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 \( ^{235}_{92}\text{U} \)

The total mass changes during this fission process. The mass defect \( \Delta m \) is the difference between the initial mass and the mass of the fission products.

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

The initial mass comprises the \( ^{235}_{92}\text{U} \) nucleus and a neutron:

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

Using the provided values:

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

The final mass consists of the \( ^{140}_{54}\text{Xe} \) and \( ^{94}_{38}\text{Sr} \) nuclei along with two neutrons:

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

Substituting the given values:

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

The mass defect is calculated as:

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

The energy released is determined using the mass-energy equivalence \( E = \Delta m \times 931 \, \text{MeV}/c^2 \):

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

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