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 given information
We are given the wavelength of the first line of the Lyman series, which is \( \lambda_L = 1215 \, \text{Å} \).

### Step 2: Use the Rydberg formula for the Lyman series
The Rydberg formula for the wavelength of spectral lines in hydrogen is given by:

\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

For the first line of the Lyman series, \( n_1 = 1 \) and \( n_2 = 2 \):

\[ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]

### Step 3: Calculate \( R \) using \( \lambda_L \)
From the above equation, we can express \( R \):

\[ \frac{1}{\lambda_L} = \frac{3R}{4} \]

Rearranging gives:

\[ R = \frac{4}{3\lambda_L} \]

Substituting \( \lambda_L = 1215 \, \text{Å} \):

\[ R = \frac{4}{3 \times 1215} = \frac{4}{3645} \, \text{Å}^{-1} \]

### Step 4: Use the Rydberg formula for the Balmer series
For the first line of the Balmer series, \( n_1 = 2 \) and \( n_2 = 3 \):

\[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]

### Step 5: Calculate \( \frac{1}{\lambda_B} \)
Finding a common denominator (36):

\[ \frac{1}{\lambda_B} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \]

### Step 6: Substitute \( R \) into the equation for \( \lambda_B \)
Substituting \( R = \frac{4}{3 \times 1215} \):

\[ \frac{1}{\lambda_B} = \frac{4}{3 \times 1215} \cdot \frac{5}{36} \]

### Step 7: Calculate \( \lambda_B \)
Taking the reciprocal gives:

\[ \lambda_B = \frac{3 \times 1215 \times 36}{4 \times 5} \]

Calculating the values:

\[ \lambda_B = \frac{3 \times 1215 \times 36}{20} \]

Calculating \( 3 \times 1215 = 3645 \):

\[ \lambda_B = \frac{3645 \times 36}{20} \]

Calculating \( 3645 \times 36 = 131220 \):

\[ \lambda_B = \frac{131220}{20} = 6561 \, \text{Å} \]

### Final Answer
The wavelength of the first line of the Balmer series is \( \lambda_B = 6561 \, \text{Å} \).

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