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: {Using the Rydberg Formula}
The shortest wavelength occurs when $n_2 = \infty$ and $n_1$ corresponds to 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: {Finding the Ratio}
$$\frac{\lambda_B}{\lambda_L} = \frac{\frac{1}{RZ^2 \times \frac{1}{4}}}{\frac{1}{RZ^2 \times 1}}$$ $$= \frac{1/4}{1} = \frac{1}{4}$$ Thus, the ratio is $4:1$.