Noesis
Level 2 — RecallModern Physics

Modern Physics

⏱ 30 minutes40 marksprintable — key stays hidden on paper

Level: 2 (Recall / Standard textbook problems) Time limit: 30 minutes Total marks: 40

Useful constants: h=6.63×10−34 J⋅sh = 6.63\times10^{-34}\,\text{J·s}, c=3.00×108 m/sc = 3.00\times10^{8}\,\text{m/s}, me=9.11×10−31 kgm_e = 9.11\times10^{-31}\,\text{kg}, 1 eV=1.60×10−19 J1\,\text{eV} = 1.60\times10^{-19}\,\text{J}, mec2=0.511 MeVm_e c^2 = 0.511\,\text{MeV}.

Q1. (4 marks) State Einstein's photoelectric equation. Sodium has a work function of 2.28 eV2.28\,\text{eV}. Calculate the maximum kinetic energy (in eV) of photoelectrons emitted when light of wavelength 400 nm400\,\text{nm} falls on it.

Q2. (4 marks) Calculate the de Broglie wavelength of an electron accelerated through a potential difference of 100 V100\,\text{V}. (Use non-relativistic relation.)

Q3. (5 marks) An electron is confined to a one-dimensional box of width L=1.0×10−10 mL = 1.0\times10^{-10}\,\text{m}. (a) Write the expression for the allowed energy levels EnE_n. (1) (b) Calculate the ground-state energy in eV. (3) (c) State the value of nn for the first excited state. (1)

Q4. (4 marks) A photon of wavelength 0.0500 nm0.0500\,\text{nm} undergoes Compton scattering through an angle of 90∘90^\circ from a free electron. Calculate the wavelength shift Δλ\Delta\lambda and the scattered wavelength. (Compton wavelength =2.43×10−12 m= 2.43\times10^{-12}\,\text{m}.)

Q5. (4 marks) For the hydrogen atom, En=−13.6/n2 eVE_n = -13.6/n^2\,\text{eV}. (a) Compute the energy of the n=2n=2 level. (1) (b) Calculate the wavelength of the photon emitted in the transition n=3→n=2n=3 \to n=2 (Balmer). (3)

Q6. (4 marks) State the Heisenberg uncertainty principle for position and momentum. An electron's position is known to within Δx=1.0×10−10 m\Delta x = 1.0\times10^{-10}\,\text{m}. Estimate the minimum uncertainty in its momentum (ℏ=1.055×10−34 J⋅s\hbar = 1.055\times10^{-34}\,\text{J·s}).

Q7. (5 marks) The half-life of a radioactive isotope is 8.0 days8.0\,\text{days}. (a) Calculate the decay constant λ\lambda (in day−1\text{day}^{-1}). (2) (b) What fraction of the original sample remains after 24 days24\,\text{days}? (3)

Q8. (4 marks) Define binding energy and mass defect. Given that the mass defect of a helium-4 nucleus is 0.0304 u0.0304\,\text{u} (where 1 u=931.5 MeV/c21\,\text{u} = 931.5\,\text{MeV}/c^2), calculate the total binding energy and the binding energy per nucleon.

Q9. (3 marks) A muon travels at v=0.99cv = 0.99c. Its proper lifetime is 2.2 μs2.2\,\mu\text{s}. Calculate the lifetime observed in the laboratory frame (γ=1/1−v2/c2\gamma = 1/\sqrt{1-v^2/c^2}).

Q10. (3 marks) State the two postulates of the special theory of relativity, and name the experiment (with its null result) that motivated them.

Answer keyMark scheme & solutions

Q1. (4 marks) Photoelectric equation: Kmax=hf−ϕ=hcλ−ϕK_{max} = hf - \phi = \dfrac{hc}{\lambda} - \phi. (1)

Photon energy: E=hcλ=(6.63×10−34)(3.00×108)400×10−9=4.97×10−19 JE = \dfrac{hc}{\lambda} = \dfrac{(6.63\times10^{-34})(3.00\times10^8)}{400\times10^{-9}} = 4.97\times10^{-19}\,\text{J}. (1) =4.97×10−19/1.60×10−19=3.11 eV= 4.97\times10^{-19}/1.60\times10^{-19} = 3.11\,\text{eV}. (1) Kmax=3.11−2.28=0.83 eVK_{max} = 3.11 - 2.28 = 0.83\,\text{eV}. (1)

Why: Photon energy goes partly to overcome the work function; remainder is max KE.

Q2. (4 marks) λ=hp=h2meeV\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2m_e eV}}. (1) 2meeV=2(9.11×10−31)(1.60×10−19)(100)=2.915×10−472m_e eV = 2(9.11\times10^{-31})(1.60\times10^{-19})(100) = 2.915\times10^{-47}. (1) p=2.915×10−47=5.40×10−24 kg⋅m/sp = \sqrt{2.915\times10^{-47}} = 5.40\times10^{-24}\,\text{kg·m/s}. (1) λ=6.63×10−34/5.40×10−24=1.23×10−10 m=0.123 nm\lambda = 6.63\times10^{-34}/5.40\times10^{-24} = 1.23\times10^{-10}\,\text{m} = 0.123\,\text{nm}. (1)

Q3. (5 marks) (a) En=n2h28mL2E_n = \dfrac{n^2 h^2}{8mL^2}. (1) (b) E1=(6.63×10−34)28(9.11×10−31)(1.0×10−10)2E_1 = \dfrac{(6.63\times10^{-34})^2}{8(9.11\times10^{-31})(1.0\times10^{-10})^2} (1) =4.396×10−677.288×10−50=6.03×10−18 J= \dfrac{4.396\times10^{-67}}{7.288\times10^{-50}} = 6.03\times10^{-18}\,\text{J} (1) =6.03×10−18/1.60×10−19=37.7 eV= 6.03\times10^{-18}/1.60\times10^{-19} = 37.7\,\text{eV}. (1) (c) First excited state: n=2n = 2. (1)

