Question

Mass Defect and Energy Released in the Fission of \( ^{235}_{92}\text{U} \) 
When a neutron collides with \( ^{235}_{92}\text{U} \), the nucleus gives \( ^{140}_{54}\text{Xe} \) and \( ^{94}_{38}\text{Sr} \) as fission products, and two neutrons are ejected. Calculate the mass defect and the energy released (in MeV) in the process.
Given:

• Mass of \( ^{235}_{92}\text{U} \): \( m(^{235}_{92}\text{U}) = 235.04393 \, \text{u} \)
• Mass of \( ^{140}_{54}\text{Xe} \): \( m(^{140}_{54}\text{Xe}) = 139.92164 \, \text{u} \)
• Mass of \( ^{94}_{38}\text{Sr} \): \( m(^{94}_{38}\text{Sr}) = 93.91536 \, \text{u} \)
• Mass of neutron \( ^{1}_0n \): \( m(^{1}_0n) = 1.00866 \, \text{u} \)
• Conversion factor: \( 1 \, \text{u} = 931 \, \text{MeV}/c^2 \)

Hint:

The mass defect in a nuclear reaction is the difference in mass between the initial nucleus and the sum of the masses of the products. This mass defect is converted into energy, which can be calculated using the equation \( E = \Delta m \times 931 \, \text{MeV}/c^2 \).

Answer 1

Mass Defect and Energy Released in the Fission of \( ^{235}_{92}\text{U} \) 

In this fission process, the total mass before and after the reaction changes. The total mass defect \( \Delta m \) is the difference between the mass of the fission products and the initial mass.

\[ \Delta m = m(\text{Initial mass}) - m(\text{Final mass}) \]

The initial mass is the mass of the \( ^{235}_{92}\text{U} \) nucleus plus the mass of the neutron:

\[ m_{\text{initial}} = m(^{235}_{92}\text{U}) + m(^{1}_0n) \]

Substituting the given values:

\[ m_{\text{initial}} = 235.04393 + 1.00866 = 236.05259 \, \text{u} \]

The final mass is the mass of the fission products (the \( ^{140}_{54}\text{Xe} \) and \( ^{94}_{38}\text{Sr} \) nuclei) plus the mass of the two neutrons:

\[ m_{\text{final}} = m(^{140}_{54}\text{Xe}) + m(^{94}_{38}\text{Sr}) + 2 \times m(^{1}_0n) \]

Substituting the given values:

\[ m_{\text{final}} = 139.92164 + 93.91536 + 2 \times 1.00866 = 235.85432 \, \text{u} \]

Now, the mass defect is:

\[ \Delta m = 236.05259 - 235.85432 = 0.19827 \, \text{u} \]

To find the energy released, we use the equivalence \( E = \Delta m \times 931 \, \text{MeV}/c^2 \):

\[ E = 0.19827 \times 931 = 184.59 \, \text{MeV} \]

Thus, the energy released in the process is \( \boxed{184.59} \, \text{MeV} \).