Question

In the reaction \(^2_1H+^3_1H\) \(\rightarrow\) \(^4_2He+^1_0n,\) if the binding energies of \(^2_1H\), \(^3_1H\) and \(^4_2He\) are respectively a, b and c (in MeV), then the energy (in MeV) released in this reaction is:

• A. c+a-b
• B. c-a-b
• C. a+b+c
• D. a+b-c

Answer 1

The Correct Option is B

we have \(^2_1H+^3_1H\) \(\rightarrow\) \(^4_2He+^1_0n,\)
Energy released, E = (△m)×931MeV
atomic mass unit = 931MeV
\(\triangle\)m=mass of product− mass of reactant
\(\triangle\)m=c−a−b
E=(\(\triangle\)m)×931
i. e ,E =(c−a−b)
The binding energies of \(^2_1H, ^3_1H\) and \(^4_2He\) are respectively a,b and c

Therefore, the correct answer is (B): c-a-b