If I exciation energy for the H-like (hypothetical) sample is 24 eV, then binding energy in III excited state is :
(a)2eV
(b)3eV
(c)4eV
(d)5eV

Options

A. 2 eV
B. 3 eV
C. 4 eV
D. 5 eV

Answer

To find the binding energy in the third excited state of a hydrogen-like atom given the excitation energy of 24 eV, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Binding Energy Formula**:
The binding energy (BE) for a hydrogen-like atom can be expressed as:
\[ BE = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

2. **Identify the Excitation Energy**:
The excitation energy given is 24 eV. This is the energy required to move from the ground state (n=1) to the first excited state (n=2).

3. **Relate Excitation Energy to Binding Energy**:
The excitation energy can be expressed in terms of the binding energies of the two states:
\[ E_{excitation} = BE(n=1) - BE(n=2) \]
Substituting the binding energy formula:
\[ 24 = \left(-\frac{13.6 Z^2}{1^2}\right) - \left(-\frac{13.6 Z^2}{2^2}\right) \]
Simplifying gives:
\[ 24 = -13.6 Z^2 \left(1 - \frac{1}{4}\right) = -13.6 Z^2 \left(\frac{3}{4}\right) \]

4. **Solve for Ionization Energy**:
Rearranging the equation:
\[ 24 = -\frac{13.6 Z^2 \cdot 3}{4} \]
Multiplying both sides by -4/3:
\[ -32 = 13.6 Z^2 \]
Thus, we find:
\[ Z^2 = \frac{32}{13.6} \approx 2.35 \]

5. **Calculate Ionization Energy**:
The ionization energy (IE) for the first excited state (n=2) can be calculated as:
\[ IE = -BE(n=2) = \frac{13.6 Z^2}{2^2} = \frac{13.6 \cdot 2.35}{4} \approx 8 \text{ eV} \]

6. **Find Binding Energy in the Third Excited State**:
The third excited state corresponds to \( n=4 \):
\[ BE(n=4) = -\frac{13.6 Z^2}{4^2} = -\frac{13.6 \cdot 2.35}{16} \approx -2.0 \text{ eV} \]
Therefore, the binding energy is:
\[ BE(n=4) = 2 \text{ eV} \]

### Final Answer:
The binding energy in the third excited state is **2 eV** (Option a).