Chapter 24: Q24.78P (page 1105)

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

Short Answer

Expert verified

The energy released per nucleus is $2.63MeV/$ nucleus.

Step by step solution

Step 1:

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

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

We have to find the energy released per nucleus in $MeV$

- ${1MeV = 1.602×10}^{- 13} J$

Total number of nucleus is

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

Now, let us convert total energy released, from $kJ to MeV$

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

Therefore, the energy released per nucleus in $MeV$ is

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

Hence The energy released per nucleus is $2.63MeV/$ nucleus.

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

Short Answer

Step by step solution

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