Question

How many joules are released when$\mathbf{1}\mathbf{.}\mathbf{5}$mol of$\mathbf{239}$Pu decays, if each nucleus releases$\mathbf{5}\mathbf{.}\mathbf{243}$MeV?

Short answer

The joules released are${\text{7.587×10}}^{\text{11}}\text{J}$.

Step-by-step solution

To find how many joules are released

-${\text{1.5 mol of}}^{\text{239}}\text{Pu}$ decays

- each nucleus releases$\text{5.243MeV}$

First, we need to calculate the number of nuclei in$1.5\text{molof}{}^{\text{239}}\text{Pu}$

$\text{Numberofnuclei}=1.50\text{mol}\times \frac{{\displaystyle {\text{6.022×10}}^{\text{23}}\text{nuclei}}}{1\text{mol}}=\text{9.033×}{10}^{\text{23}}\text{nuclei}$

Total energy released in $\text{MeV}$is

$\text{Total energy}={\text{9.033×10}}^{\text{23}}\text{nuclei×5.243}\frac{\text{MeV}}{\text{nuclei}}={\text{4.736×10}}^{\text{24}}\text{MeV}$

-${\text{1MeV = 1.602×10}}^{\text{- 13}}\text{J}$

Therefore, the number of joules released is

${\text{4.736×10}}^{\text{24}}\text{MeV}={\text{4.736×10}}^{\text{24}}\text{MeV×}\frac{{\displaystyle {\text{1.602×10}}^{\text{- 13}}\text{J}}}{{\displaystyle \text{1MeV}}}={\text{7.587×10}}^{\text{11}}\text{J}$

Hence the joules released are${\text{7.587×10}}^{\text{11}}\text{J}$