Question 156: Atoms & Nuclei

Question

The energy released in the fusion of $$2 \mathrm{~kg}$$ of hydrogen deep in the sun is $$E_H$$ and the energy released in the fission of $$2 \mathrm{~kg}$$ of $${ }^{235} \mathrm{U}$$ is $$E_U$$. The ratio $$\frac{E_H}{E_U}$$ is approximately:
(Consider the fusion reaction as $$4_1^1H+2 \mathrm{e}^{-} ightarrow{ }_2^4 \mathrm{He}+2 v+6 \gamma+26.7 \mathrm{~MeV}$$, energy released in the fission reaction of $${ }^{235} \mathrm{U}$$ is $$200 \mathrm{~MeV}$$ per fission nucleus and $$\mathrm{N}_{\mathrm{A}}= 6.023 	imes 10^{23})$$

Accepted Answer

To determine the ratio $$\frac{E_H}{E_U}$$, let's first consider the energy released in the fusion of hydrogen and the energy released in the fission of $$^{235}U$$.
For the fusion reaction:
The given reaction is:
$$4_1^1H + 2 \mathrm{e}^{-} ightarrow {}_2^4 \mathrm{He} + 2 \mathrm{
u} + 6 \gamma + 26.7 \, \mathrm{MeV}$$
This reaction shows that 4 hydrogen atoms and 2 electrons produce 1 helium atom, 2 neutrinos, and 6 gamma photons while releasing 26.7 MeV of energy.
The mass of 1 mole of $$H_1^1$$ is approximately 1 gram, so 2 kg of hydrogen is equal to:
$$2 	imes 10^3 \, \mathrm{g}$$
The number of moles of hydrogen in 2 kg is:
$$\mathrm{N}_	ext{moles} = \frac{2 	imes 10^3}{1} = 2 	imes 10^3 \, \mathrm{moles}$$
The number of hydrogen atoms in 2 kg is:
$$\mathrm{N}_	ext{atoms} = N_A 	imes 2 	imes 10^3 = 6.023 	imes 10^{23} 	imes 2 	imes 10^3 = 1.2046 	imes 10^{27} \, \mathrm{atoms}$$
Since 4 hydrogen atoms release 26.7 MeV, the total energy released ($$E_H$$) is:
$$E_H = \frac{1.2046 	imes 10^{27}}{4} 	imes 26.7 \, \mathrm{MeV}$$
$$E_H = 3.0115 	imes 10^{26} 	imes 26.7 \, \mathrm{MeV}$$
$$E_H \approx 8.04 	imes 10^{27} \, \mathrm{MeV}$$
For the fission reaction:
The energy released per fission of one $$^{235}U$$ nucleus is 200 MeV.
The mass of 1 mole of $$^{235}U$$ is approximately 235 grams, so 2 kg of $${}^{235}U$$ is equal to:
$$2 	imes 10^3 \, \mathrm{g}$$
The number of moles of $$^{235}U$$ in 2 kg is:
$$\mathrm{N}_	ext{moles} = \frac{2 	imes 10^3}{235} \approx 8.51 \, \mathrm{moles}$$
The number of $$^{235}U$$ nuclei is:
$$\mathrm{N}_	ext{nuclei} = N_A 	imes 8.51 \approx 6.023 	imes 10^{23} 	imes 8.51 \approx 5.12 	imes 10^{24} \, \mathrm{nuclei}$$
The total energy released ($$E_U$$) is:
$$E_U = 200 \, \mathrm{MeV} 	imes 5.12 	imes 10^{24} \approx 1.024 	imes 10^{27} \, \mathrm{MeV}$$
Finally, the ratio $$\frac{E_H}{E_U}$$ is:
$$\frac{E_H}{E_U} \approx \frac{8.04 	imes 10^{27}}{1.024 	imes 10^{27}} \approx 7.85$$
Therefore, the ratio is closest to option A: 7.62.

Answer

7.62

Answer

25.6

Answer

9.13

Answer

15.04

Atoms & Nuclei

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