Question

What will be the ratio of the wavelength of the 3rd line of the Paschen Series to the 2nd line of the Balmer series of H-atom?

• A. \( \frac{9}{4} \)
• B. \( \frac{3}{2} \)
• C. \( \frac{2}{3} \)
• D. \( \frac{16}{4} \)

Hint:

For spectral lines in the hydrogen atom, use the Rydberg formula to find the wavelengths of different series. The ratio of wavelengths can be easily calculated by comparing the differences in energy levels.

Answer 1

The Correct Option is A

Step 1: Use the Rydberg formula for wavelengths.
The wavelength of a spectral line in the hydrogen atom 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, - \( R_H \) is the Rydberg constant, - \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final energy levels, respectively.
Step 2: Apply to the Paschen series.
For the 3rd line of the Paschen series, the electron transitions from \( n_2 = 5 \) to \( n_1 = 3 \). Using the Rydberg formula: \[ \frac{1}{\lambda_{\text{P3}}} = R_H \left( \frac{1}{3^2} - \frac{1}{5^2} \right) \] \[ \frac{1}{\lambda_{\text{P3}}} = R_H \left( \frac{1}{9} - \frac{1}{25} \right) \] \[ \frac{1}{\lambda_{\text{P3}}} = R_H \times \frac{16}{225} \] \[ \lambda_{\text{P3}} = \frac{225}{16R_H} \]
Step 3: Apply to the Balmer series.
For the 2nd line of the Balmer series, the electron transitions from \( n_2 = 4 \) to \( n_1 = 2 \). Using the Rydberg formula: \[ \frac{1}{\lambda_{\text{B2}}} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \] \[ \frac{1}{\lambda_{\text{B2}}} = R_H \left( \frac{1}{4} - \frac{1}{16} \right) \] \[ \frac{1}{\lambda_{\text{B2}}} = R_H \times \frac{3}{16} \] \[ \lambda_{\text{B2}} = \frac{16}{3R_H} \]
Step 4: Find the ratio of the wavelengths.
Now, we can find the ratio of the wavelengths of the 3rd line of the Paschen series to the 2nd line of the Balmer series: \[ \frac{\lambda_{\text{P3}}}{\lambda_{\text{B2}}} = \frac{\frac{225}{16R_H}}{\frac{16}{3R_H}} = \frac{225}{16} \times \frac{3}{16} = \frac{9}{4} \] Thus, the ratio of the wavelengths is \( \frac{9}{4} \).