1. Interpret the given masses

The parent nucleus has mass $M+\Delta m$ and is initially at rest.

It decays into two identical daughter nuclei, each of mass $\frac{M}{2}.$ So total mass of products is $\frac{M}{2}+\frac{M}{2}=M.$

Since the initial mass is greater than the final mass by $\Delta m$, the decay releases energy $Q = \Delta m\,c^2.$

Because the parent was initially at rest, this released energy appears as kinetic energy of the two daughter nuclei. Thus the decay is energetically possible only if $\Delta m>0.$ Hence, $M+\Delta m > M.$

2. Relate nuclear mass and binding energy

For any nucleus, $m_{\text{nucleus}} = Zm_p + Nm_n - \frac{B}{c^2},$ where $B$ is the total binding energy.

Greater binding energy means smaller nuclear mass.

Let the parent nucleus have mass number $A$. Then each daughter nucleus has mass number $A/2$.

If binding energy per nucleon of parent is $E_1$, then total binding energy of parent is $B_1 = AE_1.$

If binding energy per nucleon of each daughter is $E_2$, then total binding energy of one daughter is $B_2 = \frac{A}{2}E_2.$ So total binding energy of both daughters is $2B_2 = 2\left(\frac{A}{2}E_2\right)=AE_2.$

3. Compare parent and daughter total binding energies

The daughters have smaller total mass than the parent: $M < M+\Delta m.$

For the same total numbers of protons and neutrons before and after decay, this means the final system is more tightly bound than the initial system. Therefore, $\text{total binding energy of daughters} > \text{binding energy of parent}.$ That is, $AE_2 > AE_1.$ Cancelling $A$, $E_2 > E_1.$

4. Check options

• A: $E_2 = 2E_1$ — not necessarily true.
• B: $E_1 > E_2$ — false.
• C: $E_2 > E_1$ — true.
• D: $E_1 = 2E_2$ — false.

5. Final answer

The correct option is $\boxed{C}$ with $\boxed{E_2 > E_1}.$