Question

Find out the shortest wavelength of paschen series for H- atom

• A. \(\frac{9}{R}\)
• B. \(\frac{16}{R}\)
• C. \(\frac{144}{7R}\)
• D. \(\frac{7R}{144}\)

Answer 1

The Correct Option is A

To solve the problem of finding the shortest wavelength in the Paschen series for the hydrogen atom, we must first understand the concept of spectral series. The Paschen series occurs when electrons transition from higher energy levels to the third energy level (\(n_1 = 3\)) in a hydrogen atom.

The wavelength of the photon emitted during these transitions can be calculated using the Rydberg formula:

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

where:

• \(\lambda\) is the wavelength of the emitted photon,
• \(R_H\) is the Rydberg constant for hydrogen (\(1.097 \times 10^7 \, \text{m}^{-1}\)),
• \(n_1\) is the lower energy level (for Paschen series, \(n_1 = 3\)),
• \(n_2\) is the higher energy level (\(n_2 > n_1\)).

To find the shortest wavelength, we should focus on the transition that results in the largest energy difference, which occurs when \(n_2\) approaches infinity.

Substituting \(\infty\) for \(n_2\), the Rydberg formula becomes:

\(\frac{1}{\lambda} = R_H \left(\frac{1}{3^2} - \frac{1}{\infty^2}\right) = R_H \left(\frac{1}{9} - 0\right) = \frac{R_H}{9}\)

Thus, the shortest wavelength (\(\lambda_{\text{shortest}}\)) is:

\(\lambda_{\text{shortest}} = \frac{9}{R_H}\)

The correct answer is hence \(\frac{9}{R}\), which matches the given correct answer in the question options.