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Found in: Page 1331

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# 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${}^{{\mathbf{238}}}{\mathbf{U}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{238}}{\mathbf{.}}{\mathbf{05079}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{}^{{\mathbf{140}}}{\mathbf{Ce}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{139}}{\mathbf{.}}{\mathbf{90543}}{\mathbf{u}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{n}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{1}}{\mathbf{.}}{\mathbf{00866}}{\mathbf{u}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{}^{{\mathbf{99}}}{\mathbf{Ru}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{9890594}}{\mathbf{u}}$

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

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

See the step by step solution

## Step 1: 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 ${}^{99}Ru,{m}_{Ru}=98.90594u$

## Step 2: 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.

${\mathbit{Q}}{\mathbf{=}}{\mathbf{-}}{\mathbf{∆}}{\mathbit{m}}{{\mathbit{c}}}^{{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}{\mathbit{Q}}{\mathbf{=}}\left({m}_{u}+{m}_{n}-{m}_{Ce}-{m}_{Ru}-N{m}_{e}\right){{\mathbit{c}}}^{{\mathbf{2}}}$ ...(i)

Here,N is the number of the beta decay.

## Step 3: (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.

$N\left(1\right)+44+58=92\phantom{\rule{0ex}{0ex}}N+102=92\phantom{\rule{0ex}{0ex}}N=92-102\phantom{\rule{0ex}{0ex}}N=-10$

Hence the number of the beta decay is 10.

## Step 4: (b) Calculate the energy for the fission process.

Recall equation (i).

$Q=\left({\mathrm{m}}_{\mathrm{u}}+{\mathrm{m}}_{\mathrm{n}}-{\mathrm{m}}_{\mathrm{Ce}}-{\mathrm{m}}_{\mathrm{Ru}}-{\mathrm{Nm}}_{\mathrm{e}}\right)×{c}^{2}\phantom{\rule{0ex}{0ex}}Q=\left({\mathrm{m}}_{\mathrm{u}}+{\mathrm{m}}_{\mathrm{n}}-{\mathrm{m}}_{\mathrm{Ce}}-{\mathrm{m}}_{\mathrm{Ru}}\right){c}^{2}-\left({m}_{e}{c}^{2}\right)$

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" $Q=\left(238.05078+1.00866-139.90543-98.90594\right)931.5\mathrm{MeV}-10×0.511\mathrm{MeV}\phantom{\rule{0ex}{0ex}}Q=0.24809×931.5\mathrm{MeV}-10×0.511\mathrm{MeV}\phantom{\rule{0ex}{0ex}}Q=231.095835\mathrm{MeV}-5.11\mathrm{MeV}\phantom{\rule{0ex}{0ex}}Q=225.99\mathrm{MeV}$

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