Ionization poțential of hydrogen is `13.6V`. If it is excited by a photon of energy `12.1 eV`, then the number of lines in the enission spectrum will be

Answer

To solve the problem, we need to determine how many energy levels the hydrogen atom can be excited to when it absorbs a photon of energy \(12.1 \, \text{eV}\) and then find the number of lines in the emission spectrum.

### Step-by-Step Solution:

1. **Understand Ionization Potential**: The ionization potential of hydrogen is \(13.6 \, \text{eV}\). This means that to completely remove an electron from the ground state (n=1), we need \(13.6 \, \text{eV}\) of energy.

2. **Photon Energy**: The photon has an energy of \(12.1 \, \text{eV}\). This energy is less than the ionization potential, which means the electron will not be completely removed but can still be excited to a higher energy level.

3. **Calculate the Excitation Energy**: The energy required to excite the electron from the ground state (n=1) to a higher energy level (n=n) can be calculated using the formula:
\[ E_n = -\frac{13.6}{n^2} \]
The energy difference between the ground state and the nth state is:
\[ E_n - E_1 = -\frac{13.6}{n^2} - (-13.6) = 13.6 \left(1 - \frac{1}{n^2}\right) \]
Setting this equal to the photon energy:
\[ 13.6 \left(1 - \frac{1}{n^2}\right) = 12.1 \]

4. **Solve for n**:
\[ 1 - \frac{1}{n^2} = \frac{12.1}{13.6} \]
\[ \frac{1}{n^2} = 1 - \frac{12.1}{13.6} = \frac{1.5}{13.6} \]
\[ n^2 = \frac{13.6}{1.5} \approx 9.0667 \]
\[ n \approx 3.01 \]
Since \(n\) must be a whole number, the maximum energy level the electron can be excited to is \(n = 3\).

5. **Determine the Number of Lines in the Emission Spectrum**: The number of lines in the emission spectrum can be calculated using the formula:
\[ \text{Number of lines} = \frac{n(n-1)}{2} \]
For \(n = 3\):
\[ \text{Number of lines} = \frac{3(3-1)}{2} = \frac{3 \times 2}{2} = 3 \]

### Final Answer:
The number of lines in the emission spectrum will be **3**.