Nuclear Physics QWC practice question

OCR A A2 standard 6 marks 6.4.1(a) 6.4.1(b) 6.4.1(c) 6.4.1(d)

Question

The graph of binding energy per nucleon against nucleon number rises steeply, peaks near iron, and then falls slowly. Using the idea of binding energy per nucleon, explain why energy is released in both nuclear fusion and nuclear fission, and why iron is the most stable nucleus.

Student guidance

Write in clear, linked sentences using correct physics terms.

Expected physics points

A higher binding energy per nucleon means a more stable nucleus.
Iron lies at the peak of the curve, so it has the greatest binding energy per nucleon and is the most stable.
In fusion, light nuclei join to form a heavier nucleus with a higher binding energy per nucleon.
In fission, a heavy nucleus splits into lighter nuclei with a higher binding energy per nucleon.
In both cases the products are more tightly bound, so energy is released.
The released energy corresponds to the loss of mass (mass defect) via E = mc².

Mark scheme points

A higher binding energy per nucleon means a more stable nucleus.
Iron lies at the peak of the curve, so it has the greatest binding energy per nucleon and is the most stable.
In fusion, light nuclei join to form a heavier nucleus with a higher binding energy per nucleon.
In fission, a heavy nucleus splits into lighter nuclei with a higher binding energy per nucleon.
In both cases the products are more tightly bound, so energy is released.
The released energy corresponds to the loss of mass (mass defect) via E = mc².

Indicative content

A higher binding energy per nucleon means a more stable nucleus.
Iron lies at the peak of the curve, so it has the greatest binding energy per nucleon and is the most stable.
In fusion, light nuclei join to form a heavier nucleus with a higher binding energy per nucleon.
In fission, a heavy nucleus splits into lighter nuclei with a higher binding energy per nucleon.
In both cases the products are more tightly bound, so energy is released.
The released energy corresponds to the loss of mass (mass defect) via E = mc².

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