If the binding energy of `2^(nd)` excited state of a hydrogen like sample os `24 eV` approximately, then the ionisation energy of the sample is approximately

Options

A. `54.4 eV`
B. `24 eV`
C. `122.4 eV`
D. `216 eV`

Explanation

To solve the problem, we need to find the ionization energy of a hydrogen-like atom given the binding energy of its second excited state. Here’s how we can approach this step by step:

### Step-by-Step Solution:

1. **Understanding Binding Energy**:
The binding energy (BE) of an electron in a hydrogen-like atom is given by the formula:
\[ BE = \frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

2. **Identifying the Given Information**:
We are given that the binding energy of the second excited state is 24 eV. The second excited state corresponds to \( n = 3 \).

3. **Setting Up the Equation**:
Substituting the known values into the binding energy formula:
\[ 24 \, \text{eV} = \frac{13.6 \, \text{eV} \cdot Z^2}{3^2} \]
Simplifying \( 3^2 \):
\[ 24 \, \text{eV} = \frac{13.6 \, \text{eV} \cdot Z^2}{9} \]

4. **Solving for \( Z^2 \)**:
Rearranging the equation to solve for \( Z^2 \):
\[ 24 \cdot 9 = 13.6 \cdot Z^2 \]
\[ 216 = 13.6 \cdot Z^2 \]
Dividing both sides by 13.6:
\[ Z^2 = \frac{216}{13.6} \approx 15.88 \]

5. **Calculating Ionization Energy**:
The ionization energy (IE) for a hydrogen-like atom is given by:
\[ IE = 13.6 \, \text{eV} \cdot Z^2 \]
Substituting the value of \( Z^2 \):
\[ IE = 13.6 \cdot 15.88 \approx 216 \, \text{eV} \]

### Final Answer:
The ionization energy of the sample is approximately **216 eV**.