Consider the case of bombardment of `U^(235)` nucleus with a thermal neutron. The fission products are `Mo^(95)` & `La^(139)` and two neutrons. Calculate the energy released by one `U^(235)` nucleus. (Rest masses of the nuclides are `U^(235)=235.0439 u, ._(0)^(1)n=1.0087u, Mo^(95)=94.9058 u, La^(139).9061u)`.

Explanation

To calculate the energy released by one \( U^{235} \) nucleus during fission, we will follow these steps:

### Step 1: Write down the reaction
The fission reaction can be represented as:
\[ U^{235} + n \rightarrow Mo^{95} + La^{139} + 2n \]

### Step 2: Identify the masses involved
We need to identify the rest masses of the reactants and products:
- Mass of \( U^{235} \) = 235.0439 u
- Mass of 1 neutron \( (n) \) = 1.0087 u
- Mass of \( Mo^{95} \) = 94.9058 u
- Mass of \( La^{139} \) = 138.9061 u

### Step 3: Calculate the total mass of reactants
The total mass of the reactants (initial mass) is:
\[ \text{Initial mass} = \text{mass of } U^{235} + \text{mass of } n = 235.0439 \, u + 1.0087 \, u = 236.0526 \, u \]

### Step 4: Calculate the total mass of products
The total mass of the products (final mass) is:
\[ \text{Final mass} = \text{mass of } Mo^{95} + \text{mass of } La^{139} + \text{mass of 2 neutrons} \]
Since we have two neutrons produced, we need to add the mass of two neutrons:
\[ \text{Final mass} = 94.9058 \, u + 138.9061 \, u + 2 \times 1.0087 \, u \]
Calculating this gives:
\[ \text{Final mass} = 94.9058 \, u + 138.9061 \, u + 2.0174 \, u = 235.8293 \, u \]

### Step 5: Calculate the change in mass
The change in mass (\( \Delta m \)) is given by:
\[ \Delta m = \text{Initial mass} - \text{Final mass} \]
Substituting the values:
\[ \Delta m = 236.0526 \, u - 235.8293 \, u = 0.2233 \, u \]

### Step 6: Calculate the energy released
Using Einstein's equation \( E = \Delta m \cdot c^2 \), where \( 1 \, u \) corresponds to \( 931 \, MeV \):
\[ E = 0.2233 \, u \times 931 \, MeV/u \]
Calculating this gives:
\[ E \approx 207.9 \, MeV \]

### Final Answer
The energy released by one \( U^{235} \) nucleus during fission is approximately \( 207.9 \, MeV \).

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