The ionization energy of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by electromagnetic radiation of energy 12.1 eV. How many spectral lines will be emitted by the hydrogen atoms

Options

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

Answer

To solve the problem of how many spectral lines will be emitted by hydrogen atoms that are excited by electromagnetic radiation of energy 12.1 eV, we can follow these steps:

### Step 1: Determine the initial state of the hydrogen atom
The hydrogen atom is initially in the ground state, which corresponds to \( n = 1 \).

### Step 2: Calculate the energy of the excited state
The energy of the ground state of hydrogen is given as:
\[ E_1 = -13.6 \text{ eV} \]
When the hydrogen atom absorbs energy of 12.1 eV, the total energy of the atom becomes:
\[ E_{excited} = E_1 + 12.1 \text{ eV} = -13.6 \text{ eV} + 12.1 \text{ eV} = -1.5 \text{ eV} \]

### Step 3: Find the excited state \( n \)
The energy of the nth excited state of hydrogen is given by:
\[ E_n = -\frac{13.6}{n^2} \text{ eV} \]
Setting this equal to the energy we found in step 2:
\[ -\frac{13.6}{n^2} = -1.5 \]
This simplifies to:
\[ \frac{13.6}{n^2} = 1.5 \]
Rearranging gives:
\[ n^2 = \frac{13.6}{1.5} \approx 9.0667 \]
Taking the square root:
\[ n \approx 3 \]

### Step 4: Determine possible transitions
The electron can be excited to \( n = 3 \). From this state, it can transition back to lower energy levels. The possible transitions are:
1. From \( n = 3 \) to \( n = 2 \)
2. From \( n = 2 \) to \( n = 1 \)
3. From \( n = 3 \) to \( n = 1 \)

### Step 5: Count the spectral lines
Each transition corresponds to a spectral line. The transitions are:
- \( 3 \rightarrow 2 \)
- \( 2 \rightarrow 1 \)
- \( 3 \rightarrow 1 \)

Thus, the total number of spectral lines emitted will be 3.

### Final Answer:
The total number of spectral lines emitted by the hydrogen atoms is **3**.

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