Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the `2^(nd)` Balmer line (n = 4 to n = 2) will be :

Options

A. `889.2 nm`
B. `488.9 nm`
C. `642.7 nm`
D. `388.9 nm`

Answer

To find the wavelength of the second Balmer line (from n = 4 to n = 2) given the wavelength of the first Balmer line (from n = 3 to n = 2) as 660 nm, we can use the Rydberg formula for hydrogen:

### Step-by-Step Solution:

1. **Understand the Rydberg Formula**: The Rydberg formula for the wavelength of light emitted during electron transitions in hydrogen is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \( R \) is the Rydberg constant, \( n_f \) is the final energy level, and \( n_i \) is the initial energy level.

2. **Calculate for the First Balmer Line**: For the first Balmer line (n = 3 to n = 2):
- Here, \( n_f = 2 \) and \( n_i = 3 \).
- The wavelength \( \lambda_1 \) is given as 660 nm.
- Plugging into the formula:
\[ \frac{1}{660} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
- Calculating the right-hand side:
\[ \frac{1}{2^2} = \frac{1}{4}, \quad \frac{1}{3^2} = \frac{1}{9} \]
- Therefore,
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
- Thus, we have:
\[ \frac{1}{660} = R \cdot \frac{5}{36} \]

3. **Calculate for the Second Balmer Line**: For the second Balmer line (n = 4 to n = 2):
- Here, \( n_f = 2 \) and \( n_i = 4 \).
- Using the Rydberg formula again:
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
- Calculating the right-hand side:
\[ \frac{1}{2^2} = \frac{1}{4}, \quad \frac{1}{4^2} = \frac{1}{16} \]
- Therefore,
\[ \frac{1}{4} - \frac{1}{16} = \frac{4 - 1}{16} = \frac{3}{16} \]
- Thus, we have:
\[ \frac{1}{\lambda_2} = R \cdot \frac{3}{16} \]

4. **Relating the Two Equations**: Now we have two equations:
\[ \frac{1}{660} = R \cdot \frac{5}{36} \quad \text{(1)} \]
\[ \frac{1}{\lambda_2} = R \cdot \frac{3}{16} \quad \text{(2)} \]
- From (1), we can express \( R \):
\[ R = \frac{1}{660} \cdot \frac{36}{5} \]
- Substitute \( R \) into (2):
\[ \frac{1}{\lambda_2} = \left( \frac{1}{660} \cdot \frac{36}{5} \right) \cdot \frac{3}{16} \]

5. **Calculate \( \lambda_2 \)**:
- Simplifying the right-hand side:
\[ \frac{1}{\lambda_2} = \frac{36 \cdot 3}{5 \cdot 16 \cdot 660} \]
- Calculate \( \lambda_2 \):
\[ \lambda_2 = \frac{5 \cdot 16 \cdot 660}{36 \cdot 3} \]
- Performing the calculations:
\[ \lambda_2 = \frac{5 \cdot 16 \cdot 660}{108} = \frac{52800}{108} \approx 489.44 \text{ nm} \]

### Final Answer:
The wavelength of the second Balmer line (n = 4 to n = 2) is approximately **489.44 nm**.