A summary of the nuclear reactions that power our sun is $4 \mathrm { p } \rightarrow 4 \mathrm { He } + 2 \mathrm { e } ^ { - }$, with masses of $938.272$ $\mathrm { MeV } / \mathrm { c } ^ { 2 }$ for a proton, $3727.38 \mathrm { MeV } / \mathrm { c } ^ { 2 }$ for the helium nucleus, and $0.511 \mathrm { MeV } / \mathrm { c } ^ { 2 }$ for each electron. How much energy is released by each of these reactions?
A) $28.3 \mathrm { MeV }$
B) $2790.13 \mathrm { MeV }$
C) $279.01 \mathrm { MeV }$
D) $24.69 \mathrm { MeV }$

Question

Quizplus · Accepted Answer

The energy released is calculated using the mass difference before and after the reaction, converted to energy via $E=mc^2$. The initial mass is $4 \times 938.272 \, \mathrm{MeV}/c^2$ for the protons. The final mass is $3727.38 \, \mathrm{MeV}/c^2$ for the helium nucleus plus $2 \times 0.511 \, \mathrm{MeV}/c^2$ for the electrons, totaling $3728.402 \, \mathrm{MeV}/c^2$. The mass difference is $3751.088 \, \mathrm{MeV}/c^2 - 3728.402 \, \mathrm{MeV}/c^2 = 22.686 \, \mathrm{MeV}/c^2$, which is not an option provided, indicating a calculation error in the explanation. Correcting this, the mass of 4 protons is $4 \times 938.272 = 3753.088 \, \mathrm{MeV}/c^2$. Subtracting the final mass from the initial gives $3753.088 - 3728.402 = 24.686 \, \mathrm{MeV}$, which rounds to $24.69 \, \mathrm{MeV}$, matching option D.

A Summary of the Nuclear Reactions That Power Our Sun 4p→4He+2e−4 \mathrm { p } \rightarrow 4 \mathrm { He } + 2 \mathrm { e } ^ { - }4p→4He+2e−

A Summary of the Nuclear Reactions That Power Our Sun $4 \mathrm { p } \rightarrow 4 \mathrm { He } + 2 \mathrm { e } ^ { - }$