A-LevelPhysicsNuclear physics9702/23 · Oct/Nov 2019 · Q77 Marks
A stationary nucleus of a radioactive isotope X decays by emitting an α-particle to produce a nucleus of neptunium-237 and 5.5MeV of energy. The decay is represented by X → ⁹³₂³⁷Np + α + 5.5MeV.
- (a)Calculate the number of protons and the number of neutrons in a nucleus of X.[2]
- (b)Explain why the energy transferred to the α-particle as kinetic energy is less than the 5.5MeV of energy released in the decay process.[1]
- (c)(i)Calculate the average current produced by the beam of α-particles.[2]
- (c)(ii)Determine the total power, in W, that is produced by the decay of 6.9 × 10¹¹ nuclei of X in a time of 30 s.[2]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.
- [1] number of protons = 95
- [1] number of neutrons = 146
- [1] Np/neptunium (nucleus) has kinetic energy or gamma/γ-radiation produced
- [1] I = NQ/t
- [1] I = (6.9 × 1011 × 2 × 1.60 × 10-19)/30 = 7.4 x 10-9 A
- [1] P = (6.9 × 1011 × 5.5 × 106 × 1.60 × 10-19)/30
- [1] = 0.020 W
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