DP Physics (first assessment 2025)

Question SPM.2.HL.TZ0.d

[Maximum mark: 3]

SPM.2.HL.TZ0.d

The initial activity of the source is 42 kBq. 33% of the alpha particles emitted by this source enter the chamber and form an ion pair.

Each alpha particle has an initial kinetic energy of 5.5 MeV.

The energy required to form one ion pair is 15 eV.

Calculate the maximum current in the chamber due to the electrons when there is no smoke in the chamber.

[3]

Markscheme

Each alpha gives rise to $\frac{5.5 \times 10^{6}}{15} = 3.67 \times 10^{5}$ ion pairs ✓

So $\frac{3.67 \times 10^{5} \times 42000}{3} = 5.13 \times 10^{9}$ ion pairs per second ✓

current $= 1.6 \times 10^{- 19} \times 5.13 \times 10^{9} = 0.82 \times 10^{- 9}$ «A» ✓

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