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. (a)one
B. (b)two
C. (c)four
D. (d)ten

Answer

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

### Step-by-step Solution:

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

2. **Calculate the Energy after Excitation**:
When the atom absorbs a photon of energy 13.056 eV, the total energy of the electron after excitation becomes:
\[ E_{\text{final}} = E_1 + \text{photon energy} = -13.6 \, \text{eV} + 13.056 \, \text{eV} = -0.544 \, \text{eV} \]

3. **Determine the Principal Quantum Number (n)**:
The energy of an electron in the nth orbit of a hydrogen atom is given by:
\[ E_n = -\frac{13.6}{n^2} \, \text{eV} \]
Setting this equal to the final energy we calculated:
\[ -\frac{13.6}{n^2} = -0.544 \]
Rearranging gives:
\[ \frac{13.6}{n^2} = 0.544 \quad \Rightarrow \quad n^2 = \frac{13.6}{0.544} \quad \Rightarrow \quad n^2 = 25 \]
Thus, we find:
\[ n = 5 \]

4. **Calculate the Number of Emission Lines**:
When an electron transitions from a higher energy level to a lower energy level, it can emit a photon. The number of possible emission lines (or transitions) from the excited state (n=5) to the ground state (n=1) can be calculated using the combination formula:
\[ \text{Number of emission lines} = nC2 = \frac{n(n-1)}{2} \]
For n = 5:
\[ \text{Number of emission lines} = \frac{5 \times 4}{2} = 10 \]

5. **Final Answer**:
The number of emission lines produced is **10**.