Question

If the wavelength of the first member of the Lyman series of hydrogen is \( \lambda \). The wavelength of the second member will be:

• A. \( \frac{27}{32} \lambda \)
• B. \( \frac{32}{27} \lambda \)
• C. \( \frac{27}{5} \lambda \)
• D. \( \frac{5}{27} \lambda \)

Answer 1

The Correct Option is A

To calculate the wavelength of the second spectral line in the Lyman series, we employ the Rydberg formula for hydrogen. The Lyman series involves electron transitions to the n=1 energy level from higher levels (n=2, n=3, etc.).

The Rydberg formula for spectral line wavelength \( \lambda \) in hydrogen is:

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

Where:

• \(R\) is the Rydberg constant (\(1.097 \times 10^7 \, \text{m}^{-1}\))
• \(n_1\) is the final energy level (\(n_1 = 1\) for the Lyman series)
• \(n_2\) is the initial energy level

The first line of the Lyman series results from a transition from \(n_2 = 2\) to \(n_1 = 1\). Its wavelength \( \lambda \) is calculated as:

\(\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \cdot \frac{3}{4}\)

The second line of the Lyman series originates from a transition from \(n_2 = 3\) to \(n_1 = 1\). Its wavelength is denoted \( \lambda_2 \) and given by:

\(\frac{1}{\lambda_2} = R \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R \left( 1 - \frac{1}{9} \right) = R \cdot \frac{8}{9}\)

To relate \( \lambda \) and \( \lambda_2 \), we divide the equation for \( \lambda \) by the equation for \( \lambda_2 \):

\(\frac{\frac{1}{\lambda}}{\frac{1}{\lambda_2}} = \frac{R \cdot \frac{3}{4}}{R \cdot \frac{8}{9}}\)

This simplifies to:

\(\frac{\lambda_2}{\lambda} = \frac{\frac{8}{9}}{\frac{3}{4}} = \frac{8}{9} \times \frac{4}{3} = \frac{32}{27}\)

Therefore, the wavelength of the second member, \( \lambda_2 \), is:

\(\lambda_2 = \frac{27}{32} \lambda\)

The result is: \( \frac{27}{32} \lambda \).