In a mixture of `He^(o+)` gas H atom and `He^(o+)` ions Are excited to three respective first excited subsepuenly , H atom transfers its total excitation energy to `He^(o+)`ions by collision .Assuming that Bohr model of an atom is applicable , answer the following question
If each hydrogen atom in the ground state of `1.0 mol ` of H atom is excited by absorbing photon of energy `8.4 eV, 12.09 eV` and `15.0 eV` of energy then the number of spectral lines emitted is equal to

Options

A. 5
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to determine the number of spectral lines emitted when hydrogen atoms in the ground state are excited by absorbing photons of specific energies. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Energy Levels of Hydrogen
According to the Bohr model, the energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
Where \( n \) is the principal quantum number (1, 2, 3, ...).

### Step 2: Calculate the Energy Levels
- For \( n = 1 \):
\[ E_1 = -13.6 \, \text{eV} \]
- For \( n = 2 \):
\[ E_2 = -3.4 \, \text{eV} \]
- For \( n = 3 \):
\[ E_3 = -1.51 \, \text{eV} \]
- For \( n = 4 \):
\[ E_4 = -0.85 \, \text{eV} \]

### Step 3: Determine the Energy Required for Transitions
The energy required to excite an electron from one level to another can be calculated as the difference between the energy levels:
- From \( n = 1 \) to \( n = 2 \):
\[ E_{1 \to 2} = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \, \text{eV} \]
- From \( n = 1 \) to \( n = 3 \):
\[ E_{1 \to 3} = E_3 - E_1 = -1.51 - (-13.6) = 12.09 \, \text{eV} \]
- From \( n = 1 \) to \( n = 4 \):
\[ E_{1 \to 4} = E_4 - E_1 = -0.85 - (-13.6) = 12.75 \, \text{eV} \]

### Step 4: Analyze the Given Photon Energies
The energies of the photons absorbed are:
- \( 8.4 \, \text{eV} \)
- \( 12.09 \, \text{eV} \)
- \( 15.0 \, \text{eV} \)

1. **For \( 8.4 \, \text{eV} \)**: This energy is less than \( 10.2 \, \text{eV} \), so it cannot excite the electron.
2. **For \( 12.09 \, \text{eV} \)**: This energy corresponds to the transition from \( n = 1 \) to \( n = 3 \).
3. **For \( 15.0 \, \text{eV} \)**: This energy is greater than \( 13.6 \, \text{eV} \) and will ionize the hydrogen atom.

### Step 5: Determine the Number of Spectral Lines
When an electron transitions from \( n = 3 \) to \( n = 1 \), it can emit spectral lines corresponding to possible transitions:
- \( n = 3 \) to \( n = 2 \)
- \( n = 3 \) to \( n = 1 \)
- \( n = 2 \) to \( n = 1 \)

The number of spectral lines emitted can be calculated using the formula:
\[ \text{Number of spectral lines} = \frac{n(n-1)}{2} \]
Where \( n \) is the number of energy levels involved in the transitions.

In this case, we have:
- \( n = 3 \) (levels 1, 2, and 3)
\[ \text{Number of spectral lines} = \frac{3(3-1)}{2} = \frac{3 \times 2}{2} = 3 \]

### Final Answer
The number of spectral lines emitted is **3**.