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Found in: Page 1329

### Fundamentals Of Physics

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

# The fission properties of the plutonium isotope ${}^{{\mathbf{239}}}{\mathbf{Pu}}$ are very similar to those of ${}^{{\mathbf{235}}}{\mathbf{U}}$. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1.0 kg of pure ${}^{{\mathbf{239}}}{\mathbf{Pu}}$ undergo fission?

The released energy is $4.534×{10}^{26}\mathrm{MeV}$.

See the step by step solution

## Step 1: Given data

The mass of the plutonium isotope,m = 1 kg

The average energy released per fission,Q = 180 MeV

The molar mass of plutonium, M = 239

## Step 2: Determine the formulas to calculate the released energy.

The equation to calculate the number of the atoms in the sample is given as follows.

${\mathbit{N}}{\mathbf{=}}\frac{\mathbf{m}{\mathbf{N}}_{\mathbf{A}}}{\mathbf{M}}$ ...(i)

Here,${{\mathbf{N}}}_{{\mathbf{A}}}$ is the Avogadro number $\left(6.022×{10}^{23}{\mathrm{mol}}^{-1}\right)$and M is the molar number.

The expression to calculate the released energy is given as follows.

${\mathbf{E}}{\mathbf{=}}{\mathbf{NQ}}$ ...(ii)

## Step 3: Calculate the released energy in the fission.

Calculate the number of the atoms.

Substitute $239forM,1kgformand6.022×{10}^{23mo{l}^{-1}}for{N}_{A}$ into equation (i).

$N=\frac{1000×6.022×{10}^{23}}{239}\phantom{\rule{0ex}{0ex}}N=\frac{6022×{10}^{23}}{239}\phantom{\rule{0ex}{0ex}}N=25.19×{10}^{23}\phantom{\rule{0ex}{0ex}}N=2.519×{10}^{24}$

Calculate the released energy.

Substitute $2.519×{10}^{24}forNand180MeVforQ$into equation (ii).

$E=2.519×{10}^{24}×180\phantom{\rule{0ex}{0ex}}E=453.42×{10}^{24}\mathrm{MeV}\phantom{\rule{0ex}{0ex}}E=4.534×{10}^{26}\mathrm{MeV}$

Hence the released energy is $4.534×{10}^{26}\mathrm{MeV}$.