Consider one of fission reactions of `^(235)U` by thermal neutrons `._(92)^(235)U +n rarr ._(38)^(94)Sr +._(54)^(140)Xe+2n` . The fission fragments are however unstable and they undergo successive `beta`-decay until `._(38)^(94)Sr` becomes `._(40)^(94)Zr` and `._(54)^(140)Xe` becomes `._(58)^(140)Ce`. The energy released in this process is
Given: `m(.^(235)U) =235.439u,m(n)=1.00866 u, m(.^(94)Zr)=93.9064 u, m(.^(140)Ce) =139.9055 u,1u=931 MeV]` .

Options

A. `156 MeV`
B. `208 MeV`
C. `465 MeV`
D. cannot be computed

Explanation

To solve the problem, we need to calculate the energy released during the fission of Uranium-235 by thermal neutrons, considering the masses of the reactants and products involved in the reaction.

**Step 1: Identify the masses of the reactants.**
The reactants in the fission reaction are:
- Uranium-235: \( m(^{235}U) = 235.439 \, u \)
- Neutron: \( m(n) = 1.00866 \, u \)

So, the total mass of the reactants is:
\[ m_{\text{reactants}} = m(^{235}U) + m(n) = 235.439 \, u + 1.00866 \, u = 236.44766 \, u \]

**Step 2: Identify the masses of the products.**
The products of the fission reaction after beta decay are:
- Strontium-94: \( m(^{94}Sr) = 93.9064 \, u \)
- Xenon-140: \( m(^{140}Xe) = 139.9055 \, u \)
- 2 Neutrons: \( 2 \times m(n) = 2 \times 1.00866 \, u = 2.01732 \, u \)

So, the total mass of the products is:
\[ m_{\text{products}} = m(^{94}Sr) + m(^{140}Xe) + 2 \times m(n) = 93.9064 \, u + 139.9055 \, u + 2.01732 \, u = 235.82922 \, u \]

**Step 3: Calculate the mass defect (\( \Delta m \)).**
The mass defect is given by the difference between the mass of the reactants and the mass of the products:
\[ \Delta m = m_{\text{reactants}} - m_{\text{products}} = 236.44766 \, u - 235.82922 \, u = 0.61844 \, u \]

**Step 4: Convert the mass defect to energy (Q).**
Using the conversion factor \( 1 \, u = 931 \, \text{MeV} \):
\[ Q = \Delta m \times 931 \, \text{MeV} = 0.61844 \, u \times 931 \, \text{MeV/u} = 575.54 \, \text{MeV} \]

**Final Answer:**
The energy released in this fission reaction is approximately \( 575.54 \, \text{MeV} \).

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