Question

The nucleus finally formed in fusion of the proton in a proton cycle is that of:

• A. Helium
• B. Carbon
• C. Deuterium
• D. Hydrogen

Answer 1

Answer: (A)

Proton - Proton cycle :
The thermonuclear reactions involved are :
$2{(}_1^{1}H)+2{(}_1^{1}H)\Rightarrow 2{(}_1^{2}H)+2{(}_1^{0}e)+Q_1$
$2{(}_1^{2}H)+2{(}_1^{1}H)\Rightarrow 2{(}_2^{3}He)+Q_2$
$2{(}_2^{3}He)\Rightarrow {(}_2^{4}He)+2{(}_1^{2}H)+Q_3$
On adding up these reactions, we obtain
$4{(}_1^{1}H){\Rightarrow }_2^{4}He+2{(}_1^{0}e)+Q$
where, $Q=Q_1+Q_2+Q_3$ is the total energy involved in the fusion of 4 hydrogen nuclei to form Helium nucleus. The value of Q as calculated from mass defect comes out to be 26.7 MeV.