In H–spectrum wavelength of `1^(st)` line of Balmer series is `lambda= 6561Å`. Find out wavelength of `2^(nd)` line of same series in nm.

Explanation

To find the wavelength of the second line of the Balmer series in the hydrogen spectrum, we will follow these steps:

### Step 1: Understand the Balmer Series
The Balmer series corresponds to electron transitions in a hydrogen atom where the final energy level (n_final) is 2. The first line of the Balmer series corresponds to the transition from n = 3 to n = 2.

### Step 2: Given Data
We are given the wavelength of the first line of the Balmer series:
- Wavelength (λ₁) = 6561 Å

### Step 3: Use the Rydberg Formula
The Rydberg formula for the wavelength of spectral lines in hydrogen is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2} \right) \]
where R is the Rydberg constant (approximately \(1.097 \times 10^7 \, \text{m}^{-1}\)), \(n_{final}\) is 2 for the Balmer series, and \(n_{initial}\) is the principal quantum number of the higher energy level.

### Step 4: Calculate for the First Line (n_initial = 3)
For the first line (transition from n = 3 to n = 2):
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Calculating the right side:
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) \]

### Step 5: Calculate for the Second Line (n_initial = 4)
For the second line (transition from n = 4 to n = 2):
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
Calculating the right side:
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4 - 1}{16} \right) = R \left( \frac{3}{16} \right) \]

### Step 6: Relate the Two Wavelengths
From the two equations:
\[ \frac{1}{\lambda_1} = R \left( \frac{5}{36} \right) \quad \text{and} \quad \frac{1}{\lambda_2} = R \left( \frac{3}{16} \right) \]
We can relate the two wavelengths:
\[ \frac{\lambda_2}{\lambda_1} = \frac{(5/36)}{(3/16)} = \frac{5 \cdot 16}{3 \cdot 36} = \frac{80}{108} = \frac{20}{27} \]

### Step 7: Calculate λ₂
Now, substituting λ₁ = 6561 Å into the equation:
\[ \lambda_2 = \frac{20}{27} \cdot 6561 \, \text{Å} \]
Calculating λ₂:
\[ \lambda_2 = \frac{20 \cdot 6561}{27} = 4860 \, \text{Å} \]

### Step 8: Convert to Nanometers
To convert from angstroms to nanometers, we use the conversion:
\[ 1 \, \text{Å} = 0.1 \, \text{nm} \]
Thus,
\[ \lambda_2 = 4860 \, \text{Å} = 486 \, \text{nm} \]

### Final Answer
The wavelength of the second line of the Balmer series is:
\[ \lambda_2 = 486 \, \text{nm} \]