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Date	November 2018	Marks available	3	Reference code	18N.2.HL.TZ0.6
Level	Higher level	Paper	Paper 2	Time zone	0 - no time zone
Command term	Calculate	Question number	6	Adapted from	N/A

Question

$_{15}^{32} P$ ${}_{15}^{32}{\text{P}}$ is formed when a nucleus of deuterium ( $_{1}^{2} H$ ${}_{1}^{2}{\text{H}}$ ) collides with a nucleus of $_{15}^{31} P$ ${}_{15}^{31}{\text{P}}$ . The radius of a deuterium nucleus is 1.5 fm.

State how the density of a nucleus varies with the number of nucleons in the nucleus.

[1]

a.i.

Show that the nuclear radius of phosphorus-31 ( $_{15}^{31} P$ ${}_{15}^{31}{\text{P}}$ ) is about 4 fm.

[1]

a.ii.

State the maximum distance between the centres of the nuclei for which the production of $_{15}^{32} P$ ${}_{15}^{32}{\text{P}}$ is likely to occur.

[1]

b.i.

Determine, in J, the minimum initial kinetic energy that the deuterium nucleus must have in order to produce $_{15}^{32} P$ ${}_{15}^{32}{\text{P}}$ . Assume that the phosphorus nucleus is stationary throughout the interaction and that only electrostatic forces act.

[2]

b.ii.

$_{15}^{32} P$ ${}_{15}^{32}{\text{P}}$ undergoes beta-minus (β^–) decay. Explain why the energy gained by the emitted beta particles in this decay is not the same for every beta particle.

[2]

State what is meant by decay constant.

[2]

d.i.

In a fresh pure sample of $_{15}^{32} P$ ${}_{15}^{32}{\text{P}}$ the activity of the sample is 24 Bq. After one week the activity has become 17 Bq. Calculate, in s^–1, the decay constant of $_{15}^{32} P$ ${}_{15}^{32}{\text{P}}$ .

[3]

d.ii.

Markscheme

it is constant ✔

a.i.

R = $1 .20 \times 10^{- 15} \times 31^{\frac{1}{3}} = 3.8 \times 10^{- 15}$ ${\text{1}}{\text{.20}} \times {10^{ - 15}} \times {31^{\frac{1}{3}}} = 3.8 \times {10^{ - 15}}$ «m» ✔

Must see working and answer to at least 2SF

a.ii.

separation for interaction = 5.3 or 5.5 «fm» ✔

b.i.

energy required = $\frac{15 e^{2}}{4 π ε_{0} \times 5.3 \times 10^{- 15}}$ $\frac{{15{e^2}}}{{4\pi {\varepsilon _0} \times 5.3 \times {{10}^{ - 15}}}}$ ✔

= 6.5 / 6.6 ×10⁻¹³ OR 6.3 ×10⁻¹³«J» ✔

Allow ecf from (b)(i)

b.ii.

«electron» antineutrino also emitted ✔

energy split between electron and «anti»neutrino ✔

probability of decay of a nucleus ✔

the fraction of the number of nuclei that decay

in one/the next second

per unit time ✔

d.i.

1 week = 6.05 × 10⁵ «s»

17 = $24 e^{- λ \times 6.1 \times 10^{5}}$ $24{{\text{e}}^{ - \lambda \times 6.1 \times {{10}^5}}}$ ✔

5.7 × 10⁻⁷«s^–1» ✔

Award [2 max] if answer is not in seconds

If answer not in seconds and no unit quoted award [1 max] for correct substitution into equation (MP2)

d.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

d.i.

[N/A]

d.ii.

Syllabus sections

Additional higher level (AHL) » Topic 12: Quantum and nuclear physics » 12.2 – Nuclear physics

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Additional higher level (AHL) » Topic 12: Quantum and nuclear physics

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