If each hydrogen atom in the ground state `1.0 mol of H ` atom are excited by axeited by absorbing photon of energy `8.4 eV, 12.09 eV and 15.0 eV` of energy, then number of spectral lines emitted is equal to

Options

A. 1
B. 2
C. 3
D. 4

Explanation

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

### Step 1: Determine the Energy Levels of Hydrogen
The energy levels for a hydrogen atom can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

- For \( n = 1 \):
\[ E_1 = -\frac{13.6}{1^2} = -13.6 \, \text{eV} \]

- 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 \( n = 4 \):
\[ E_4 = -\frac{13.6}{4^2} = -0.85 \, \text{eV} \]

### Step 2: Calculate the Energy Differences
Next, we calculate the energy differences between the levels that correspond to the energies of the absorbed photons.

- **For 8.4 eV**:
- The energy difference between \( n=1 \) and \( n=2 \):
\[ E_2 - E_1 = (-3.4) - (-13.6) = 10.2 \, \text{eV} \]
- No transitions match 8.4 eV.

- **For 12.09 eV**:
- The energy difference between \( n=1 \) and \( n=3 \):
\[ E_3 - E_1 = (-1.51) - (-13.6) = 12.09 \, \text{eV} \]
- This transition is valid.

- **For 15.0 eV**:
- The energy difference between \( n=1 \) and \( n=4 \):
\[ E_4 - E_1 = (-0.85) - (-13.6) = 12.75 \, \text{eV} \]
- This energy exceeds the ionization energy (13.6 eV), so it will ionize the atom.

### Step 3: Determine the Spectral Lines
Now, we determine the spectral lines emitted from the transitions:

1. **For the transition from \( n=1 \) to \( n=3 \)** (12.09 eV):
- Possible transitions:
- \( n=3 \) to \( n=2 \)
- \( n=3 \) to \( n=1 \)
- \( n=2 \) to \( n=1 \)
- This results in 3 spectral lines.

2. **For the transition from \( n=1 \) to \( n=4 \)** (15.0 eV):
- The atom is ionized and can emit spectral lines from \( n=4 \) to \( n=1 \) and \( n=2 \).
- Possible transitions:
- \( n=4 \) to \( n=3 \)
- \( n=4 \) to \( n=2 \)
- \( n=4 \) to \( n=1 \)
- \( n=3 \) to \( n=2 \)
- \( n=3 \) to \( n=1 \)
- \( n=2 \) to \( n=1 \)
- This results in 6 spectral lines.

### Step 4: Total Spectral Lines
Adding the spectral lines from both cases:
- From \( 12.09 \, \text{eV} \): 3 lines
- From \( 15.0 \, \text{eV} \): 6 lines

Total spectral lines emitted:
\[ 3 + 6 = 9 \]

### Final Answer
The number of spectral lines emitted is **9**.

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