A-LevelPhysicsNuclear physics9702/42 · May/June 2017 · Q1210 Marks

One nuclear reaction that can take place in a nuclear reactor may be represented, in part, by the equation ²³⁵₉₂U + ¹₀n → ⁹⁵₄₂Mo + ¹³⁹₅₇La + 2¹₀n + ______ + energy Data for a nucleus and some particles are given in Fig. 12.1. [Figure 12.1]

(a)Complete the nuclear reaction shown above.[1]
(b)(i)Show that the energy equivalent to 1.00 u is 934 MeV.[3]
(b)(ii)Calculate the binding energy per nucleon, in MeV, of lanthanum-139 (¹³⁹La). binding energy per nucleon = .............................................. MeV [3][3]
(c)State and explain whether the binding energy per nucleon of uranium-235 (²³⁵U) will be greater, equal to or less than your answer in (b)(ii).[3]

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Mark scheme

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

[A1] ⁷_(-1)e
[C1] E = mc²
[M1] = 1.66 × 10⁻²⁷ × (3.00 × 10⁸)²
[A1] = 1.494 × 10⁻¹⁰ J division by 1.60 × 10⁻¹³ clear to give 934 MeV
[C1] ∆m = (82 × 1.00863u) + (57 × 1.00728u) – 138.955u
[C1] = (-) 1.16762 (u) energy = 1.16762 × 934 energy per nucleon = (1.16762 × 934)/139
[A1] = 7.85 MeV
[B1] above A = 56, binding energy per nucleon decreases as A increases
[M1] U-235 has larger nucleon number
[A1] so less (binding energy per nucleon)
(B1) fission takes place with uranium
(M1) fission reaction releases energy
(A1) binding energy per nucleon less (for uranium than for products)

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