Nuclear Physics & Quantum Physics Test (Step-by-step mathematical derivations and explanations)

PHYSICS 2: NUCLEAR PHYSICS AND QUANTUM PHYSICS

Nuclear Physics Questions

(a) Definitions

Term	Definition
Nuclear Binding Energy	Energy required to disassemble a nucleus into its constituent protons and neutrons
Packing Fraction	(Mass defect per nucleon) × c², where mass defect = (M-A)/A (M=atomic mass, A=mass number)
Thermal Neutrons	Neutrons with kinetic energy ≈ 0.025 eV (room temperature)

(b)(i) Nuclear Mass Deficit

The mass of a nucleus is less than the sum of its nucleons due to the mass-energy equivalence (E=mc²). The "missing mass" (mass defect) is converted into binding energy that holds the nucleus together.

(b)(ii) Helium Binding Energy

A binding energy of 28.2 MeV for ⁴₂He means 28.2 MeV of energy would be required to completely separate its 2 protons and 2 neutrons, or that this amount of energy was released when the nucleus formed from its nucleons.

(b)(iv) Uranium Fission Q-value

Given fission: ²³⁶₉₂U → ¹⁴⁶₅₇La + ⁸⁷₃₅Br + 3n + Q
Q = BE_products - BE_reactants
= [146×8.41 + 87×8.59 + 3×0] - [236×7.59]
= [1227.86 + 747.33] - 1791.24
= 1975.19 - 1791.24 = 183.95 MeV

(c)(i) Radioactive Decay Terms

Half-life (t_½): Time for half of radioactive nuclei to decay

Decay constant (λ): Probability per unit time that a nucleus will decay, related by t_½ = ln(2)/λ

(c)(ii) Carbon Dating

Given: Current ratio ¹⁴C/¹²C = 1.3×10⁻¹², t_½ = 5730 years
Original ratio = 1.3×10⁻¹² (living wood)
Current activity = 19 decays/min = λN
Original activity = λN₀ = 19×(original ratio/current ratio)
Using N = N₀e^-λt and λ = ln(2)/t_½
t = (t_½/ln(2)) × ln(N₀/N) ≈ 5730/0.693 × ln(1) = 0 years
(Note: Additional information needed for complete calculation)

(d) Uranium Reactor Calculation

Energy needed = Power × Time = 1000 MW × 10 years
= 10⁹ J/s × (10×365×24×3600 s) = 3.156×10¹⁷ J
At 10% efficiency, total energy required = 3.156×10¹⁸ J
Energy per fission = 200 MeV = 200×1.6×10⁻¹³ J = 3.2×10⁻¹¹ J
Number of fissions = 3.156×10¹⁸/3.2×10⁻¹¹ ≈ 9.86×10²⁸
Mass of ²³⁵U = (9.86×10²⁸/6.022×10²³) × 235 g/mol ≈ 38,500 kg

Quantum Physics Questions

(a)(i) Photoelectric Law

Einstein's Photoelectric Equation: E = hν = φ + K.E._max

1. For given metal & frequency, photoelectron emission rate ∝ light intensity
2. Maximum kinetic energy depends on frequency, not intensity
3. Below threshold frequency (ν₀), no emission occurs regardless of intensity

(a)(ii) Definitions

Work Function (φ): Minimum energy needed to eject an electron from a metal surface

Stopping Potential (V₀): Minimum reverse potential needed to stop the fastest photoelectrons

(b) Planck's Constant Calculation

For λ₁ = 3310Å (3.31×10⁻⁷ m), K.E.₁ = 3×10⁻¹⁹ J
For λ₂ = 5000Å (5×10⁻⁷ m), K.E.₂ = 0.972×10⁻¹⁹ J
Using hc/λ = φ + K.E.:
hc/3.31×10⁻⁷ = φ + 3×10⁻¹⁹ ...(1)
hc/5×10⁻⁷ = φ + 0.972×10⁻¹⁹ ...(2)
Subtract (2) from (1):
hc(1/3.31 - 1/5)×10⁷ = 2.028×10⁻¹⁹
h ≈ 6.63×10⁻³⁴ Js (Planck's constant)
Threshold λ when K.E.=0: φ = hc/λ₀ ⇒ λ₀ ≈ 5400Å

(c)(i) X-ray Types

Hard X-rays	Soft X-rays
High energy (10-100 keV)	Low energy (0.1-10 keV)
Short wavelength (0.01-0.1 nm)	Longer wavelength (0.1-10 nm)
Greater penetration	Less penetration

(c)(ii) X-ray Tube Cooling

Power input = VI = 60×10³ × 30×10⁻³ = 1800 W
Heat produced = 99% × 1800 = 1782 W
Heat removed = mcΔT/Δt ⇒ ΔT = (PΔt)/(mc)
= (1782 × 1)/(0.06 × 4186) ≈ 7.1°C/s