If the total binding energies of `._(1)H^(2),._(2)He^(4),._(26)Fe^(56)` and `._(92)U^(235)` nuclei are `2.22, 28.3, 492` and `1786MeV` respectively, identify the most stable nucleus out of the following

Options

A. `._(26)Fe^(56)`
B. `._(1)H^(2)`
C. `._(92)U^(235)`
D. `._(2)He^(4)`

Explanation

To determine the most stable nucleus among the given options, we need to calculate the binding energy per nucleon for each nucleus. The nucleus with the highest binding energy per nucleon is considered the most stable.

### Step-by-Step Solution:

1. **Identify the total binding energies and nucleon counts**:
- For Hydrogen \( _{1}^{2}H \): Total binding energy = 2.22 MeV, Number of nucleons = 2
- For Helium \( _{2}^{4}He \): Total binding energy = 28.3 MeV, Number of nucleons = 4
- For Iron \( _{26}^{56}Fe \): Total binding energy = 492 MeV, Number of nucleons = 56
- For Uranium \( _{92}^{235}U \): Total binding energy = 1786 MeV, Number of nucleons = 235

2. **Calculate binding energy per nucleon for each nucleus**:
- For Hydrogen:
\[ \text{Binding Energy per Nucleon} = \frac{2.22 \text{ MeV}}{2} = 1.11 \text{ MeV/nucleon} \]
- For Helium:
\[ \text{Binding Energy per Nucleon} = \frac{28.3 \text{ MeV}}{4} = 7.075 \text{ MeV/nucleon} \]
- For Iron:
\[ \text{Binding Energy per Nucleon} = \frac{492 \text{ MeV}}{56} = 8.7857 \text{ MeV/nucleon} \approx 8.78 \text{ MeV/nucleon} \]
- For Uranium:
\[ \text{Binding Energy per Nucleon} = \frac{1786 \text{ MeV}}{235} = 7.6 \text{ MeV/nucleon} \]

3. **Compare the binding energies per nucleon**:
- Hydrogen: 1.11 MeV/nucleon
- Helium: 7.075 MeV/nucleon
- Iron: 8.78 MeV/nucleon
- Uranium: 7.6 MeV/nucleon

4. **Identify the most stable nucleus**:
The nucleus with the highest binding energy per nucleon is Iron with 8.78 MeV/nucleon.

### Conclusion:
The most stable nucleus among the given options is **Iron \( _{26}^{56}Fe \)**.

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