91Ó°ÊÓ

1. The half-life ${\mathrm{of}}^{137}\mathrm{Cs},{\mathrm{T}}_{1/2^{(137}\mathrm{Cs})}=30.2\mathrm{y}$,
2. The cesium isotope ${}^{137}\mathrm{Cs}$is decayed into barium isotope${}^{137}\mathrm{Ba}$.
3. The atomic mass of cesium isotope${}^{137}\mathrm{Cs}.{\mathrm{m}}_{\left({137}_{\mathrm{Cs}}\right)}=136.9071\mathrm{u}$ ,
4. The atomic mass of the barium isotope${}^{137}\mathrm{Ba}.{\mathrm{m}}_{\left({137}_{\mathrm{Ba}}\right)}=136.9058\mathrm{u}$ ,

The disintegration energy, also known as the Q-value, is the energy that is absorbed or released when a nuclear reaction takes place. Thus, we can consider the energy released as the disintegration energy in the decay process of the cesium to barium nucleus.

Formula:

The disintegration energy of a nuclear reaction,

$\mathbf{Q}\mathbf{=}\left(m_{\mathrm{parent}\mathrm{nucleus}}-\left(\underset{}{∑}{m}_{\mathrm{daughter}\mathrm{nuclei}}\right)\right){\mathbf{c}}^{\mathbf{2}}$ (i)

Let $m_{\left(137Cs\right)}$be the mass of one atom of ${}_{55}^{137}\mathrm{Cs}$and ${\mathrm{m}}_{\left(137\mathrm{Ba}\right)}$be the mass of one atom of ${}_{55}^{137}Ba$. The energy released is the change in mass due to the decay and can be given using equation (i) considering the disintegration energy as follows:

$$\begin{gathered} \mathrm{Q}=\left({\mathrm{m}}_{\left({137}_{\mathrm{Cs}}\right)}-{\mathrm{m}}_{\left({137}_{\mathrm{Ba}}\right)}\right){\mathrm{c}}^2 \\ =\left(136.9071\mathrm{u}-136.9058\mathrm{u}\right)\left(931.5\mathrm{MeV}/\mathrm{u}\right) \\ =\left(0.0013\mathrm{u}\right)\left(931.5\mathrm{MeV}/\mathrm{u}\right) \\ =1.21\mathrm{MeV} \end{gathered}$$

Hence, the total energy released is 1.21 MeV .