Question

Bombardment of lithium with protons gives rise to the following reaction.
${}_3{\text{H}}^7{+}_1{\text{H}}^1\to 2{(}_2{\text{He}}^4)+Q$.
Find the $Q$ value of the reaction.

The atomic masses of lithium, proton and helium are $7.016u,1.008u$ and $4.004u$ respectively.

• A. $16.904\text{MeV}$
• B. $14.904\text{MeV}$
• C. $12.904\text{MeV}$
• D. $10.904\text{MeV}$

Answer 1

The correct option is B $14.904\text{MeV}$

Total mass of reactants,

$M_R=M_{\text{Li}}+M_p$

$\Rightarrow M_R=7.016u+1.008u$

$\Rightarrow M_R=8.024u$

Total mass of products,

$M_{\text{product}}=2\left(M_{\text{He}}\right)$

$\Rightarrow M_{\text{product}}=2\times 4.004u=8.008u$

Mass defect, $\Delta m=M_R-M_{\text{product}}$

$\Rightarrow \Delta m=8.024u-8.008u$

$\Rightarrow \Delta m=0.016u$

$Q=\Delta m\times c^2=0.016u{c}^2$

$\Rightarrow Q=0.016\times 931.5\text{MeV}$

$\Rightarrow Q=14.904\text{MeV}$

Hence, option $\left(B\right)$ is correct.