Question

If four atoms of hydrogen combine to form an $2^{4}He$ atom, then the amount of energy released is :

• A. $26.7MeV$
• B. $25.3MeV$
• C. $22MeV$
• D. $24.5MeV$

Answer 1

Answer: (A)

$26.7MeV$

The proton-proton cycle by which this occurs is represented by the following sets of reactions
${}_1^{1}H{+}_1^{1}H{\to }_1^{2}H+e^{+}+v+0.42MeV$ .....(i)
$e^{+}+e^{-}\to \gamma +\gamma +1.02MeV$ ....(ii)
${}_1^{2}H{+}_1^{1}H{\to }_2^{3}He+\gamma +5.49MeV$ .....(iii)
${}_2^{3}He{+}_2^{3}He{\to }_2^{4}He{+}_1^{1}H{+}_1^{1}H+12.86MeV$ ....(iv)
For the fourth reaction to occur, the first three reactions must occur twice, if we consider the combination $2$(i) $+2$(ii) $+2$(iii) $+2$(iv), the net effect is
$4_1^{1}H+2e^{-1}{\to }_2^{4}He+2v+6v+26.7MeV$
Thus, four hydrogen atom combines to form an ${}_2^{4}He$ atom with a release of $26.7MeV$ of energy.