Question

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?

Short answer

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

Step-by-step solution

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

Determine the formulas to calculate the released energy.

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

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

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

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

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

Calculate the released energy in the fission.

Calculate the number of the atoms.

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

$$\begin{gathered} N=\frac{1000\times 6.022\times {10}^{23}}{239} \\ N=\frac{6022\times {10}^{23}}{239} \\ N=25.19\times {10}^{23} \\ N=2.519\times {10}^{24} \end{gathered}$$

Calculate the released energy.

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

$$\begin{gathered} E=2.519\times {10}^{24}\times 180 \\ E=453.42\times {10}^{24}\mathrm{MeV} \\ E=4.534\times {10}^{26}\mathrm{MeV} \end{gathered}$$

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