Problem

Calculate the total energy released if 1 kg of $\mathrm{U}^{235}$ undergoes fission, given that the disintegration energy per event is $Q = 208 \, \mathrm{MeV}$.

Data

• Mass of uranium sample ($M$): $1 \, \mathrm{kg}$
• Disintegration energy per event ($Q$): $208 \, \mathrm{MeV}$
• Atomic mass of uranium ($A$): $235 \, \mathrm{kg/mol}$
• Avogadro’s number ($N_A$): $6.023 \times 10^{23} \, \mathrm{atoms/mol}$

Prerequisite Concepts

1. Energy Released in Fission:
   The total energy released ($E_{\text{total}}$) is calculated as the product of the number of atoms ($N$) in the sample and the energy released per fission event ($Q$):

$$E_{\text{total}} = N \cdot Q$$

2. Number of Atoms in the Sample:
   The number of atoms ($N$) is determined by dividing the total mass of the sample by the mass of a single mole of the isotope:

$$N = \frac{M \cdot N_A}{A}$$

3. Conversions:
   1 MeV = $1.602 \times 10^{-13} \, \mathrm{J}$.

Solution

Step 1: Calculate the Number of Atoms ($N$) in the Sample

Using the formula:

$$N = \frac{M \cdot N_A}{A}$$

Substitute the values:

$$N = \frac{1 \cdot 6.023 \times 10^{23}}{235}$$ $$N = 2.56 \times 10^{24} \, \text{atoms of } \mathrm{U}^{235}$$

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Step 2: Calculate the Total Energy Released ($E_{\text{total}}$)

Using the formula:

$$E_{\text{total}} = N \cdot Q$$

Substitute $N = 2.56 \times 10^{24}$ and $Q = 208 \, \mathrm{MeV}$:

$$E_{\text{total}} = 2.56 \times 10^{24} \cdot 208$$ $$E_{\text{total}} = 5.3248 \times 10^{26} \, \mathrm{MeV}$$

Convert to joules:

$$E_{\text{total}} = 5.3248 \times 10^{26} \cdot 1.602 \times 10^{-13}$$ $$E_{\text{total}} = 8.53 \times 10^{13} \, \mathrm{J}$$

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Answer

The total energy released from the fission of 1 kg of $\mathrm{U}^{235}$ is:

• $5.3248 \times 10^{26} \, \mathrm{MeV}$ or
• $8.53 \times 10^{13} \, \mathrm{J}$.