Question

Consider the nuclear reaction,

${\text{X}}^{200}\to {\text{A}}^{110}+{\text{B}}^{90}$

If the binding energy per nucleon for $\text{X},\text{A}$ and $\text{B}$ is $7.4\text{MeV},8.2\text{MeV}$ and $8.2\text{MeV}$ respectively, what is the energy released?

• A. $200\text{MeV}$
• B. $90\text{MeV}$
• C. $110\text{MeV}$
• D. $160\text{MeV}$

Answer 1

The correct option is D $160\text{MeV}$

Given,
$\text{Binding energy of X}=7.4\text{MeV}\text{Binding energy of A}=8.2\text{MeV}\text{Binding energy of B}=8.2\text{MeV}$

Energy released $Q$ = BE of Products − BE of Reactants

$=(8.2\times 110+8.2\times 90)-(7.4\times 200)$

$=1640-1480=160\text{MeV}$

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