Question

The binding energies per nucleon are $5.3\text{MeV},6.2\text{MeV}$ and $7.4\text{MeV}$ for the nuclei with mass numbers $3,4$ and $5$ respectively. If one nucleus of mass number $3$ combines with one nucleus of mass number $5$ to give two nuclei of mass number $4$, then

• A. $3.3\text{MeV}$ of energy is absorbed.
• B. $0.3\text{MeV}$ of energy is released.
• C. $0.3\text{MeV}$ of energy is absorbed.
• D. $3.3\text{MeV}$ of energy is released.

Answer 1

The correct option is A $3.3\text{MeV}$ of energy is absorbed.

Let $X,Y$ and $Z$ be the nuclei of mass number $3,4$ and $5$ respectively.

The reaction be,

$X+Z=2Y$

B.E. of reactants,

$E_r=(3\times 5.3)+(5\times 7.4)=52.9\text{MeV}$

B.E. of Products,

$E_p=2\times 4\times 6.2=49.6\text{MeV}$

So, the energy absorbed in this reaction will be,

$\Delta E=52.9-49.6=3.3\text{MeV}$

Hence, $\left(\text{A}\right)$ is the correct answer.