Question

How many MeV are released per nucleus when ${\mathbf{\text{3.2×10}}}^{\mathbf{\text{- 3}}}\mathbf{\text{mol}}$of chromium-$\mathbf{49}$releases${\mathbf{\text{8.11×10}}}^{\mathbf{\text{5}}}\mathbf{\text{kJ}}$?

Short answer

The energy released per nucleus is $\text{2.63MeV/}$nucleus.

Step-by-step solution

Step 1:

- The number of moles of chromium-$49$ is${\text{3.2×10}}^{\text{- 3}}\text{mol}$

- Total energy released is${\text{8.11×10}}^{\text{5}}\text{kJ}$

We have to find the energy released per nucleus in$\text{MeV}$

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

Total number of nucleus is

Number of nucleus $\text{=}{\text{3.2×10}}^{\text{- 3}}\text{mol×}\frac{{\displaystyle {\text{6.022×10}}^{\text{23}}\text{nucleus}}}{1{\displaystyle \text{mol}}}={\text{1.927×10}}^{\text{21}}$nucleus

Now, let us convert total energy released, from$\text{kJ to MeV}$

$\text{Total energy released =}{\text{8.11×10}}^{\text{5}}\text{kJ×}\frac{1000\text{J}}{1{\displaystyle \text{kJ}}}\text{×}\frac{{\displaystyle \text{1MeV}}}{{\displaystyle {\text{1.602×10}}^{\text{- 13}}\text{J}}}={\text{5.06×10}}^{\text{21}}\text{MeV}$

Therefore, the energy released per nucleus in $\text{MeV}$is

$\text{Energy per nucleus}\text{=}\frac{{\displaystyle {\text{5.06×10}}^{\text{21}}\text{MeV}}}{{\displaystyle {\text{1.927×10}}^{\text{21}}\text{nucleus}}}=\text{2.63MeV/ nucleus}$

Hence The energy released per nucleus is $\text{2.63MeV/}$nucleus.