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

Answer: (B)

c-a-b

Answer 2

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