Question

Question: Consider the fission of ${}^{\mathbf{238}}\mathbf{U}$by fast neutrons. In one fission event, no neutrons are emitted and the final stable end products, after the beta decay of the primary fission fragments, are ${}^{140}Ce\mathrm{and}{}^{99}\mathrm{Ru}$. (a) What is the total of the beta-decay events in the two beta-decay chains? (b) Calculate for this fission process. The relevant atomic and particle masses are

$$\begin{gathered} {}^{\mathbf{238}}\mathbf{U}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{238}\mathbf{.}\mathbf{05079}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{}^{\mathbf{140}}\mathbf{Ce}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{139}\mathbf{.}\mathbf{90543}\mathbf{u} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{n}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{00866}\mathbf{u}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{}^{\mathbf{99}}\mathbf{Ru}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{9890594}\mathbf{u} \end{gathered}$$

Short answer

(a) The number of the beta decay is 10.

(b) The energy release in fission process is 225.98Me V.

Step-by-step solution

Given data

The mass of ${}^{238}U,m_u=238.05079u$

The mass of ${}^{140}Ce,m_{Ce}=139.90543u$

The mass of $n,m_n=1.00866u$

The mass of${}^{99}Ru,m_{Ru}=98.90594u$

Determine the formulas to calculate the number of the beta decay and energy for the fission process.

The expression to calculate released energy is given as follows.

$$\begin{gathered} \mathit{Q}\mathbf{=}\mathbf{-}\mathbf{∆}\mathit{m}{\mathit{c}}^{\mathbf{2}} \\ \mathit{Q}\mathbf{=}\left(m_u+m_n-m_{Ce}-m_{Ru}-N{m}_e\right){\mathit{c}}^{\mathbf{2}} \end{gathered}$$ ...(i)

Here,N is the number of the beta decay.

(a) Calculate the number of the beta decay.

Consider the reaction given below.

${}^{238}U+n\to {}^{140}Ce+{}^{99}Ru+Ne$

Here, is the number of the beta decay.

To calculate the number of the beta decay, equals the atomic number of both the sides of the equation.

Atomic number of uranium,$Z_u=92$

Atomic number of cerium,$Z_{Ce}=58$

Atomic number of Ruthenium,$Z_{Ru}=44$

Calculate the number of beta decay.

$Ne+Z_{Ru}+Z_{Ce}=Z_u$

Substitute for 92 for $Z_u$, 58 for $Z_{Ce}$, 44 for$Z_{Ru}$ and 1 for e into above equation.

$$\begin{gathered} N\left(1\right)+44+58=92 \\ N+102=92 \\ N=92-102 \\ N=-10 \end{gathered}$$

Here, negative sign indicates that beta decay reduces the atomic number.

Hence the number of the beta decay is 10.

(b) Calculate the energy for the fission process.

Recall equation (i).

$$\begin{gathered} Q=({\mathrm{m}}_{\mathrm{u}}+{\mathrm{m}}_{\mathrm{n}}-{\mathrm{m}}_{\mathrm{Ce}}-{\mathrm{m}}_{\mathrm{Ru}}-{\mathrm{Nm}}_{\mathrm{e}})\times c^2 \\ Q=({\mathrm{m}}_{\mathrm{u}}+{\mathrm{m}}_{\mathrm{n}}-{\mathrm{m}}_{\mathrm{Ce}}-{\mathrm{m}}_{\mathrm{Ru}})c^2-\left(m_e{c}^2\right) \end{gathered}$$

Substitute f $238.05079u\mathrm{for}{\mathrm{m}}_{\mathrm{u}}$, $139.90543u\mathrm{for}{\mathrm{m}}_{\mathrm{Ce}}$, $1.00866u\mathrm{for}{\mathrm{m}}_{\mathrm{n}}$, $98.90594u\mathrm{for}{\mathrm{m}}_{\mathrm{Ru}}$, $931.5\mathrm{MeV}\mathrm{for}{\mathrm{c}}^2$, $0.511\mathrm{MeV}\mathrm{for}{\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2$and 10 for N into equation (i).

role="math" localid="1661924965359" $$\begin{gathered} Q=(238.05078+1.00866-139.90543-98.90594)931.5\mathrm{MeV}-10\times 0.511\mathrm{MeV} \\ Q=0.24809\times 931.5\mathrm{MeV}-10\times 0.511\mathrm{MeV} \\ Q=231.095835\mathrm{MeV}-5.11\mathrm{MeV} \\ Q=225.99\mathrm{MeV} \end{gathered}$$

Hence the energy release in fission process is $225.99\mathrm{MeV}$.