Q4. (4 marks) Compton: Δλ=hmec(1−cos⁡θ)=λC(1−cos⁡θ)\Delta\lambda = \dfrac{h}{m_e c}(1-\cos\theta) = \lambda_C(1-\cos\theta). (1) At 90∘90^\circ, cos⁡90∘=0\cos 90^\circ = 0, so Δλ=λC=2.43×10−12 m=0.00243 nm\Delta\lambda = \lambda_C = 2.43\times10^{-12}\,\text{m} = 0.00243\,\text{nm}. (2) Scattered λ′=0.0500+0.00243=0.0524 nm\lambda' = 0.0500 + 0.00243 = 0.0524\,\text{nm}. (1)

Q5. (4 marks) (a) E2=−13.6/4=−3.40 eVE_2 = -13.6/4 = -3.40\,\text{eV}. (1) (b) ΔE=E3−E2=−13.6/9−(−13.6/4)=−1.511+3.40=1.889 eV\Delta E = E_3 - E_2 = -13.6/9 - (-13.6/4) = -1.511 + 3.40 = 1.889\,\text{eV}. (1) =1.889×1.60×10−19=3.02×10−19 J= 1.889\times1.60\times10^{-19} = 3.02\times10^{-19}\,\text{J}. λ=hcΔE=(6.63×10−34)(3.00×108)3.02×10−19\lambda = \dfrac{hc}{\Delta E} = \dfrac{(6.63\times10^{-34})(3.00\times10^8)}{3.02\times10^{-19}} (1) =6.58×10−7 m=658 nm= 6.58\times10^{-7}\,\text{m} = 658\,\text{nm} (H-alpha line). (1)

Q6. (4 marks) Statement: Δx Δp≥ℏ2\Delta x\,\Delta p \geq \dfrac{\hbar}{2} — position and momentum cannot both be known to arbitrary precision. (2) Δp≥ℏ2Δx=1.055×10−342(1.0×10−10)=5.3×10−25 kg⋅m/s\Delta p \geq \dfrac{\hbar}{2\Delta x} = \dfrac{1.055\times10^{-34}}{2(1.0\times10^{-10})} = 5.3\times10^{-25}\,\text{kg·m/s}. (2)

Q7. (5 marks) (a) λ=ln⁡2T1/2=0.6938.0=0.0866 day−1\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{8.0} = 0.0866\,\text{day}^{-1}. (2) (b) 24 days=324\,\text{days} = 3 half-lives, so fraction =(1/2)3=1/8=0.125= (1/2)^3 = 1/8 = 0.125. (3) (Or N/N0=e−λt=e−0.0866×24=e−2.079=0.125N/N_0 = e^{-\lambda t} = e^{-0.0866\times24} = e^{-2.079} = 0.125.)

Q8. (4 marks) Mass defect Δm\Delta m = difference between sum of constituent nucleon masses and actual nuclear mass. Binding energy = energy equivalent of mass defect (BE=Δm c2BE = \Delta m\,c^2), the energy to separate the nucleus. (2) BE=0.0304×931.5=28.3 MeVBE = 0.0304 \times 931.5 = 28.3\,\text{MeV}. (1) Per nucleon: 28.3/4=7.08 MeV/nucleon28.3/4 = 7.08\,\text{MeV/nucleon}. (1)

Q9. (3 marks) γ=1/1−0.992=1/0.0199=1/0.141=7.09\gamma = 1/\sqrt{1-0.99^2} = 1/\sqrt{0.0199} = 1/0.141 = 7.09. (2) t=γt0=7.09×2.2=15.6 μst = \gamma t_0 = 7.09 \times 2.2 = 15.6\,\mu\text{s}. (1)

Q10. (3 marks) Postulate 1: The laws of physics are the same in all inertial frames (principle of relativity). (1) Postulate 2: The speed of light in vacuum is constant (cc) for all observers, independent of source/observer motion. (1) Experiment: Michelson–Morley experiment — null result (no fringe shift, no detectable ether wind). (1)

[
  {"claim": "Q1 photoelectron max KE ≈ 0.83 eV", "code": "E=6.63e-34*3.00e8/400e-9/1.60e-19; K=E-2.28; result=abs(K-0.83)<0.02"},
  {"claim": "Q2 de Broglie wavelength ≈ 0.123 nm", "code": "import sympy as sp; p=sp.sqrt(2*9.11e-31*1.60e-19*100); lam=6.63e-34/p; result=abs(float(lam)-1.23e-10)<3e-12"},
  {"claim": "Q5 Balmer 3->2 wavelength ≈ 658 nm", "code": "dE=(-13.6/9-(-13.6/4))*1.60e-19; lam=6.63e-34*3.00e8/dE; result=abs(lam-6.58e-7)<1e-8"},
  {"claim": "Q7 fraction remaining after 24 days = 1/8", "code": "import sympy as sp; lam=sp.log(2)/8; frac=sp.exp(-lam*24); result=abs(float(frac)-0.125)<1e-6"},
  {"claim": "Q9 muon dilated lifetime ≈ 15.6 us", "code": "import sympy as sp; g=1/sp.sqrt(1-0.99**2); t=g*2.2; result=abs(float(t)-15.6)<0.2"}
]