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Question

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

A
Helium
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B
Deuterium
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C
Carbon
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D
Hydrogen
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Solution

The correct option is A Helium
Proton - Proton cycle :
The thermonuclear reactions involved are :
2(11H)+2(11H)2(21H)+2(01e)+Q1
2(21H)+2(11H)2(32He)+Q2
2(32He)(42He)+2(21H)+Q3
On adding up these reactions, we obtain
4(11H)42He+2(01e)+Q
where, Q=Q1+Q2+Q3 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.

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