(a) Calculate the energy released in the neutron-induced fission reaction\(n{ + ^{235}}U{ \to ^{92}}Kr{ + ^{142}}Ba + 2n\),

Given \(m{(^{92}}Kr) = 91.926269{\rm{ }}u\) and

\(m{(^{142}}Ba) = 141.916361{\rm{ }}u\).

(b) Confirm that the total number of nucleons and total charge are conserved in this reaction.

The power kept in the centre of an atom is known as nuclear energy. All stuff in the cosmos is made up of tiny particles called atoms. The centre of an atom's nucleus is typically where most of its mass is concentrated. Neutrons and protons are the two subatomic particles that make up the nucleus. Atomic bonds that hold them together carry a lot of energy.

The energy released in a neutron-induced fission reaction is\(E = \Delta m{c^2}\),where\(\Delta m\)is the difference between the mass of the parent nucleus plus themass of the neutron, minus the sum of the masses of the products.

\(n = 1.008665u\), the difference in mass is –

\(\begin{align}{}\Delta m & = \left[ {m\left( {^{235}U} \right) + m(n)} \right] - \left[ {m\left( {^{92}Kr} \right) + m\left( {^{142}Ba} \right) + 2m(n)} \right]\\ & = [235.043924u + 1.008665u] - [91.926269u + 141.916361u + 2(1.008665)u]\\ & = 0.19262911\end{align}\)

Hence, the energy is –

\(\begin{align}{}E &= 0.192629n\frac{{931.5\frac{{McV}}{{{c^2}}}}}{u}{c^2}\\ &= 179.4MeV\end{align}\)

Therefore, the value for energy is obtained as \(E = 179.4MeV\).

The total number of nucleons and the total charge are conserved since,

The number of nucleons on the right side and on the left side of the reaction are equal –

\(\begin{align}{}A &= {(235 + 1)_{left{\rm{ }}}}\\ &= {(92 + 142 + 2)_{right{\rm{ }}}}\\ &= 236\end{align}\)

The total charge on the right side and on the left side of the reaction are equal –

\(\begin{align}{}Z & = {(92 + 0)_{left{\rm{ }}}}\\ &= {(36 + 56 + 0)_{right{\rm{ }}}}\\ &= 92\end{align}\)

Therefore, all the quantities are conserved.