Question

The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is

• A. 2
• B. 1
• C. 4
• D. 0.5

Answer 1

The Correct Option is C

 The Balmer series and the Lyman series are two of the spectral line series for the hydrogen atom, described by 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 (\(1.097 \times 10^7 \, \text{m}^{-1}\)).
• \(n_1\) and \(n_2\) are integers with \(n_2 > n_1\).

For the Lyman series, \(n_1 = 1\) and for the Balmer series, \(n_1 = 2\).

The "last line" refers to the transition approaching the shortest wavelength (highest energy transition possible). For the Balmer series, this is when \(n_2 = \infty\), and for the Lyman series, \(n_2 = \infty\).

Calculation of Wavelengths:

1. Last line of Balmer series:

Using \(n_1 = 2\) and \(n_2 = \infty\),

\(\frac{1}{\lambda_{\text{Balmer}}} = R \left(\frac{1}{2^2} - \frac{1}{\infty}\right) = R \left(\frac{1}{4}\right)\)

Thus, \(\lambda_{\text{Balmer}} = \frac{4}{R}\)

2. Last line of Lyman series:

Using \(n_1 = 1\) and \(n_2 = \infty\),

\(\frac{1}{\lambda_{\text{Lyman}}} = R \left(\frac{1}{1^2} - \frac{1}{\infty}\right) = R\)

Thus, \(\lambda_{\text{Lyman}} = \frac{1}{R}\)

Ratio of Wavelengths:

The ratio of the wavelengths of the last line of Balmer series to the last line of Lyman series can be calculated as:

\(\frac{\lambda_{\text{Balmer}}}{\lambda_{\text{Lyman}}} = \frac{\frac{4}{R}}{\frac{1}{R}} = 4\)

Thus, the correct answer is 4.