Question

Assuming that four protons combine to form a helium nucleus and two positrons each of mass $0.000549a.m.u.$ , calculate the energy released. Given, the mass of ${}_1{H}^1=1.007825$ $a.m.u.$ and mas of ${}_{2}H{e}^4$ $=4.002603a.m.u.$

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

Answer 1

Answer: (A)

The nuclear fusion reaction can be written as:
${}_1{H}^1+{}_1{H}^1+{}_1{H}^1+{}_1{H}^1\to {}_{2}H{e}^4+2{}_1{e}^0+Q$
If $m_N\left({}_1{H}^1\right)$ and $m_N\left({}_{2}H{e}^4\right)$ represent the masses of ${}_1{H}^1$ and ${}_{2}H{e}^4$ nuclei respectively, then
$Q=\left[14m_N\left({}_1{H}^1\right)-m_N\left({}_{2}H{e}^4\right)-2m_e\right]\times 931.5MeV\dots \dots$ (i)
Where $m_e$ represents the mass of the positron $\left(1e^0\right)$.
Here, $m_N\left({}_{2}H{e}^4\right)4.002603$ a. $m.u.;m_e=0.0005499a.m.u$
Substituting for $m_N\left({}_{1}H{e}^1\right),m_N\left({}_{2}H{e}^4\right)$ and $m_e$ in the equation (i), we have
$Q=[4\times 1.007825-4.002603-2\times 0.0000549]\times 931.5$
$=0.027599\times 931.5=25.71MeV$