If the wavelength of the first line of the Balmer series of hydrogen is `6561 Å`, the wavelngth of the second line of the series should be

Explanation

To find the wavelength of the second line of the Balmer series of hydrogen, we can use the Rydberg formula for hydrogen spectral lines:

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

where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant,
- \( n_f \) is the final energy level (for Balmer series, \( n_f = 2 \)),
- \( n_i \) is the initial energy level.

### Step 1: Determine the first line of the Balmer series
For the first line of the Balmer series, \( n_i = 3 \) (transition from \( n=3 \) to \( n=2 \)).

Using the formula:
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

Calculating the values:
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
Finding a common denominator (36):
\[ \frac{1}{\lambda_1} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \]

### Step 2: Determine the second line of the Balmer series
For the second line of the Balmer series, \( n_i = 4 \) (transition from \( n=4 \) to \( n=2 \)).

Using the formula:
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

Calculating the values:
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{4} - \frac{1}{16} \right) \]
Finding a common denominator (16):
\[ \frac{1}{\lambda_2} = R \left( \frac{4}{16} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right) \]

### Step 3: Relate the two wavelengths
Now we have two equations:
1. \(\frac{1}{\lambda_1} = R \left( \frac{5}{36} \right)\)
2. \(\frac{1}{\lambda_2} = R \left( \frac{3}{16} \right)\)

Dividing the two equations:
\[ \frac{\lambda_2}{\lambda_1} = \frac{\frac{5}{36}}{\frac{3}{16}} = \frac{5 \times 16}{36 \times 3} = \frac{80}{108} = \frac{20}{27} \]

### Step 4: Calculate \( \lambda_2 \)
Given that \( \lambda_1 = 6561 \, \text{Å} \):
\[ \lambda_2 = \frac{20}{27} \times 6561 \]

Calculating:
\[ \lambda_2 = \frac{20 \times 6561}{27} = 4860 \, \text{Å} \]

Thus, the wavelength of the second line of the Balmer series is \( \lambda_2 = 4860 \, \text{Å} \).

### Summary
The wavelength of the second line of the Balmer series is \( 4860 \, \text{Å} \).