Question

The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number $Z$ of hydrogen like ion is

• A. 3
• B. 4
• C. 1
• D. 2

Answer 1

The Correct Option is D

To solve this problem, we need to equate the wavelengths of the Lyman series of hydrogen and the Balmer series of a hydrogen-like ion.

The Lyman series corresponds to transitions where the final state is n_f = 1. The wavelength of the first line in the Lyman series (transition from n = 2 to n = 1) is given by the Rydberg formula:

\[ \dfrac{1}{\lambda_{1,\text{Lyman}}} = R_H \left( 1 - \dfrac{1}{4} \right) \]

where R_H is the Rydberg constant for hydrogen.

For the Balmer series of a hydrogen-like ion, the series corresponds to transitions ending at n_f = 2. The wavelength of the second line (transition from n = 4 to n = 2) for a hydrogen-like ion with atomic number Z is given by:

\[ \dfrac{1}{\lambda_{2,\text{Balmer}}} = R_H Z^2 \left( \dfrac{1}{4} - \dfrac{1}{16} \right) \]

Since the wavelength of the first line of the Lyman series of hydrogen equals the second line of the Balmer series of the hydrogen-like ion, we equate the expressions:

\[ R_H \left( \dfrac{3}{4} \right) = R_H Z^2 \left( \dfrac{3}{16} \right) \]

Cancel R_H from both sides and solve for Z^2:

\[ \dfrac{3}{4} = Z^2 \dfrac{3}{16} \implies Z^2 = 4 \implies Z = 2 \]

Thus, the atomic number Z of the hydrogen-like ion is 2. The correct answer is 2.