Question

In hydrogen spectrum, the ratio of wavelengths of the last line of Lyman series and that of the last line of Balmer series is

• A. 1
• B. 0.5
• C. 0.25
• D. 0.2

Hint:

Shortest wavelength (last line) ratio is always the ratio of the squares of the lower orbits: $\frac{\lambda_1}{\lambda_2} = \frac{n_1^2}{n_2^2}$.

Answer 1

The Correct Option is C

Step 1: Concept
The wavelength $\lambda$ of spectral lines is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.

Step 2: Meaning
The "last line" or series limit occurs when the electron transitions from $n_2 = \infty$ to the series ground state $n_1$.

Step 3: Analysis
For Lyman series: $n_1 = 1, n_2 = \infty \implies \frac{1}{\lambda_L} = R(1/1^2 - 0) = R \implies \lambda_L = 1/R$.
For Balmer series: $n_1 = 2, n_2 = \infty \implies \frac{1}{\lambda_B} = R(1/2^2 - 0) = R/4 \implies \lambda_B = 4/R$.
Ratio $\frac{\lambda_L}{\lambda_B} = \frac{1/R}{4/R} = \frac{1}{4}$.

Step 4: Conclusion
The ratio is $0.25$. Final Answer: (C)