A-LevelPhysicsNuclear physics9702/41 · May/June 2017 · Q129 Marks

One possible nuclear reaction that takes place in a nuclear reactor is given by the equation ²³⁵₉₂U + ¹₀n → ¹³⁹₅₇La + ⁹⁵₄₂Mo + 2¹₀n + x⁰₋₁e Data for the nuclei and particles are given in Fig. 12.1. [Figure 12.1]

(a)Determine, for this nuclear reaction, the value of x.[1]
(b)(i)Show that the energy equivalent to 1.00 u is 934 MeV.[3]
(b)(ii)Calculate the energy, in MeV, released in this reaction. Give your answer to three significant figures.[3]
(c)Suggest the forms of energy into which the energy calculated in (b)(ii) is transformed.[2]

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Mark scheme

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[A1] x = 7
[C1] E = mc²
[C1] = 1.66 × 10⁻²⁷ × (3.0 × 10⁸)²
[C1] = 1.494 × 10⁻¹⁰ J
[A1] division by 1.6 × 10⁻¹³ clear to give 934 MeV
[C1] ∆m = (235.123 + 1.00863) – (94.945 + 138.955 + 2 × 1.00863 + 7 × 5.49 × 10⁻⁴)
[C1] or ∆m = 235.123 – (94.945 + 138.955 + 1 × 1.00863 + 7 × 5.49 × 10⁻⁴)
[C1] = 0.21053 u
[A1] energy = 0.21053 × 934
[A1] = 197 MeV
[B1] kinetic energy of nuclei/particles/products/fragments
[B1] γ-ray photon energy

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