Wavelength of first line in Lyman series is `lambda`. What is wavelength of first line in Balmer series?

Options

A. (a)`(5)/(27)lambda`
B. (b)`(36)/(5)lambda`
C. (c)`(27)/(5)lambda`
D. (d)`(5)/(36)lambda`

Answer

To solve the problem of finding the wavelength of the first line in the Balmer series given that the wavelength of the first line in the Lyman series is λ, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Series and Transitions:**
- For the Lyman series, the transition is from n2 = 2 to n1 = 1.
- For the Balmer series, the transition is from n2 = 3 to n1 = 2.

2. **Use the Rydberg Formula:**
The Rydberg formula for the wavelength (λ) of spectral lines is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where R is the Rydberg constant.

3. **Calculate Wavelength for Lyman Series:**
For the Lyman series (first line):
- n1 = 1, n2 = 2
\[ \frac{1}{\lambda_{L}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
Thus,
\[ \lambda_{L} = \frac{4}{3R} \]

4. **Calculate Wavelength for Balmer Series:**
For the Balmer series (first line):
- n1 = 2, n2 = 3
\[ \frac{1}{\lambda_{B}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
To combine the fractions:
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus,
\[ \frac{1}{\lambda_{B}} = R \left( \frac{5}{36} \right) \implies \lambda_{B} = \frac{36}{5R} \]

5. **Relate Wavelengths of Lyman and Balmer Series:**
We know that:
\[ \lambda_{L} = \frac{4}{3R} \quad \text{and} \quad \lambda_{B} = \frac{36}{5R} \]
To find the relationship between λ (wavelength of Lyman series) and λ_B (wavelength of Balmer series):
\[ \frac{\lambda_{L}}{\lambda_{B}} = \frac{\frac{4}{3R}}{\frac{36}{5R}} = \frac{4 \cdot 5}{3 \cdot 36} = \frac{20}{108} = \frac{5}{27} \]
Thus,
\[ \lambda_{B} = \frac{27}{5} \lambda_{L} \]

6. **Substituting λ:**
Since λ = λ_L, we have:
\[ \lambda_{B} = \frac{27}{5} \lambda \]

### Final Answer:
The wavelength of the first line in the Balmer series is:
\[ \lambda_B = \frac{27}{5} \lambda \]