the same set of four quantum numbers n, l, mₗ and mₛ. So an orbital can hold at most two electrons, with opposite spins.

Q2. Write the possible j values and spectral terms for a single electron with l = 2.

j = l ± s = 2 ± ½ = 5/2 or 3/2. With S = ½ (doublet), the terms are ²D₅/₂ and ²D₃/₂.

Q3. Calculate the Landé g factor for the ²P₃/₂ and ²P₁/₂ states.

1. For both, l = 1, s = ½; l(l + 1) = 2, s(s + 1) = 3/4.
2. ²P₃/₂: j(j + 1) = 15/4. g = 1 + (15/4 + 3/4 − 2)/(2 × 15/4) = 1 + (10/4)/(30/4) = 1 + 1/3 = 4/3.
3. ²P₁/₂: j(j + 1) = 3/4. g = 1 + (3/4 + 3/4 − 2)/(2 × 3/4) = 1 + (−1/2)/(3/2) = 1 − 1/3 = 2/3.
4. Different g values explain why the D₁ and D₂ lines split into different patterns (anomalous Zeeman effect).

Q4. Find the wavelength of the first Balmer line (Hα) and the normal Zeeman splitting in a 1 T field.

1. 1/λ = R(1/2² − 1/3²) = 1.097 × 10⁷ × (1/4 − 1/9) = 1.097 × 10⁷ × 5/36.
2. 1/λ ≈ 1.524 × 10⁶ m⁻¹ → λ ≈ 6.563 × 10⁻⁷ m = 656.3 nm (red).
3. Δν = eB/(4πm) = (1.6 × 10⁻¹⁹ × 1)/(4π × 9.11 × 10⁻³¹).
4. Denominator = 1.145 × 10⁻²⁹; Δν ≈ 1.40 × 10¹⁰ Hz.

Q5. Explain the fine structure of sodium D lines using the vector atom model.

1. Sodium has one valence electron (outside closed shells), so L = l and S = ½.
2. Ground state 3s: l = 0, j = ½ → ²S₁/₂.
3. First excited state 3p: l = 1, j = 3/2 or ½ → ²P₃/₂ and ²P₁/₂.
4. Spin–orbit interaction makes ²P₃/₂ slightly higher than ²P₁/₂.
5. Allowed transitions (Δl = −1, Δj = 0, ±1): ²P₃/₂ → ²S₁/₂ gives D₂ (5890 Å); ²P₁/₂ → ²S₁/₂ gives D₁ (5896 Å).
6. Hence the yellow line is a close doublet.

Exam tip

Draw the energy-level diagram of sodium with term symbols and the two transitions labelled. Practise the Landé g formula — it is a quick, high-scoring numerical.

Quick summary

Molecules have energy not only in their electrons but also in rotation and vibration of the nuclei. Changes in these energies give molecular spectra: rotational spectra in the microwave region, vibrational (vibration–rotation) spectra in the infrared, and electronic band spectra in the visible and ultraviolet. This topic covers the rigid rotator, the harmonic and anharmonic oscillator, selection rules, P and R branches, and how bond length and force constant are found from spectra.

Key points

• Order of energy spacings: electronic (a few eV) much greater than vibrational (about 0.1 eV) much greater than rotational (about 0.001 eV).
• Molecular spectra are band spectra (many close lines), unlike the line spectra of atoms.
• Rigid rotator model of a diatomic molecule: energy E_J = (ħ²/2I) J(J + 1) = hcB J(J + 1), J = 0, 1, 2 …
• Rotational constant B = h/(8π²Ic) (in cm⁻¹); I = μr² where μ is the reduced mass and r the bond length.
• Selection rule for pure rotation: ΔJ = ±1, and the molecule must have a permanent dipole moment (HCl yes; H₂, N₂ no).
• Rotational lines are equally spaced by 2B; line positions ν̄ = 2B(J + 1).
• Harmonic oscillator: E_v = (v + ½)hν₀, v = 0, 1, 2 …; even at v = 0 there is zero-point energy ½hν₀. Selection rule Δv = ±1, so there is one fundamental line.
• Real molecules are anharmonic (Morse potential): levels get closer as v rises, overtones (Δv = ±2, ±3) appear weakly, and the molecule can dissociate.
• Vibration–rotation spectrum: each vibrational line splits into a P branch (ΔJ = −1) and an R branch (ΔJ = +1), with a gap at the band centre (no Q branch for most diatomics).
• The Franck–Condon principle explains the intensity of electronic band lines: electronic transitions are so fast that the nuclei do not move during them.

