The binding energy per nucleon are 5.3 Mev, 6.2 MeV and 7.4 MeV for the nucleus with mass number ,3,4 and 5 respectively . If one nucleus of mass number 3 combines with one nucleus of mass number 5 to give two nuclei of mass number 4, then

Options

A. (a)0.3 MeV energy is absorbed
B. (b)0.3 MeV energy is released
C. (c)28.1 MEV energy is abosrbed
D. (d)3.3 MeV energy is absorbed

Answer

To solve the problem, we need to calculate the binding energies of the given nuclei and determine the change in energy when one nucleus of mass number 3 combines with one nucleus of mass number 5 to form two nuclei of mass number 4.

### Step 1: Calculate the binding energy of each nucleus

1. **For nucleus with mass number 3 (X)**:
- Binding energy per nucleon = 5.3 MeV
- Total binding energy for nucleus X:
\[ BE_X = 5.3 \, \text{MeV} \times 3 = 15.9 \, \text{MeV} \]

2. **For nucleus with mass number 4 (Y)**:
- Binding energy per nucleon = 6.2 MeV
- Total binding energy for nucleus Y:
\[ BE_Y = 6.2 \, \text{MeV} \times 4 = 24.8 \, \text{MeV} \]

3. **For nucleus with mass number 5 (Z)**:
- Binding energy per nucleon = 7.4 MeV
- Total binding energy for nucleus Z:
\[ BE_Z = 7.4 \, \text{MeV} \times 5 = 37.0 \, \text{MeV} \]

### Step 2: Combine the nuclei

According to the problem, one nucleus of mass number 3 (X) combines with one nucleus of mass number 5 (Z) to form two nuclei of mass number 4 (Y).

### Step 3: Calculate the total binding energy before and after the reaction

1. **Total binding energy before the reaction (E1)**:
\[ E1 = BE_X + BE_Z = 15.9 \, \text{MeV} + 37.0 \, \text{MeV} = 52.9 \, \text{MeV} \]

2. **Total binding energy after the reaction (E2)**:
Since two nuclei of mass number 4 are formed:
\[ E2 = 2 \times BE_Y = 2 \times 24.8 \, \text{MeV} = 49.6 \, \text{MeV} \]

### Step 4: Calculate the change in binding energy

The change in binding energy (ΔE) is given by:
\[ \Delta E = E1 - E2 = 52.9 \, \text{MeV} - 49.6 \, \text{MeV} = 3.3 \, \text{MeV} \]

### Step 5: Determine if energy is absorbed or released

Since ΔE is positive (3.3 MeV), it indicates that energy is absorbed in the process.

### Conclusion

The final result is that the energy absorbed during the reaction is **3.3 MeV**.