Rest Mass Energy Comparison in Nuclear Fission

To determine the correct relation among the rest mass energies, consider the nuclear fission reaction:
${ }_{92}^{236} \mathrm{U} \longrightarrow{ }_{53}^{137} \mathrm{I}+{ }_{39}^{97} \mathrm{Y}+2{ }_{0}^{1} \mathrm{n}$

Let us verify the conservation of nucleon number $A$ and atomic number $Z$:

• Atomic number (charge):
  $Z_{\text{initial}} = 92$
  $Z_{\text{final}} = 53 + 39 + 2(0) = 92$
• Mass number (nucleons):
  $A_{\text{initial}} = 236$
  $A_{\text{final}} = 137 + 97 + 2(1) = 236$

The reaction represents a possible fission mode of uranium-236. Nuclear fission is an exothermic process, meaning it releases energy ($Q > 0$) primarily in the form of kinetic energy of the daughter nuclei and emitted neutrons.

The $Q$-value of the nuclear reaction is defined as the difference between the rest mass energy of the reactants and the rest mass energy of the products:
$Q = E\left({ }_{92}^{236} \mathrm{U}\right) - \left[ E\left({ }_{53}^{137} \mathrm{I}\right) + E\left({ }_{39}^{97} \mathrm{Y}\right) + 2 E(n) \right]$

Since the process releases energy, $Q > 0$:
$E\left({ }_{92}^{236} \mathrm{U}\right) - \left[ E\left({ }_{53}^{137} \mathrm{I}\right) + E\left({ }_{39}^{97} \mathrm{Y}\right) + 2 E(n) \right] > 0$

$\implies E\left({ }_{92}^{236} \mathrm{U}\right) > E\left({ }_{53}^{137} \mathrm{I}\right) + E\left({ }_{39}^{97} \mathrm{Y}\right) + 2 E(n)$

Similarly, for the fission reaction yielding barium and krypton:
${ }_{92}^{236} \mathrm{U} \longrightarrow{ }_{56}^{140} \mathrm{Ba}+{ }_{36}^{94} \mathrm{Kr}+2{ }_{0}^{1} \mathrm{n}$
The $Q$-value is also strictly positive, which requires:
$E\left({ }_{92}^{236} \mathrm{U}\right) > E\left({ }_{56}^{140} \mathrm{Ba}\right) + E\left({ }_{36}^{94} \mathrm{Kr}\right) + 2 E(n)$
Thus, options (B), (C), and (D) are incorrect.

Therefore, the correct option is A.