Question

The bond energy per nucleon for deuteron and an $\alpha -particle$ are $X_1$ and $X_2$ respectively. The energy released in the reaction will be:

${}_1{H}^2+{}_1{H}^2\to {}_{2}H{e}^4$

• A. $X_2-X_1$
• B. $2\left(X_2-X_1\right)$
• C. $4\left(X_2-X_1\right)$
• D. $8\left(X_2-X_1\right)$

Answer 1

The correct option is C $4\left(X_2-X_1\right)$

$H_1^2+H_1^2\to H{e}_2^4+Q_2$

$\Delta H=\sum B{E}_{Product}-\sum B{E}_{Reactant}$

$=4X_2-2\times 2X_1$

$=4(X_2-X_1)$