Question

The smallest wavelength of Lyman series is \(91\ \text{nm}\). The difference between the largest wavelengths of Paschen and Balmer series is nearly _________ nm.

• A. 1784
• B. 1217
• C. 1875
• D. 1550

Hint:

Largest wavelength in a spectral series corresponds to transition between nearest energy levels.

Answer 1

The Correct Option is B

Step 1: Use the Rydberg relation for hydrogen spectrum.
Smallest wavelength of Lyman series: \[ \lambda_{\min} = \frac{1}{R} = 91\ \text{nm} \]
Step 2: Find largest wavelengths of Balmer and Paschen series.
Largest wavelength of Balmer series corresponds to transition \(n=3 \to 2\): \[ \lambda_B = \frac{1}{R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)} = \frac{36}{5R} \] Largest wavelength of Paschen series corresponds to transition \(n=4 \to 3\): \[ \lambda_P = \frac{1}{R\left(\frac{1}{3^2}-\frac{1}{4^2}\right)} = \frac{144}{7R} \]
Step 3: Calculate the difference.
\[ \lambda_P - \lambda_B = \frac{144}{7R} - \frac{36}{5R} = \frac{468}{35R} \] \[ = \frac{468}{35}\times 91 \approx 1217\ \text{nm} \]
Final Answer: \[ \boxed{1217\ \text{nm}} \]