Question

Calculate the energy released in the fission reaction

${}^{\mathbf{235}}\mathbf{U}\mathbf{+}\mathbf{n}\mathbf{\to }{}^{\mathbf{141}}\mathbf{Cs}\mathbf{+}{}^{\mathbf{93}}\mathbf{Rb}\mathbf{+}\mathbf{2}\mathbf{n}$

Here are some atomic and particle masses.

$\begin{array}{cccc}{}^{\mathbf{235}}\mathbf{U}& \mathbf{235}\mathbf{.}\mathbf{04392}\mathbf{u}& {}^{\mathbf{93}}\mathbf{Rb}& \mathbf{92}\mathbf{.}\mathbf{92157}\mathbf{u}\\ {}^{\mathbf{141}}\mathbf{Cs}& \mathbf{140}\mathbf{.}\mathbf{91964}\mathbf{u}& \mathbf{n}& \mathbf{1}\mathbf{.}\mathbf{00866}\mathbf{u}\end{array}$

Short answer

The released energy in fission reaction is 181. MeV.

Step-by-step solution

Given data

The mass of ${}^{235}U$,$m_u=235.04392u$

The mass of ${}^{93}Rb$,$m_{Rb}=92.92157u$

The mass of ${}^{147}Cs$,$m_{Cs}=140.91963u$

The mass of n,n=1.00866

Determine the energy released in the fission reaction.

The expression to calculate released energy is given as follows.

$$\begin{gathered} \mathbf{Q}\mathbf{=}\mathbf{-}\mathbf{∆}{\mathbf{mc}}^{\mathbf{2}} \\ \mathbf{Q}\mathbf{=}\left(m_u+m_n-m_{\mathrm{Cs}}-m_{\mathrm{Rb}}-2m_n\right){\mathbf{c}}^{\mathbf{2}} \end{gathered}$$ ...(i)

Calculate the released energy in the fission reaction.

Calculate the released energy.

Substitute 235.04392u for $m_2=92.92157u$, for $m_{Rb}$,140.91963u for $m_{Cs}$,1.00866u forn and 931.5MeV/u for$c^2$ into equation (i).

$$\begin{gathered} Q=(235.04392u+1.00866u-140.91963u-92.92157-2\times 1.00866u)\times 931.5MeV/u \\ Q=(235.04392+1.00866-140.91963-92.92157-2.01732)\times 931.5MeV \\ Q=0.19406\times 931.5MeV \\ Q=181\hspace{0.33em}MeV \end{gathered}$$

Hence the released energy in fission reaction is 181..MeV