Question

The ratio of the wavelengths of the spectral lines emitted due to transitions \( 3 \rightarrow 2 \) and \( 2 \rightarrow 1 \) orbits in the hydrogen atom is

• A. \( 3:1 \)
• B. \( 9:17 \)
• C. \( 27:5 \)
• D. \( 25:9 \)

Hint:

The Rydberg formula is essential for calculating wavelengths (or frequencies/energies) of spectral lines in hydrogen-like atoms. The formula is \( \frac{1}{\lambda} = R_H \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \), where \( n_i \) is the initial higher energy level and \( n_f \) is the final lower energy level. For ratio problems, it's often helpful to keep the Rydberg constant (\( R_H \)) as a variable and cancel it out at the end, simplifying calculations. Ensure correct identification of \( n_i \) and \( n_f \) for each transition.

Answer 1

The Correct Option is C

Step 1: Recall the Rydberg formula for wavelength.
The wavelength (\( \lambda \)) of a spectral line emitted during a transition from an initial energy level \( n_i \) to a final energy level \( n_f \) in a hydrogen atom is given by the Rydberg formula:
\( \frac{1}{\lambda} = R_H \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \)
where \( R_H \) is the Rydberg constant.
Step 2: Calculate \( \frac{1}{\lambda} \) for the \( 3 \rightarrow 2 \) transition.
For the transition from \( n_i = 3 \) to \( n_f = 2 \):
\( \frac{1}{\lambda_{3 \rightarrow 2}} = R_H \left(\frac{1}{2^2} - \frac{1}{3^2}\right) \)
\( \frac{1}{\lambda_{3 \rightarrow 2}} = R_H \left(\frac{1}{4} - \frac{1}{9}\right) \)
\( \frac{1}{\lambda_{3 \rightarrow 2}} = R_H \left(\frac{9 - 4}{36}\right) \)
\( \frac{1}{\lambda_{3 \rightarrow 2}} = R_H \left(\frac{5}{36}\right) \)
So, \( \lambda_{3 \rightarrow 2} = \frac{36}{5R_H} \).
Step 3: Calculate \( \frac{1}{\lambda} \) for the \( 2 \rightarrow 1 \) transition.
For the transition from \( n_i = 2 \) to \( n_f = 1 \):
\( \frac{1}{\lambda_{2 \rightarrow 1}} = R_H \left(\frac{1}{1^2} - \frac{1}{2^2}\right) \)
\( \frac{1}{\lambda_{2 \rightarrow 1}} = R_H \left(\frac{1}{1} - \frac{1}{4}\right) \)
\( \frac{1}{\lambda_{2 \rightarrow 1}} = R_H \left(\frac{4 - 1}{4}\right) \)
\( \frac{1}{\lambda_{2 \rightarrow 1}} = R_H \left(\frac{3}{4}\right) \)
So, \( \lambda_{2 \rightarrow 1} = \frac{4}{3R_H} \).
Step 4: Find the ratio of the wavelengths.
We need to find the ratio \( \frac{\lambda_{3 \rightarrow 2}}{\lambda_{2 \rightarrow 1}} \).
\( \frac{\lambda_{3 \rightarrow 2}}{\lambda_{2 \rightarrow 1}} = \frac{\frac{36}{5R_H}}{\frac{4}{3R_H}} \)
\( \frac{\lambda_{3 \rightarrow 2}}{\lambda_{2 \rightarrow 1}} = \frac{36}{5R_H} \times \frac{3R_H}{4} \)
Cancel out \( R_H \):
\( \frac{\lambda_{3 \rightarrow 2}}{\lambda_{2 \rightarrow 1}} = \frac{36 \times 3}{5 \times 4} \)
\( \frac{\lambda_{3 \rightarrow 2}}{\lambda_{2 \rightarrow 1}} = \frac{9 \times 3}{5 \times 1} \) (by dividing 36 by 4)
\( \frac{\lambda_{3 \rightarrow 2}}{\lambda_{2 \rightarrow 1}} = \frac{27}{5} \).
The ratio is \( 27:5 \).