If first excitation potential of a hydrogen-like atom is `V` electron volt, then the ionization energy of this atom will be:

Options

A. V electron volt
B. `(3V)/(4)` electron volt
C. `(4V)/(3)` electron volt
D. Cannot be calculated by given information

Answer

To find the ionization energy of a hydrogen-like atom given its first excitation potential \( V \) in electron volts, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Energy Levels**:
For a hydrogen-like atom, the energy levels are given by the formula:
\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

2. **First Excitation Potential**:
The first excitation potential corresponds to the transition from the ground state (n=1) to the first excited state (n=2). The energy difference (excitation potential) can be expressed as:
\[ E_2 - E_1 = \left(-\frac{Z^2 \cdot 13.6}{2^2}\right) - \left(-\frac{Z^2 \cdot 13.6}{1^2}\right) \]
Simplifying this gives:
\[ E_2 - E_1 = -\frac{Z^2 \cdot 13.6}{4} + \frac{Z^2 \cdot 13.6}{1} = Z^2 \cdot 13.6 \left(1 - \frac{1}{4}\right) = Z^2 \cdot 13.6 \cdot \frac{3}{4} \]

3. **Relating to Given Potential \( V \)**:
According to the problem, this energy difference is equal to \( V \):
\[ V = \frac{3}{4} \cdot \text{Ionization Energy (IE)} \]

4. **Finding Ionization Energy**:
Rearranging the equation to find the ionization energy:
\[ \text{Ionization Energy (IE)} = \frac{4}{3} V \]

5. **Conclusion**:
Therefore, the ionization energy of the hydrogen-like atom is:
\[ \text{Ionization Energy} = \frac{4}{3} V \, \text{eV} \]

### Final Answer:
The ionization energy of the atom is \( \frac{4}{3} V \) eV.

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