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The mass of the plutonium isotope, m =2 megaton

The 8% of the total plutonium is fissioned.

megaton releases $2.6脳{10}^{26}\mathrm{MeV}$energy.

The mass of plutonium, M =0.239 kg/mol

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

$\mathit{N}\mathbf{=}\frac{\mathbf{Y}}{\mathbf{Q}}$ ...(i)

Here,Y is the yield andM is the molar number.

The expression to calculate the mass is given as follows.

$\mathit{m}\mathbf{=}\mathit{M}{\mathit{N}}_{\mathbf{0}}$ ...(ii)

Here,${\mathit{N}}_{\mathbf{0}}$ is the actual number of the atom.

Calculate the yielded equivalent of the explosion.

$$\begin{gathered} Y=2脳2.6脳{10}^{28} \\ Y=5.2脳{10}^{28}\mathrm{MeV} \end{gathered}$$

The energy released in the typical fission process is 200 MeV

Calculate the number of the atom,

Substitute$5.2脳{10}^{28}\mathrm{MeV}$ for Y and 200 MeV for Q into equation (i).

$$\begin{gathered} N=\frac{5.2脳{10}^{28}\mathrm{MeV}}{200\mathrm{MeV}} \\ N=2.6脳{10}^{26} \end{gathered}$$

The Actual number of the atoms are 8% actually fissioned of the total mass of plutonium.

$$\begin{gathered} N_0=\frac{2.6脳{10}^{26}}{0.08}脳\frac{1}{6.022脳0^{23}} \\ {N}_0=\frac{3.25脳{10}^{27}}{6.022脳{10}^{23}} \\ {N}_0=5.4脳{10}^{23}\mathrm{mol} \end{gathered}$$

Calculate the total mass.

Substitute$5.4脳{10}^{23}$ forN and 0.239 kg/mol into equation (ii).

$$\begin{gathered} m=0.239脳5.4脳{10}^3 \\ m=1.3脳{10}^3\mathrm{kg} \end{gathered}$$

Hence the total mass is $1.3脳{10}^{23}\mathrm{kg}$.