Question

In hydrogen spectrum, if the longest wavelength of the spectral line in Balmer series is \(\lambda\), then the shortest wavelength of the spectral line in Paschen series is

• A. \(5\lambda\)
• B. \(3.75\lambda\)
• C. \(2.5\lambda\)
• D. \(1.25\lambda\)

Hint:

For hydrogen spectrum, \[ \boxed{ \frac1\lambda = R\left(\frac1{n_1^2}-\frac1{n_2^2}\right) } \] Longest wavelength corresponds to the smallest energy transition, while shortest wavelength corresponds to the series limit.

Answer 1

The Correct Option is D

Step 1: Longest wavelength in Balmer series. The longest wavelength corresponds to the transition \[ 3\rightarrow2. \] Using Rydberg's formula, \[ \frac{1}{\lambda} = R\left(\frac14-\frac19\right) = \frac{5R}{36}. \] Thus, \[ \lambda = \frac{36}{5R}. \]

Step 2: Shortest wavelength in Paschen series. The shortest wavelength corresponds to the transition \[ \infty\rightarrow3. \] Hence, \[ \frac{1}{\lambda_P} = R\left(\frac19\right), \] \[ \lambda_P = \frac{9}{R}. \]

Step 3: Find the ratio. \[ \frac{\lambda_P}{\lambda} = \frac{9/R}{36/(5R)} = \frac54. \] Therefore, \[ \lambda_P = \frac54\lambda = 1.25\lambda. \] Hence, \[ \boxed{1.25\lambda} \] Therefore, \[ \boxed{(D)} \] is the correct answer.