Question

Assuming that four protons combine to form a helium nucleus and two positrons each of mass $0.000549 \, a.m.u.$ , calculate the energy released. Given, the mass of $_{1}H^{1}=1.007825 \, $ $a.m.u.$ and mas of $_{2}He^{4}$ $=4.002603 \, a.m.u.$

• A. $25.71MeV$
• B. $20.71MeV$
• C. $22.75MeV$
• D. $23.50MeV$

Answer 1

Answer: (A)

The nuclear fusion reaction can be written as:
${ }_{1} H ^{1}+{ }_{1} H ^{1}+{ }_{1} H ^{1}+{ }_{1} H ^{1} \rightarrow{ }_{2} He ^{4}+2{ }_{1} e ^{0}+ Q$
If $m_{N}\left({ }_{1} H ^{1}\right)$ and $m_{N}\left({ }_{2} He ^{4}\right)$ represent the masses of ${ }_{1} H ^{1}$ and ${ }_{2} He ^{4}$ nuclei respectively, then
$Q=\left[14 m_{N}\left({ }_{1} H ^{1}\right)-m_{N}\left({ }_{2} He ^{4}\right)-2 m_{e}\right] \times 931.5 MeV \ldots \ldots$ (i)
Where $m_{e}$ represents the mass of the positron $\left(1 e^{0}\right)$.
Here, $m_{N}\left({ }_{2} He ^{4}\right) 4.002603$ a. $m . u . ; m_{e}=0.0005499 a . m . u$
Substituting for $m_{N}\left({ }_{1} He ^{1}\right), m_{N}\left({ }_{2} He ^{4}\right)$ and $m_{e}$ in the equation (i), we have
$Q=[4 \times 1.007825-4.002603-2 \times 0.0000549] \times 931.5$
$=0.027599 \times 931.5=25.71 MeV$