Question

The energy of the reaction ${\text{Li}}^7+p\to 2{\text{He}}^4$ is
(the binding energy per nucleon of ${\text{Li}}^7$ and ${\text{He}}^4$ nuclei are $5.60\text{MeV}$ and $7.06\text{MeV}$ respectively).

• A. $17.3\text{MeV}$
• B. $1.73\text{MeV}$
• C. $1.46\text{MeV}$
• D. Depends on binding energy of the proton

Answer 1

The correct option is A $17.3\text{MeV}$

Given,
${\text{BE}}_{\text{Li}}=5.60\text{MeV};{\text{BE}}_{\text{He}}=7.06\text{MeV}$

The energy of reaction means $Q-\text{value of the reaction}.$

$Q-\text{value}=\text{BE of products}-\text{B.E of reactants}$

$=2{\text{BE}}_{\text{He}}-{\text{BE}}_{\text{Li}}$

$=2\times (7.06\times 4)-(5.60\times 7)$

$=17.28\text{MeV}\approx 17.3\text{MeV}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.