Consider the fission of

Solution:

In the fission of $_{92}^{238} \mathrm{U}, 10 \beta^{-}$particles decay from the parent nucleus. The nuclear reaction can be written as:

${ }_{92}^{238} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{58}^{140} \mathrm{Ce}+{ }_{44}^{99} \mathrm{Ru}+10{ }_{-1}^{0} e$

It is given that:

Mass of a nucleus ${ }_{92}^{238} \mathrm{U}, m_{1}=238.05079 \mathrm{u}$

Mass of a nucleus ${ }_{58}^{140} \mathrm{Ce}, m_{2}=139.90543 \mathrm{u}$

Mass of a nucleus ${ }_{44}^{99} \mathrm{Ru}, m_{3}=98.90594 \mathrm{u}$

Mass of a neutron ${ }_{0}^{1} \mathrm{n}, m_{4}=1.008665 \mathrm{u}$

Q-value of the above equation,

$Q=\left[m^{\prime}\left({ }_{92}^{238} \mathrm{U}\right)+m\left({ }_{0}^{1} \mathrm{n}\right)-m^{\prime}\left({ }_{58}^{140} \mathrm{Ce}\right)-m^{\prime}\left({ }_{44}^{90} \mathrm{Ru}\right)-10 m_{e}\right] c^{2}$

Where,

m’ = Represents the corresponding atomic masses of the nuclei

$m^{\prime}\left({ }_{92}^{238} \mathrm{U}\right)=m_{1}-92 m_{e}$

$m^{\prime}\left(\begin{array}{c}140 \\ 58\end{array}\right.$ Ce $)=m_{2}-58 m_{e}$

$m^{\prime}\left({ }_{44}^{99} \mathrm{Ru}\right)=m_{3}-44 m_{e}$

$m\left(\begin{array}{l}1 \\ 0\end{array} \mathrm{n}\right)=m_{4}$

$Q=\left[m_{1}-92 m_{e}+m_{4}-m_{2}+58 m_{e}-m_{3}+44 m_{e}-10 m_{e}\right] c^{2}$

$=\left[m_{1}+m_{4}-m_{2}-m_{3}\right] c^{2}$

$=[238.0507+1.008665-139.90543-98.90594] c^{2}$

$=\left[0.247995 c^{2}\right] \mathrm{u}$

But $1 \mathrm{u}=931.5 \mathrm{MeV} / c^{2}$

$\therefore Q=0.247995 \times 931.5=231.007 \mathrm{MeV}$

Hence, the Q-value of the fission process is 231.007 MeV.