the wavelength of the first line of lyman series is `1215 Å`, the wavelength of first line of balmer series will be

Options

A. `4545 Å`
B. `5295 Å`
C. `6561 Å`
D. `6750 Å`

Explanation

To find the wavelength of the first line of the Balmer series given the wavelength of the first line of the Lyman series, we can follow these steps:

### Step 1: Understand the Lyman Series
The Lyman series corresponds to electron transitions in a hydrogen atom where the final energy level (n_final) is 1. The first line of the Lyman series occurs when the electron transitions from n_initial = 2 to n_final = 1.

### Step 2: Use the Rydberg Formula for Lyman Series
The Rydberg formula for the wavelength (λ) of the spectral lines is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2} \right) \]
For the first line of the Lyman series:
- n_final = 1
- n_initial = 2

Substituting these values into the formula:
\[ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
Thus,
\[ \lambda_L = \frac{4}{3R} \]

### Step 3: Understand the Balmer Series
The Balmer series corresponds to transitions where the final energy level (n_final) is 2. The first line of the Balmer series occurs when the electron transitions from n_initial = 3 to n_final = 2.

### Step 4: Use the Rydberg Formula for Balmer Series
For the first line of the Balmer series:
- n_final = 2
- n_initial = 3

Substituting these values into the Rydberg formula:
\[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
Finding a common denominator (36):
\[ \frac{1}{\lambda_B} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \]
Thus,
\[ \lambda_B = \frac{36}{5R} \]

### Step 5: Relate the Two Wavelengths
From the previous steps, we have:
\[ \lambda_L = \frac{4}{3R} \quad \text{and} \quad \lambda_B = \frac{36}{5R} \]
To find the ratio of the two wavelengths:
\[ \frac{\lambda_B}{\lambda_L} = \frac{\frac{36}{5R}}{\frac{4}{3R}} = \frac{36}{5} \cdot \frac{3}{4} = \frac{27}{5} \]
Thus, we can express λ_B in terms of λ_L:
\[ \lambda_B = \frac{27}{5} \lambda_L \]

### Step 6: Substitute the Given Wavelength
Given that the wavelength of the first line of the Lyman series, λ_L = 1215 Å, we can substitute this value:
\[ \lambda_B = \frac{27}{5} \times 1215 \]
Calculating this:
\[ \lambda_B = \frac{27 \times 1215}{5} = \frac{32805}{5} = 6561 \text{ Å} \]

### Final Answer
The wavelength of the first line of the Balmer series is:
\[ \lambda_B = 6561 \text{ Å} \]