Question

In hydrogen type atom, shortest wavelength in Lyman series is given as 91 nm. Then the longest wavelength in Paschen series of this atom shall be

• A. 31.82 nm
• B. 113.3 nm
• C. 1.87 μm
• D. 2.31 μm

Hint:

The longest wavelength in a series corresponds to the transition between the lowest possible \( n_2 \) and \( n_1 \).

Answer 1

The Correct Option is C

Step 1: Use the Rydberg formula.
The Rydberg formula for the wavelengths of spectral lines in hydrogen atom is: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right), \] where \( R_H \) is the Rydberg constant, \( n_1 \) and \( n_2 \) are the principal quantum numbers for the two energy levels. Step 2: Shortest wavelength in Lyman series.
For Lyman series, the shortest wavelength corresponds to the transition from \( n_2 = 2 \) to \( n_1 = 1 \), which gives the given wavelength of 91 nm. Step 3: Longest wavelength in Paschen series.
For the longest wavelength in the Paschen series, the transition is from \( n_2 = 4 \) to \( n_1 = 3 \), which results in a wavelength of 1.87 μm. Final Answer: \[ \boxed{1.87 \, \mu\text{m}}. \]