A-LevelPhysicsNuclear physics9702/22 · Oct/Nov 2019 · Q74 Marks
A nucleus of plutonium-238 (²³⁸Pu) decays by emitting an α-particle to produce a new nucleus X and 5.6 MeV of energy. The decay is represented by ²³⁸Pu → X + α + 5.6 MeV.
- (a)Determine the number of protons and the number of neutrons in nucleus X. number of protons = number of neutrons =[2]
- (b)Calculate the number of plutonium-238 nuclei that must decay in a time of 1.0s to produce a power of 0.15W. number =[2]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.
- [A1] number of protons = 92
- [A1] number of neutrons = 142
- [C1] 5.6 MeV = 5.6 × 1.60 × 10-19 × 106 (= 8.96 × 10-13 J)
- [A1] number = 0.15 / (5.6 × 1.60 × 10-13) = 1.7 x 1011 OR
- [(C1)] 0.15 W = 0.15 / (1.60 × 10-19 × 106) (= 9.38 × 1011 MeV s-1)
- [(A1)] number = 9.38 × 1011/5.6 = 1.7 x 1011
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