Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Consider the reaction,
${ }_{1}^{2} H +{ }_{1}^{2} H \longrightarrow{ }_{2}^{3} He +{ }_{0}^{1} n$
$m\left({ }_{1}^{2} H \right)=2.014082\, u , m\left({ }_{2}^{3} He \right)=3.016029\, u , m\left({ }_{0}^{1} n\right)=1.008665\, u$ Then, mark the correct option.

Nuclei

Solution:

As, $m\left({ }_{1}^{2} H \right)=2.014082 u$
and $m\left({ }_{2}^{3} He \right)=3.016029$
and $m\left({ }_{0}^{1} n\right)=1.008665$
$\Delta m=$ mass defect
$=2 m\left({ }_{1}^{2} H \right)-m 3\left({ }_{2}^{3} He \right)-m\left({ }_{0}^{1} n\right)$
$=2 \times 2.014082-3.016029-1.008665$
$\Rightarrow Q$-value =Energy liberated in reaction
$=[\text { mass }(\text { reactants })-\text{mass}(\text { products })] \times 931.5$
$=\Delta m \times 931.5\, MeV =3.232\, MeV$
As $Q$-value is positive, so no threshold energy is required.