Question

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

• A. 1
• B. 4
• C. 0.5
• D. 2

Answer 1

The Correct Option is B

 To solve the question, we need to understand the Balmer and Lyman series in the hydrogen atom:

• The Lyman series corresponds to electron transitions from higher energy levels (\(n_2\) > 1) to \(n_1 = 1\).
• The Balmer series corresponds to electron transitions from higher energy levels (\(n_2\) > 2) to \(n_1 = 2\).

The wavelength of a spectral line in hydrogen can be calculated using the Rydberg formula:

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

where \(R\) is the Rydberg constant.

For the last line of the series (i.e., as \(n_2 \to \infty\)):

• Last line of Lyman series: \(\frac{1}{\lambda_{Lyman}} = R \left( \frac{1}{1^2} - 0 \right) = R\)
• Last line of Balmer series: \(\frac{1}{\lambda_{Balmer}} = R \left( \frac{1}{2^2} - 0 \right) = \frac{R}{4}\)

The ratio of their wavelengths is:

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

Therefore, the ratio of wavelengths of the last line of the Balmer series to the last line of the Lyman series is 4. This means the correct answer is:

• 4

Hence, the correct answer is option 4.