Important terms

• Moment of inertia (I) – μr² for a diatomic molecule.
• Wave number (ν̄) – 1/λ, in cm⁻¹.
• Force constant (k) – stiffness of the bond, ν₀ = (1/2π)√(k/μ).
• Anharmonicity constant (x_e) – measures departure from SHM.
• Dissociation energy – energy needed to break the molecule from its lowest level.

Formulas

• Rotational energy: E_J = hcB J(J + 1); B = h/(8π²cI) (cm⁻¹ if c in cm s⁻¹)
• Rotational lines: ν̄ = 2B(J + 1); spacing = 2B
• I = μr², μ = m₁m₂/(m₁ + m₂)
• Vibrational energy (harmonic): E_v = (v + ½)hν₀; ν₀ = (1/2π)√(k/μ)
• Anharmonic: E_v = (v + ½)hν̄_e c − (v + ½)² x_e hν̄_e c
• Vibration–rotation: R branch ν̄ = ν̄₀ + 2B(J + 1); P branch ν̄ = ν̄₀ − 2BJ

5 important questions

Q1. Why do H₂ and N₂ show no pure rotational spectrum?

A molecule can absorb or emit microwave radiation during rotation only if it has a permanent electric dipole moment that rotates with it. Homonuclear molecules like H₂, N₂ and O₂ are symmetric and have no dipole moment, so they are microwave-inactive. (Their rotation can be studied by the Raman effect instead.)

Q2. The rotational lines of CO are spaced by 3.86 cm⁻¹. Find B and the moment of inertia.

2B = 3.86 cm⁻¹ → B = 1.93 cm⁻¹.
I = h/(8π²cB) = 6.626 × 10⁻³⁴/(8 × 9.87 × 3 × 10¹⁰ × 1.93) = 6.626 × 10⁻³⁴/(4.572 × 10¹²) ≈ 1.45 × 10⁻⁴⁶ kg m².

Q3. For CO, μ = 1.14 × 10⁻²⁶ kg. Using Q2, find the bond length.

r = √(I/μ) = √(1.45 × 10⁻⁴⁶/1.14 × 10⁻²⁶) = √(1.272 × 10⁻²⁰) ≈ 1.13 × 10⁻¹⁰ m = 0.113 nm, which matches the known C–O bond length.

Q4. Derive the rotational energy levels and show that the lines are equally spaced.

1. Classical rotational energy E = L²/(2I).
2. In quantum mechanics L² = J(J + 1)ħ², so E_J = J(J + 1)ħ²/(2I) = hcB J(J + 1).
3. For a jump J → J + 1: ΔE = hcB[(J + 1)(J + 2) − J(J + 1)] = 2hcB(J + 1).
4. So ν̄ = 2B(J + 1) = 2B, 4B, 6B, … — equally spaced by 2B. From the spacing we get B, hence I and the bond length.

Q5. The fundamental vibration of HCl is at 2886 cm⁻¹ (μ = 1.63 × 10⁻²⁷ kg). Find the force constant and the zero-point energy.

1. ν₀ = cν̄ = 3 × 10¹⁰ × 2886 = 8.66 × 10¹³ Hz.
2. k = 4π²ν₀²μ = 39.48 × (8.66 × 10¹³)² × 1.63 × 10⁻²⁷ = 39.48 × 7.50 × 10²⁷ × 1.63 × 10⁻²⁷ ≈ 483 N m⁻¹.
3. Zero-point energy = ½hν₀ = 0.5 × 6.626 × 10⁻³⁴ × 8.66 × 10¹³ ≈ 2.87 × 10⁻²⁰ J ≈ 0.18 eV.

Exam tip

Keep c in cm s⁻¹ (3 × 10¹⁰) when B and ν̄ are in cm⁻¹ — mixing units is the commonest mistake. Remember: rotation → microwave, vibration → infrared, electronic → visible/UV.

Quick summary

When monochromatic light passes through a transparent substance, most of the scattered light has the same frequency as the incident light, but a very small part has slightly changed frequencies. This is the Raman effect, discovered by C. V. Raman in Kolkata on 28 February 1928 (Nobel Prize 1930). This topic covers Rayleigh and Raman scattering, Stokes and anti-Stokes lines, the quantum and classical explanations, selection rules, the experimental set-up, and applications of Raman spectroscopy.

