An H- atom in the ground state is excited by monochromatic radiation of photon energy 13.056 eV. The number of emission lines will be (given its ionisation energy is 13.6 eV)

Options

A. one
B. two
C. four
D. ten

Answer

To solve the problem, we need to determine the number of emission lines produced when an H-atom in the ground state is excited by a photon of energy 13.056 eV. The ionization energy of the hydrogen atom is given as 13.6 eV.

### Step-by-Step Solution:

1. **Identify the Ground State Energy**:
The energy of the hydrogen atom in the ground state (n=1) is given by:
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

2. **Calculate the Energy Required for Ionization**:
To ionize the hydrogen atom from the ground state, we need to supply energy equal to the ionization energy:
\[ \text{Ionization Energy} = 13.6 \, \text{eV} \]

3. **Determine the Energy Supplied by the Photon**:
The energy of the photon used to excite the atom is:
\[ E_{\text{photon}} = 13.056 \, \text{eV} \]

4. **Calculate the Energy Difference**:
The energy difference between the ionization energy and the photon energy is:
\[ \Delta E = E_{\text{ionization}} - E_{\text{photon}} = 13.6 \, \text{eV} - 13.056 \, \text{eV} = 0.544 \, \text{eV} \]

5. **Determine the Excitation Level**:
The remaining energy (0.544 eV) corresponds to the energy of the excited state. We can find the energy levels for n=2, n=3, n=4, and n=5:
- For n=2: \( E_2 = -\frac{13.6}{2^2} = -3.4 \, \text{eV} \)
- For n=3: \( E_3 = -\frac{13.6}{3^2} = -1.51 \, \text{eV} \)
- For n=4: \( E_4 = -\frac{13.6}{4^2} = -0.85 \, \text{eV} \)
- For n=5: \( E_5 = -\frac{13.6}{5^2} = -0.544 \, \text{eV} \)

Since the remaining energy of 0.544 eV allows the electron to reach n=5, the atom can be excited to the n=5 state.

6. **Calculate the Number of Emission Lines**:
The number of spectral lines (emission lines) produced when an electron transitions from a higher energy level to lower energy levels is given by the formula:
\[ \text{Number of emission lines} = \frac{n(n-1)}{2} \]
where \( n \) is the highest energy level reached. Here, \( n = 5 \):
\[ \text{Number of emission lines} = \frac{5(5-1)}{2} = \frac{5 \times 4}{2} = 10 \]

### Final Answer:
The number of emission lines will be **10**.