8.4 Nuclear energy and E = mc² — Practise

1. The energy released

Answer: $E = mc^2 = 2.0 \times 10^{-30} \times (3.00 \times 10^{8})^2 = 1.8 \times 10^{-13}\,\text{J}$.

2. The binding energy

Answer: The energy equivalent of one atomic mass unit is 931.5 MeV, so $E = 0.050 \times 931.5 = 46.6\,\text{MeV}$.

3. The energy released

Answer: $E = mc^2 = 3.0 \times 10^{-29} \times 9.00 \times 10^{16} = 2.7 \times 10^{-12}\,\text{J} = \dfrac{2.7 \times 10^{-12}}{1.60 \times 10^{-13}} = 16.9\,\text{MeV}$.

4. The binding energy and its value per nucleon

Answer: The binding energy is $0.0304 \times 931.5 = 28.3\,\text{MeV}$. The nucleus has 4 nucleons, so the binding energy per nucleon is $\dfrac{28.3}{4} = 7.08\,\text{MeV}$.

5. The mass defect

Answer: $E = 200 \times 10^{6} \times 1.60 \times 10^{-19} = 3.20 \times 10^{-11}\,\text{J}$, so $\Delta m = \dfrac{E}{c^2} = 3.6 \times 10^{-28}\,\text{kg}$.

6. The total energy

Answer: $E_{\text{total}} = 2 \times 0.511 = 1.022\,\text{MeV} = 1.022 \times 1.60 \times 10^{-13} = 1.64 \times 10^{-13}\,\text{J}$.

7. The photon energy and wavelength

Answer: Each photon carries $0.511\,\text{MeV} = 8.18 \times 10^{-14}\,\text{J}$. Then $\lambda = \dfrac{hc}{E} = \dfrac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{8.18 \times 10^{-14}} = 2.43 \times 10^{-12}\,\text{m}$.

8. The energy released

Answer: $E = mc^2 = 2.0 \times 10^{-29} \times 9.00 \times 10^{16} = 1.8 \times 10^{-12}\,\text{J} = 11.3\,\text{MeV}$.

9. Explain why energy is released when a very heavy nucleus undergoes fission.

• A very heavy nucleus has a lower binding energy per nucleon than a nucleus near the peak of the curve.
• Its fission products are closer to the peak, near iron-56.
• The products therefore have a higher binding energy per nucleon.
• A more tightly bound set of nucleons has less mass.
• The missing mass is released as the kinetic energy of the products, through $E = mc^2$.

10. Explain why energy is released when two light nuclei fuse.

• Two light nuclei have a low binding energy per nucleon, on the steep left-hand side of the curve.
• The heavier product is closer to the peak, so it has a higher binding energy per nucleon.
• The nucleons are more tightly bound in the product, so the product has less mass than the two nuclei together.
• The mass difference appears as energy, through $E = mc^2$.

11. The mass defect

Answer: To three significant figures the mass defect is 0.0304 u.

12. The percentage uncertainty

Answer: The energy is $E = mc^2$. The speed of light is exact, so the percentage uncertainty in the energy is the same as the mass: 4.0%. The absolute uncertainty is $3.0 \times 10^{-29} \times \dfrac{4.0}{100} = 1.2 \times 10^{-30}\,\text{kg}$.

13. The mass

Answer: $E = 938 \times 10^{6} \times 1.60 \times 10^{-19} = 1.50 \times 10^{-10}\,\text{J}$, so $m = \dfrac{E}{c^2} = \dfrac{1.50 \times 10^{-10}}{9.00 \times 10^{16}} = 1.67 \times 10^{-27}\,\text{kg}$.

14. A nuclear reaction has a mass defect of 0.020 u. State which equation you would use to find the energy released in joules, and why.

• Find the mass defect in kilograms, or use the energy equivalent of the atomic mass unit to get the energy in MeV first.
• Use $E = mc^2$ with the mass in kilograms.
• If you found the energy in MeV, change it to joules by multiplying by $1.60 \times 10^{-13}$.

Answer: Use $E = mc^2$ with the mass defect in kilograms, or find the energy in MeV from the 931.5 MeV per atomic mass unit and change it to joules.