E: Atom, quantum physics, nuclear decay, fission, fusion & stars – IB Physics exam questions with mark scheme

1Core3 marks

A nucleus of uranium-235 is represented by ²³⁵₉₂U.

(a) State the number of protons and the number of neutrons in this nucleus. [2]

(b) Define the term isotopes. [1]

(a) 92 protons(1)
(a) 235 − 92 = 143 neutrons(1)
(b) Nuclei/atoms of the same element with the same number of protons (Z) but a different number of neutrons (accept: same proton/atomic number, different nucleon/mass number; do not accept “different number of protons” or “different mass” alone)(1)
Answer: (a) 92 protons, 143 neutrons; (b) same Z, different N
Examiner tip: Neutrons = A − Z; isotopes differ in neutrons, not in protons.

2Core4 marks

The energy of the n^th level of hydrogen is E_n = −13.6/n² eV.

(a) Explain why the emission spectrum of hydrogen consists of discrete lines. [2]

(b) Calculate the wavelength of the photon emitted when an electron falls from n = 3 to n = 2. [2]

(a) Electrons can only occupy discrete (fixed) energy levels(1)
(a) A photon is emitted when an electron drops between levels, with photon energy = difference in level energies, so only specific frequencies/wavelengths appear(1)
(b) ΔE = −1.51 − (−3.40) = 1.89 eV = 3.02 × 10⁻¹⁹ J(1)
(b) λ = hc/ΔE = 6.58 × 10⁻⁷ m (658 nm; allow 6.56–6.58 × 10⁻⁷ m)(1)
Answer: (a) discrete levels, photon energy = level difference; (b) 658 nm
Examiner tip: Convert eV to J before using λ = hc/E; say the photon energy equals the level difference.

3StandardHL only5 marks

The activity of a radioactive sample falls from 3.2 × 10⁵ Bq to 4.0 × 10⁴ Bq in 36 hours. Background radiation can be ignored.

(a) Determine the half-life of the sample. [2]

(b) Calculate the decay constant in s⁻¹. [1]

(a) Activity has fallen by a factor of 8 = 2³, so three half-lives passed(1)
(a) T½ = 36 ÷ 3 = 12 h(1)
(b) λ = ln2 ÷ (12 × 3600 s) = 1.6 × 10⁻⁵ s⁻¹ (1.60 × 10⁻⁵)(1)
(c) A = λN, so N₀ = A₀/λ = 3.2 × 10⁵ ÷ 1.6 × 10⁻⁵(1)
(c) N₀ = 2.0 × 10¹⁰ (allow 1.99 × 10¹⁰ from unrounded λ)(1)
Answer: (a) 12 h; (b) 1.6 × 10⁻⁵ s⁻¹; (c) 2.0 × 10¹⁰
Examiner tip: Half-life in hours must be converted to seconds before finding λ in s⁻¹.

4Standard4 marks

Radium-226 (²²⁶₈₈Ra) decays by alpha emission to radon (Rn).

(a) Complete the decay equation: ²²⁶₈₈Ra → Rn + α. Give the proton and nucleon numbers of both products. [2]

(b) Atomic masses: Ra-226 = 226.0254 u; Rn-222 = 222.0176 u; He-4 = 4.0026 u. Calculate the total energy released in the decay in MeV. (1 u = 931.5 MeV c⁻²) [2]

(a) Radon product written as ²²²₈₆Rn(1)
(a) Alpha particle written as ⁴₂He (or ⁴₂α)(1)
(b) Mass difference = 226.0254 − 222.0176 − 4.0026 = 0.0052 u (electron masses balance when atomic masses are used)(1)
(b) Energy = 0.0052 × 931.5 = 4.84 MeV (≈ 4.8 MeV)(1)
Answer: (a) ²²²₈₆Rn + ⁴₂He; (b) 4.8 MeV
Examiner tip: Both A and Z must balance (226 = 222 + 4; 88 = 86 + 2); then multiply the mass defect by 931.5.