Key points

• Rayleigh scattering: scattered light has the same frequency ν₀ as the incident light (elastic scattering). It is much stronger than Raman scattering.
• Raman lines: weak lines at ν₀ ± Δν appear on both sides of the Rayleigh line. The shift Δν is a property of the scattering molecule and does not depend on the exciting frequency.
• Stokes lines (ν₀ − Δν): the molecule takes energy from the photon (goes to a higher vibrational or rotational level). Stronger, because most molecules are in the ground state.
• Anti-Stokes lines (ν₀ + Δν): the molecule gives energy to the photon. Weaker; their intensity rises with temperature.
• Raman shifts equal the vibrational or rotational frequencies of the molecule, so the Raman spectrum gives the same kind of information as IR spectra, but using visible light.
• Quantum picture: a photon hν₀ collides with a molecule; energy conservation gives hν′ = hν₀ ± ΔE.
• Classical picture: the field induces a dipole p = αE. If the polarisability α changes as the molecule vibrates, the induced dipole contains frequencies ν₀ and ν₀ ± ν_vib.
• Selection rule: a vibration is Raman active if it changes the polarisability; it is IR active if it changes the dipole moment. So H₂, N₂ and O₂, which are IR-inactive, give Raman spectra.
• Rule of mutual exclusion: in molecules with a centre of symmetry (CO₂), modes active in IR are inactive in Raman and vice versa.
• Rotational Raman selection rule ΔJ = 0, ±2, giving lines spaced by 4B (first line at 6B).

Important terms

• Raman shift – difference in wave number between the incident and Raman lines, Δν̄ = ν̄₀ − ν̄_s.
• Polarisability (α) – ease with which the electron cloud of a molecule is distorted by an electric field.
• Raman active – a mode that appears in the Raman spectrum.
• Virtual state – short-lived intermediate state in the scattering process, not a real energy level.

Formulas

• hν_s = hν₀ ± hν_m (Stokes −, anti-Stokes +)
• Raman shift Δν̄ = |ν̄₀ − ν̄_s| = ν̄_m (cm⁻¹)
• Induced dipole p = αE₀ cos 2πν₀t, with α = α₀ + α₁ cos 2πν_m t → frequencies ν₀, ν₀ ± ν_m
• Intensity ratio anti-Stokes/Stokes ≈ e^(−hν_m/kT)
• Rotational Raman lines: Δν̄ = ±(4B J + 6B)

5 important questions

Q1. Distinguish between Rayleigh scattering and Raman scattering.

Q2. Why are Stokes lines more intense than anti-Stokes lines?

Stokes lines come from molecules in the ground vibrational state, and anti-Stokes lines from molecules already in an excited state. By the Boltzmann law, at room temperature far more molecules are in the ground state, so Stokes lines are much stronger. Heating the sample increases the anti-Stokes intensity.

Q3. Light of wavelength 435.8 nm gives a Stokes line at 444.7 nm. Find the Raman shift and the wavelength of the anti-Stokes line.

ν̄₀ = 1/(435.8 × 10⁻⁷ cm) = 22,946 cm⁻¹; ν̄_s = 1/(444.7 × 10⁻⁷ cm) = 22,487 cm⁻¹.
Raman shift = 22,946 − 22,487 = 459 cm⁻¹ (a vibration of CCl₄).
Anti-Stokes: ν̄_a = 22,946 + 459 = 23,405 cm⁻¹ → λ_a = 1/23,405 cm ≈ 427.2 nm.

Q4. Explain the Raman effect on the classical and quantum theories.

Quantum: an incident photon hν₀ raises the molecule to a short-lived virtual state. It falls back either to the original level (Rayleigh, hν₀), to a higher vibrational level (Stokes, h(ν₀ − ν_m)) or, if it started excited, to a lower level (anti-Stokes, h(ν₀ + ν_m)).
Classical: E = E₀ cos 2πν₀t induces p = αE. If α = α₀ + α₁ cos 2πν_m t, then p = α₀E₀ cos 2πν₀t + ½α₁E₀[cos 2π(ν₀ − ν_m)t + cos 2π(ν₀ + ν_m)t]. The oscillating dipole radiates at ν₀ (Rayleigh) and ν₀ ± ν_m (Raman). If α₁ = 0 (polarisability does not change), there is no Raman line — the selection rule. The classical model cannot explain why Stokes lines are stronger.

Q5. Describe the experimental set-up and give applications of Raman spectroscopy.

Set-up: a monochromatic source (Raman used a mercury arc with filters; today a laser) is focused into the liquid in a tube. Scattered light is collected at 90° to the beam (to avoid the direct beam), passed into a spectrograph and recorded on a plate or CCD. A dark background (black-painted tube end) cuts stray light.
Applications: identifying molecules and their structure (fingerprint spectra), studying symmetric molecules that are IR-inactive, analysing minerals and drugs, detecting stress in semiconductors, non-destructive testing of art and gems, studying biological samples in water (water is a weak Raman scatterer).

Exam tip

Remember the date: 28 February is celebrated as National Science Day in India because of the Raman effect. Convert wavelengths to wave numbers (cm⁻¹) before subtracting — never subtract wavelengths directly.