The first line in Balmer series corresponds to `n_1 = 2 and n_2 = 3` and the limiting line corresponds to `n_1 = 2 and n_2 = oo`. Calculate the wavelengths of the first and limiting lines in Balmer series.

Explanation

To calculate the wavelengths of the first and limiting lines in the Balmer series, we can use the Rydberg formula:

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

where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

### Step 1: Calculate the wavelength of the first line in the Balmer series

For the first line, we have:
- \( n_1 = 2 \)
- \( n_2 = 3 \)

Substituting these values into the Rydberg formula:

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

Calculating \( \frac{1}{2^2} \) and \( \frac{1}{3^2} \):

\[ \frac{1}{2^2} = \frac{1}{4} = 0.25 \]
\[ \frac{1}{3^2} = \frac{1}{9} \approx 0.1111 \]

Now, substituting these values back into the equation:

\[ \frac{1}{\lambda} = R \left( 0.25 - 0.1111 \right) \]
\[ \frac{1}{\lambda} = R \left( 0.1389 \right) \]

Now substituting the value of \( R \):

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times 0.1389 \]
\[ \frac{1}{\lambda} \approx 1.527 \times 10^6 \, \text{m}^{-1} \]

Now, taking the reciprocal to find \( \lambda \):

\[ \lambda \approx \frac{1}{1.527 \times 10^6} \approx 6.54 \times 10^{-7} \, \text{m} = 654 \, \text{nm} \]

### Step 2: Calculate the wavelength of the limiting line in the Balmer series

For the limiting line, we have:
- \( n_1 = 2 \)
- \( n_2 = \infty \)

Substituting these values into the Rydberg formula:

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

Since \( \frac{1}{\infty^2} = 0 \):

\[ \frac{1}{\lambda} = R \left( \frac{1}{4} \right) \]
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times 0.25 \]
\[ \frac{1}{\lambda} = 2.7425 \times 10^6 \, \text{m}^{-1} \]

Now, taking the reciprocal to find \( \lambda \):

\[ \lambda \approx \frac{1}{2.7425 \times 10^6} \approx 3.65 \times 10^{-7} \, \text{m} = 365 \, \text{nm} \]

### Final Results:
- Wavelength of the first line in the Balmer series: **654 nm**
- Wavelength of the limiting line in the Balmer series: **365 nm**