Question:
Find the energy released, if \(5\) g of \(^{235}\mathrm{U}\) is completely consumed in a chain reaction.
Find the energy released, if \(5\) g of \(^{235}\mathrm{U}\) is completely consumed in a chain reaction.
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Remember:
\[
1\ \text{fission of }^{235}\mathrm{U}
\approx 200\ \text{MeV}
\]
and
\[
1\ \text{MeV}
=
1.6\times10^{-13}\ \text{J}.
\]
First find the number of nuclei and then multiply by the energy released per fission.
Updated On: Jun 16, 2026
- \(0.4\times10^{12}\) joules
- \(0.4\times10^{12}\) MeV
- \(0.4\times10^{12}\) eV
- \(0.4\times10^{12}\) ergs
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The Correct Option is A
Solution and Explanation
Concept:
Energy released per fission of \(^{235}\mathrm{U}\) is approximately
\[
200\ \text{MeV}
=
3.2\times10^{-11}\ \text{J}.
\]
Total energy released:
\[
E=N\times E_{\text{per fission}}.
\]
Step 1: Calculate the number of uranium nuclei. Number of moles: \[ n=\frac{5}{235} \] Number of nuclei: \[ N=\frac{5}{235}N_A \] \[ =\frac{5}{235}\times6.02\times10^{23} \] \[ \approx1.28\times10^{22}. \]
Step 2: Calculate the total energy released. \[ E = 1.28\times10^{22} \times 3.2\times10^{-11} \] \[ \approx4.1\times10^{11}\ \text{J} \] \[ \approx0.4\times10^{12}\ \text{J}. \] \[\begin{aligned} \boxed{0.4\times10^{12}\ \text{J}} \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.
Step 1: Calculate the number of uranium nuclei. Number of moles: \[ n=\frac{5}{235} \] Number of nuclei: \[ N=\frac{5}{235}N_A \] \[ =\frac{5}{235}\times6.02\times10^{23} \] \[ \approx1.28\times10^{22}. \]
Step 2: Calculate the total energy released. \[ E = 1.28\times10^{22} \times 3.2\times10^{-11} \] \[ \approx4.1\times10^{11}\ \text{J} \] \[ \approx0.4\times10^{12}\ \text{J}. \] \[\begin{aligned} \boxed{0.4\times10^{12}\ \text{J}} \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.
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