Question

A spectral line in the hydrogen spectrum is due to an electron transition between energy levels whose sum and difference are respectively 5 and 3. The wavelength of it (in cm) is:

• A. \( \frac{16}{15R} \)
• B. \( \frac{15}{16R} \)
• C. \( \frac{16}{15} \)
• D. \( \frac{15}{16R} \)

Hint:

When dealing with spectral lines, always check if you need to find \( \lambda \) or \( \frac{1}{\lambda} \) based on the given or derived formula.

Answer 1

The Correct Option is A

1. Define Variables
Let the two energy levels be \( n_1 \) and \( n_2 \) (where \( n_2 > n_1 \)). We are given:
Sum of energy levels: \( n_1 + n_2 = 5 \)
Difference of energy levels: \( n_2 - n_1 = 3 \)
2. Solve for Energy Levels
Add the two equations:
\[ 2n_2 = 8 \] \[ n_2 = 4 \] Substitute \( n_2 \) back into either equation to find \( n_1 \):
\[ n_1 + 4 = 5 \] \[ n_1 = 1 \] 3. Rydberg Formula
The wavelength (\( \lambda \)) of the emitted photon when an electron transitions between energy levels \( n_1 \) and \( n_2 \) in the hydrogen atom is given by the Rydberg formula: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \( R_H \) is the Rydberg constant for hydrogen.
4. Substitute Values
Substitute \( n_1 = 1 \), \( n_2 = 4 \), and \( R_H = R \) (as given in the options) into the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{4^2} \right) \] \[ \frac{1}{\lambda} = R \left( 1 - \frac{1}{16} \right) \] \[ \frac{1}{\lambda} = R \left( \frac{15}{16} \right) \] 5. Solve for Wavelength
\[ \lambda = \frac{16}{15R} \] Therefore, the wavelength of the spectral line is \( \frac{16}{15R} \) cm.
The correct answer is (1) \( \frac{16}{15R} \).