Using Bohr's formula for energy quantization, the ionisation potential of first excited state of hydrogen atom is

Options

A. 10.2V
B. 3.4 V
C. 2.6 V
D. 1.54 V

Answer

To find the ionization potential of the first excited state of a hydrogen atom using Bohr's formula for energy quantization, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Levels**:
In the Bohr model of the hydrogen atom, the energy levels are quantized. The energy of the nth level is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

2. **Calculate the Energy of the First Excited State**:
The first excited state corresponds to \( n = 2 \). Therefore, we can calculate the energy:
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

3. **Define Ionization Energy**:
Ionization energy is the energy required to remove an electron from an atom. For the first excited state, we need to move the electron from \( n = 2 \) to \( n = \infty \) (where the energy is 0 eV).

4. **Calculate the Ionization Energy**:
The energy required to ionize the electron from the first excited state (from \( -3.4 \, \text{eV} \) to \( 0 \, \text{eV} \)) is:
\[ \text{Ionization Energy} = 0 \, \text{eV} - (-3.4 \, \text{eV}) = 3.4 \, \text{eV} \]

5. **Convert Ionization Energy to Ionization Potential**:
Ionization potential (IP) is defined as the ionization energy per unit charge (in volts). Since the charge of an electron is \( e \), the ionization potential can be expressed as:
\[ \text{Ionization Potential} = \frac{\text{Ionization Energy}}{e} = 3.4 \, \text{V} \]

### Final Answer:
The ionization potential of the first excited state of the hydrogen atom is **3.4 V**.