Question

Find the energy released in the reaction $\_{1}^{}H_{}^{2}+\_{1}^{}H_{}^{2}\overset{}{ \rightarrow }\_{2}^{}He^{4}$ if binding energy per nucleon for deuteron and $\alpha $ particle are x₁ and x₂, respectively.

• A. $4\left(\right.x_{2}-x_{1}\left.\right)$
• B. $4\left(\right.x_{1}+x_{2}\left.\right)$
• C. $\left(\right.x_{1 }-x_{2}\left.\right)$
• D. $\left(\right.x_{1 }+x_{2}\left.\right)$

Answer 1

Answer: (A)

The binding energy on reactant side ( i.e., deutrons) is $4\times x_{1}=4x_{1}$
The binding energy on product side ( i.e., $\alpha $ -particle) is $4\times x_{2}=4x_{2}$
So, energy released $Q$ is $4x_{2}-4x_{1}=4\left(x_{2} - x_{1}\right)$