DP Physics (first assessment 2025)

Question 19N.2.SL.TZ0.7

[Maximum mark: 7]

19N.2.SL.TZ0.7

A stationary nucleus of uranium-238 undergoes alpha decay to form thorium-234.

The following data are available.

Energy released in decay 4.27 MeV
Binding energy per nucleon for helium 7.07 MeV
Binding energy per nucleon for thorium 7.60 MeV

(a)

Radioactive decay is said to be “random” and “spontaneous”. Outline what is meant by each of these terms.

Random:

Spontaneous:

[2]

Markscheme

random:
it cannot be predicted which nucleus will decay
OR
it cannot be predicted when a nucleus will decay ✔

NOTE: OWTTE

spontaneous:
the decay cannot be influenced/modified in any way ✔

NOTE: OWTTE

(b(i))

Calculate the binding energy per nucleon for uranium-238.

[3]

Markscheme

234 × 7.6 OR 4 × 7.07 ✔

BE_U=« 234 × 7.6 + 4 × 7.07 – 4.27 =» $1802$ « MeV » ✔

$\frac{B E_{U}}{A} = « \frac{1802}{238} = » 7.57$ « MeV » ✔

NOTE: Allow ECF from MP2
Award [3] for bald correct answer
Allow conversion to J, final answer is 1.2 × 10^–12

(b(ii))

Calculate the ratio $\frac{kinetic energy of alpha particle}{kinetic energy of thorium nucleus}$ .

[2]

Markscheme

states or applies conservation of momentum ✔

ratio is « $\frac{E_{k α}}{E_{kTh}} = \frac{\frac{p^{2}}{2 m_{α}}}{\frac{p^{2}}{2 m_{Th}}} = \frac{234}{4}$ » 58.5 ✔

NOTE: Award [2] for bald correct answer

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