Question

The ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series for the hydrogen atom is:

• A. 4:1
• B. 1:2
• C. 1:4
• D. 2:1

Hint:

The Balmer series appears in the visible spectrum, while the Lyman series is in the ultraviolet region.

Answer 1

The Correct Option is A

Step 1: {Applying the Rydberg Formula}
The shortest wavelength is achieved when \( n_2 = \infty \) and \( n_1 \) represents the series limit. For the Balmer series (\( n_1 = 2 \)): \[ \frac{1}{\lambda_B} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{\infty} \right) \] \[ \frac{1}{\lambda_B} = RZ^2 \times \frac{1}{4} \] For the Lyman series (\( n_1 = 1 \)): \[ \frac{1}{\lambda_L} = RZ^2 \left( \frac{1}{1^2} - \frac{1}{\infty} \right) \] \[ \frac{1}{\lambda_L} = RZ^2 \times 1 \] Step 2: {Calculating the Ratio}
\[ \frac{\lambda_B}{\lambda_L} = \frac{\frac{1}{RZ^2 \times \frac{1}{4}}}{\frac{1}{RZ^2 \times 1}} \] \[ = \frac{1/4}{1} = \frac{1}{4} \] The resulting ratio is \( 4:1 \).