5Standard5 marks

Masses: hydrogen-1 atom = 1.007825 u; neutron = 1.008665 u; helium-4 atom = 4.002603 u. (1 u = 931.5 MeV c⁻²)

(a) Define the binding energy of a nucleus. [1]

(b) Calculate the mass defect of helium-4. [2]

(d) Explain why fusing light nuclei to form helium releases energy. [1]

(a) The minimum energy required to separate a nucleus completely into its individual nucleons (equivalently the energy released when it forms from free nucleons)(1)
(b) Δm = 2(1.007825) + 2(1.008665) − 4.002603(1)
(b) Δm = 0.030377 u (accept 0.0304 u, or 5.04–5.05 × 10⁻²⁹ kg, or 28.3 MeV c⁻²)(1)
(c) BE = 0.030377 × 931.5 = 28.3 MeV, so BE per nucleon = 28.3 ÷ 4 = 7.07 MeV (accept answers in the range 7.07–7.08 MeV from rounding BE to 28.3 MeV first; allow 7.1 MeV; allow ECF from (b))(1)
(d) The product (helium) has a higher binding energy per nucleon than the reactant nuclei, so the total mass decreases and the mass difference is released as energy (E = Δmc²) (accept: total binding energy of the products is greater than that of the reactants; accept: total mass of the products is less than the total mass of the reactants, so energy is released; OWTTE)(1)
Answer: (a) minimum energy to separate a nucleus into its free nucleons; (b) 0.030377 u; (c) 7.07 MeV per nucleon; (d) helium has a higher binding energy per nucleon, so mass is lost and released as energy
Examiner tip: Use atomic (H-1) masses consistently so electron masses cancel, and divide by A for 'per nucleon'.

6Hard4 marks

The Sun has a luminosity of 3.83 × 10²⁶ W. Its energy comes from fusion, in which four protons form one helium nucleus, releasing 26.7 MeV.

(a) Calculate the number of helium nuclei formed each second. [2]

(b) Calculate the mass converted to energy each second. [2]

(a) Energy per fusion in joules: 26.7 × 10⁶ × 1.60 × 10⁻¹⁹ = 4.27 × 10⁻¹² J(1)
(a) Rate = L ÷ E = 8.97 × 10³⁷ s⁻¹ (allow 8.95 × 10³⁷ to 9.0 × 10³⁷)(1)
(b) Use ΔE = Δm c², so Δm/Δt = L/c²(1)
(b) 4.26 × 10⁹ kg s⁻¹ (allow 4.3 × 10⁹)(1)
Answer: (a) 8.97 × 10³⁷ s⁻¹; (b) 4.26 × 10⁹ kg s⁻¹
Examiner tip: Convert MeV to J (×1.60 × 10⁻¹³ per MeV) before dividing into the power.

7Hard5 marks

A star has a surface temperature of 3600 K and a luminosity of 4.0 × 10³⁰ W. Assume it radiates as a black body.

(a) Calculate the peak wavelength of its emission and identify the region of the spectrum. [2]

(b) Calculate the radius of the star. [2]

(c) Suggest what type of star this is, justifying your answer using its temperature and luminosity compared with the Sun (L = 3.83 × 10²⁶ W, T = 5800 K). [1]

(a) λ_max = b/T = 2.90 × 10⁻³ ÷ 3600 = 8.06 × 10⁻⁷ m(1)
(a) Infrared (just beyond the visible red end)(1)
(b) L = σAT⁴ with A = 4πR², so R = √(L/4πσT⁴)(1)
(b) R = 1.8 × 10¹¹ m(1)
(c) Red supergiant/giant – cooler than the Sun (red) yet ~10⁴ times more luminous, so it must have a far larger surface area(1)
Answer: (a) 8.06 × 10⁻⁷ m, infrared; (b) 1.8 × 10¹¹ m; (c) red giant/supergiant
Examiner tip: Use T in kelvin and raise it to the fourth power; the area of a sphere is 4πR².

8HardHL only5 marks

Light of wavelength 400 nm is incident on a sodium surface of work function 2.28 eV.

(a) Calculate the energy of one photon in eV. [1]

(b) Calculate the maximum kinetic energy of the emitted photoelectrons in eV. [2]

(c) The emitted photoelectrons travel towards a collecting electrode. A potential difference is applied that just prevents the fastest photoelectrons from reaching the collecting electrode, so the photocurrent falls to zero. State the value of this stopping potential difference. [1]

(d) The intensity of the light is doubled at constant wavelength. State the effect on the maximum kinetic energy and on the number of photoelectrons emitted per second. [1]

(a) E = hc/λ = 4.97 × 10⁻¹⁹ J = 3.11 eV (allow 3.10 eV)(1)
(b) Einstein's equation: hf = Φ + KEmax(1)
(b) KEmax = 3.11 − 2.28 = 0.83 eV (allow 0.82–0.83 eV)(1)
(c) Stopping potential = 0.83 V (numerically equal to KEmax in eV; allow ECF from (b); unit V required)(1)
(d) KEmax unchanged; the number of photoelectrons emitted per second doubles (accept “increases”) (both needed)(1)
Answer: (a) 3.11 eV; (b) 0.83 eV; (c) 0.83 V; (d) KEmax same, rate doubles
Examiner tip: Intensity changes the number of photons per second, not the energy of each photon.

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