Ionization potential of hydrogen atom is `13.6 V`. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy `12.1 eV`. The spectral lines emitted by hydrogen atoms according to Bohr's theory will be

Options

A. One
B. Two
C. Three
D. Four

Explanation

To solve the problem, we need to determine the spectral lines emitted by hydrogen atoms when they are excited by a photon of energy 12.1 eV. The ionization potential of hydrogen is given as 13.6 V, which corresponds to the energy required to remove the electron from the ground state.

### Step-by-Step Solution:

1. **Understanding the Ground State Energy**:
The energy of the ground state (n=1) of a hydrogen atom is given by:
\[ E_1 = -13.6 \text{ eV} \]

2. **Photon Energy**:
The energy of the photon that excites the hydrogen atom is:
\[ E_{\text{photon}} = 12.1 \text{ eV} \]

3. **Calculating the Excited State Energy**:
When the hydrogen atom absorbs the photon, it gains energy. The energy of the excited state (E_n) can be calculated as:
\[ E_n = E_1 + E_{\text{photon}} = -13.6 \text{ eV} + 12.1 \text{ eV} = -1.5 \text{ eV} \]

4. **Finding the Quantum Number (n)**:
The energy of the nth level in a hydrogen atom is given by:
\[ E_n = -\frac{13.6}{n^2} \text{ eV} \]
Setting this equal to -1.5 eV, we have:
\[ -\frac{13.6}{n^2} = -1.5 \]
Rearranging gives:
\[ \frac{13.6}{n^2} = 1.5 \]
\[ n^2 = \frac{13.6}{1.5} \approx 9.0667 \]
\[ n \approx 3 \]

5. **Possible Transitions**:
The atom can transition from n=3 to lower energy levels. The possible transitions are:
- From n=3 to n=2
- From n=3 to n=1
- From n=2 to n=1

6. **Counting the Spectral Lines**:
The spectral lines correspond to the transitions:
- 3 to 2
- 3 to 1
- 2 to 1

Therefore, the total number of spectral lines emitted is 3.

### Final Answer:
The spectral lines emitted by hydrogen atoms according to Bohr's theory will be **3**.

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