Physics 2022 JAMB Past Questions - Question 4: In a nuclear plant, the final mass of the products is 6.32×10-²kg, while the initial mass of the reactant is 6.30×10-²kg, the energy released in the process is (speed of light in vacuum 3.0×10¸m/s, 1eV = 1.6×10-¹¹J)

Correct Answer

A

Explanation

The energy released in a nuclear reaction can be calculated using Einstein's mass-energy equivalence equation, which is given by:

\[ E = \Delta m \cdot c^2 \]

where:
- \( E \) is the energy released,
- \( \Delta m \) is the change in mass,
- \( c \) is the speed of light.

The change in mass (\( \Delta m \)) is given by the difference between the initial mass (\( m_{\text{initial}} \)) and the final mass (\( m_{\text{final}} \)).

\[ \Delta m = m_{\text{initial}} - m_{\text{final}} \]

Given that \( m_{\text{initial}} = 6.30 \times 10^{-27} \, \text{kg} \) and \( m_{\text{final}} = 6.32 \times 10^{-27} \, \text{kg} \), we can calculate \( \Delta m \):

\[ \Delta m = 6.30 \times 10^{-27} \, \text{kg} - 6.32 \times 10^{-27} \, \text{kg} \]

\[ \Delta m = -2 \times 10^{-29} \, \text{kg} \]

Now, we can use the mass-energy equivalence equation to calculate the energy (\( E \)):

\[ E = (-2 \times 10^{-29} \, \text{kg}) \times (3.0 \times 10^8 \, \text{m/s})^2 \]

\[ E = -2 \times 10^{-29} \, \text{kg} \times 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \]

\[ E = -18 \times 10^{-13} \, \text{J} \]

It's important to note that the negative sign indicates that energy is released in the reaction. The magnitude of the energy is \( 18 \times 10^{-13} \, \text{J} \).

If you want to convert this energy to electronvolts (eV), you can use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):

\[ E_{\text{eV}} = \frac{-18 \times 10^{-13} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \]

\[ E_{\text{eV}} \approx -1.125 \times 10^6 \, \text{eV} \]

The negative sign indicates that the reaction releases energy.