Question

Nuclei of two elements $\text{P}$ and $\text{Q}$ have binding energies ${\text{E}}_P$ and ${\text{E}}_Q.$ If four nuclei of element $\text{Q}$ combine to give one $\text{P}$ nucleus (as a daughter nucleus) releasing $e$ units of energy along then

• A. $4E_Q+e=E_P$
• B. $E_P=4E_Q$
• C. $E_P+e=4E_Q$
• D. $E_Q=4E_P$

Answer 1

The correct option is A $4E_Q+e=E_P$

From the data given in the question, it is clear that, The stability of the daughter nucleus will be more than the parent nucleus.

As we know that, higher the stability higher will be the binding energy per nucleon.

So, $E_P>E_Q$,

here,

four nuclei of $\text{Q}$ combines to form one $\text{P}$ by releasing $e$ units of energy, $4Q\to P$

In terms of binding energy, $4E_Q+e=E_P$

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