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 \( 4:1 \).

Answer 2

Step 1: Understanding the Question
We are asked to find the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series in a hydrogen atom.

Step 2: Using the Rydberg Formula
The wavelength of emitted light in hydrogen is given by the Rydberg formula: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where:

\( \lambda \) is the wavelength of emitted radiation,

\( R \) is the Rydberg constant,

\( n_1 \) is the lower energy level,

\( n_2 \) is the higher energy level.

 

Step 3: Shortest Wavelengths
The shortest wavelength corresponds to the transition from the highest possible energy level to the series’ base level.

For Lyman series:
\[ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \] \[ \Rightarrow \lambda_L = \frac{1}{R} \]

For Balmer series:
\[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \cdot \frac{1}{4} \] \[ \Rightarrow \lambda_B = \frac{4}{R} \]

Step 4: Ratio of Wavelengths
\[ \frac{\lambda_B}{\lambda_L} = \frac{\frac{4}{R}}{\frac{1}{R}} = 4 \] Hence, the required ratio is: \[ \boxed{4:1} \]

Final Answer: Option 1 — 4:1