The wavelength of first line of Balmer series is `6563Å`. The wavelength of first line of Lyman series will be

Options

A. `1215.4Å`
B. `2500Å`
C. `7500Å`
D. `600Å`

Explanation

To find the wavelength of the first line of the Lyman series given the wavelength of the first line of the Balmer series, we can use the Rydberg formula for hydrogen-like atoms.

### Step-by-Step Solution:

1. **Identify the Series and Transitions**:
- The first line of the Balmer series corresponds to the transition from \( n = 3 \) to \( n = 2 \).
- The first line of the Lyman series corresponds to the transition from \( n = 2 \) to \( n = 1 \).

2. **Rydberg Formula**:
The Rydberg formula for the wavelength of the emitted light during these transitions is given by:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)), \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

3. **Calculate for Balmer Series**:
For the Balmer series transition \( n_2 = 3 \) and \( n_1 = 2 \):
\[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
Finding a common denominator (36):
\[ \frac{1}{\lambda_B} = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) \]

4. **Calculate for Lyman Series**:
For the Lyman series transition \( n_2 = 2 \) and \( n_1 = 1 \):
\[ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]

5. **Relate the Two Wavelengths**:
Now, we can relate the two wavelengths using the equations derived:
\[ \frac{\lambda_B}{\lambda_L} = \frac{R \left( \frac{5}{36} \right)}{R \left( \frac{3}{4} \right)} = \frac{5}{36} \cdot \frac{4}{3} = \frac{20}{108} = \frac{5}{27} \]
Rearranging gives:
\[ \lambda_L = \frac{5}{27} \lambda_B \]

6. **Substitute the Value of \( \lambda_B \)**:
Given \( \lambda_B = 6563 \, \text{Å} \):
\[ \lambda_L = \frac{5}{27} \times 6563 \, \text{Å} \]

7. **Calculate \( \lambda_L \)**:
Performing the calculation:
\[ \lambda_L = \frac{5 \times 6563}{27} \approx 1215.4 \, \text{Å} \]

### Final Answer:
The wavelength of the first line of the Lyman series is approximately \( 1215.4 \, \text{Å} \).

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