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Date	November 2019	Marks available	1	Reference code	19N.3.hl.TZ0.18
Level	HL	Paper	3	Time zone	TZ0
Command term	State	Question number	18	Adapted from	N/A

Question

Red supergiant stars contain carbon-12 formed by the fusion of helium-4 nuclei with beryllium-8 nuclei.

Mass of a helium-4 nucleus = 4.002602 amu
Mass of a beryllium-8 nucleus = 8.005305 amu
Mass of a carbon-12 nucleus = 12.000000 amu

State the nuclear equation for the fusion reaction.

[1]

a(i).

Explain why fusion is an exothermic process.

[2]

a(ii).

Calculate the heat energy released, in J, by the fusion reaction producing one atom of carbon-12. Use section 2 of the data booklet and E = mc².

[3]

a(iii).

Beryllium-8 is a radioactive isotope with a half-life of 6.70 × 10⁻¹⁷ s.

Calculate the mass of beryllium-8 remaining after 2.01 × 10⁻¹⁶ s from a sample initially containing 4.00 g of beryllium-8.

[2]

b.

Markscheme

${}_2^4{\text{He + }}{}_4^8{\text{Be}} \to {}_6^{12}{\text{C}}$ ✔

NOTE: Do not penalize missing atomic numbers.

a(i).

ALTERNATIVE 1
binding energy per nucleon is larger in carbon-12/product «than beryllium-8 and helium-4/reactants» ✔

difference in «total» binding energy is released «during fusion» ✔

ALTERNATIVE 2
mass of carbon-12/product «nucleus» is less than «the sum of» the masses of helium-4 and beryllium-8 «nuclei»/reactants
OR
two smaller nuclei form a lager nucleus ✔

mass lost/difference is converted to energy «and released»
OR
E = mc² ✔

a(ii).

Δm = «12.000000 amu − (4.002602 amu + 8.005305 amu) =» −0.007907 «amu» ✔
«0.007907 amu × 1.66 × 10⁻²⁷ kg amu⁻¹ =» 1.31 × 10⁻²⁹ «kg» ✔
«E = mc² = 1.31 × 10⁻²⁹ kg × (3.00 × 108 m s⁻¹)² =» 1.18 × 10⁻¹² «J» ✔

NOTE: Accept “0.007907 «amu»”.
Award [2 max] for “7.12 x 10¹⁴ «J»”.
Award [3] for correct final answer.

a(iii).

ALTERNATIVE 1
3 half-lives ✔
0.500 g «of beryllium-8 remain» ✔

ALTERNATIVE 2
$m = 4.00{\left( {\frac{1}{2}} \right)^{\frac{{2.01 \times {{10}^{ - 16}}}}{{6.70 \times {{10}^{ - 17}}}}}}$ ✔
0.500 g «of beryllium-8 remain» ✔

ALTERNATIVE 3
λ = « $\frac{{\ln 2}}{{6.70 \times {{10}^{ - 17}}}}$ »= 1.03 × 10¹⁶ «s⁻¹» ✔
m = « $4.00{\text{ }}{{\text{e}}^{ - 1.03 \times {{10}^{16}} \times 2.01 \times {{10}^{ - 16}}}}{\text{ = }}$ » 0.500 «g» ✔

NOTE: Award [2] for correct final answer.

b.

Examiners report

[N/A]

a(i).

[N/A]

a(ii).

[N/A]

a(iii).

[N/A]

b.

Syllabus sections

Options » C: Energy » C.3 Nuclear fusion and fission

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Nuclear disasters release radioactive caesium into the atmosphere, which presents serious health risks.

Cs-137 has a half-life of 30 years.

Calculate the percentage of Cs-137 remaining in the atmosphere after 240 years.
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The half-life of ²³⁸U is $4.46 \times 10^{9}$ years. Calculate the mass of ²³⁸U that remains after $0.639 kg$ has decayed for $2.23 \times 10^{10}$ years.

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