question_answer

The binding energy of deuteron \[\begin{matrix}    2  \\    1  \\ \end{matrix}H\] is 1.112 Me V pen nucleon and an \[\alpha \]-particle \[\begin{matrix}    4  \\    2  \\ \end{matrix}He\] has a binding energy of 7.047 Me V per nucleon. Then in the fusion reaction \[\begin{matrix}    2  \\    1  \\ \end{matrix}He+\begin{matrix}    2  \\    1  \\ \end{matrix}H\to \begin{matrix}    4  \\    2  \\ \end{matrix}He+Q\], the energy Q released is

A) \[1\,MeV\]         

B) \[11.9\,MeV\]

C) \[23.8\,MeV\]    

D) \[931\,MeV\]