Question

$\mathop {92U}\nolimits^{235}$undergoes nuclear fission as follows $n + \mathop {92U}\nolimits^{235} \to \mathop {42MO}\nolimits^{98} \;\; + \mathop {54Xe}\nolimits^{136} + 2n$
Energy released in the fission of 1gm of $\mathop {92U}\nolimits^{235}$ is ( masses of n , $\mathop {92U}\nolimits^{235}$ , $\mathop {42MO}\nolimits^{98}$ and $\mathop {54Xe}\nolimits^{136}$ are $1.0087{\text{ }},{\text{ }}235.0439{\text{ }},{\text{ }}97.{\text{ }}9054{\text{ }}$ and ${\text{ }}135.9170$ all in amu respectively).
(A) $\mathop {5.06 \times 10}\nolimits^{23} \mathop J\nolimits_{}$
(B) $\mathop {5.06 \times 10}\nolimits^{26} \mathop J\nolimits_{}$
(C) $\mathop {8.1 \times 10}\nolimits^7 \mathop J\nolimits_{}$
(D) $\mathop {8.1 \times 10}\nolimits^{10} \mathop J\nolimits_{}$

Answer 1

Hint
Nuclear fission occurs usually with heavier elements. In this process subdivision of a heavy nucleus occurs into two or more fragments of roughly equal masses. The process is completed by releasing a large amount of energy. The heavy element can be that of uranium or protium. In this heavy nucleus which is unstable dissociated into two light nuclei.

Complete step by step solution
Mass defect during the fission of $1$ Uranium atom is:
$Mass{\text{ }}of{\text{ }}reactant{\text{ }}-{\text{ }}mass{\text{ }}of{\text{ }}product{\text{ }}side\;$
$U + n - \left( {MO + Xe + 2n} \right) = \Delta m$
$\Delta m = 235.0439 + 1.0087 - \left( {97.9054 + 135.9170 + 2 \times 1.0087} \right)$
\begin{array}{*{20}{l}} { = 0.2128amu} \\\ { = {\text{ }}\mathop {1.67 \times 10}\nolimits^{ - 27} \mathop {kg}\nolimits_{} } \end{array}
Energy given by Einstein = $\mathop {\Delta m}\nolimits_c^2$
$\mathop {1.67 \times 10}\nolimits^{ - 27} \mathop {kg}\nolimits_{} \times \mathop {\mathop {(3 \times 10}\nolimits^8 )}\nolimits^2$
$\mathop {3.918 \times 10}\nolimits^{ - 11} \mathop {joule}\nolimits_{}$
No. of atoms in $1gm$ =
$\dfrac{{\mathop {6.022 \times 10}\nolimits^{23} }}{{235}} = \mathop {0.0256 \times 10}\nolimits^{23}$
Now , energy released in 1 gm will be = $\mathop {0.0256 \times 10}\nolimits^{23} \times \mathop {3.918 \times 10}\nolimits^{ - 11} \mathop {joule}\nolimits_{}$
$= \mathop {8.18 \times 10}\nolimits^8 {\text{joule }}.$
Now let us match this value with given options:
Option A: this value does not match with the calculated value. Thus, this option is not correct.
Option B: this value does not match with the calculated value. Thus, this option is not correct.
Option C: this value does not match with the calculated value. Thus, this option is not correct.
Option D: this value exactly matches the calculated value. Thus, this option is correct.
Our required answer is (D) that is $= \mathop {8.18 \times 10}\nolimits^8 {\text{joule }}.$

Note
Nuclear fission reaction is usually defined as splitting of heavier nuclei into lighter ones. The mass defect in such reactions is the difference of sum of reactants and sum of products. To initiate the nuclear fission the atom is bombarded with the neutron to generate isotope which undergoes fission.