A-LevelPhysicsNuclear physics9702/42 · Feb/Mar 2017 · Q128 Marks

A stationary nucleus of uranium-238 (²³⁸U) decays to form a nucleus of thorium-234 (²³⁴Th). An α-particle and a gamma-ray photon are emitted. The equation representing the decay is ²³⁸U → ²³⁴Th + ⁴He + γ 92 90 2 The masses of the nuclei are given in Fig. 12.1. nucleus mass/u uranium-238 238.05076 thorium-234 234.04357 helium-4 4.00260 Fig. 12.1

(a)Define the binding energy of a nucleus.[2]
(b)(i)State the relationship between the binding energies of the nuclei that is consistent with this reaction being energetically possible.[1]
(b)(ii)1.Calculate, for this reaction, the change, in u, of the mass,[1]
(b)(ii)2.the total energy, in J, released.[2]
(b)(iii)State and explain whether the energy of the gamma-ray photon is equal to the energy released in the reaction.[2]

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Mark scheme

Marking points from the official mark scheme (Feb/Mar 2017). The mark or mark type is shown in brackets.

[M1] (minimum) energy required/work done to separate the nucleons (in a nucleus)
[A1] to infinity
[M1] or energy released when nucleons come together (to form a nucleus)
[A1] from infinity
[B1] (total) binding energy of thorium and helium (nuclei) greater than binding energy of uranium (nucleus)
[A1] change in mass = 238.05076 – (234.04357 + 4.00260) = 4.59 x 10⁻³ u
[C1] E = mc²
[A1] = 4.59 × 10⁻³ × 1.66 × 10⁻²⁷ × (3.00 × 10⁸)² = 6.9 × 10⁻¹³ J
[C1] or 1u = 931 MeV
[A1] E = 4.59 × 10⁻³ × 931 × 10⁶ × 1.6 × 10⁻¹⁹ = 6.8 x 10⁻¹³ J
[M1] Th nucleus/He nucleus/product nucleus has kinetic energy
[A1] energy of gamma photon must be less than energy released

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