Question

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

• A. 4
• B. 1
• C. 2
• D. 3

Hint:

When solving problems involving the Lyman and Balmer series, use the Rydberg formula and equate the wavelengths of corresponding lines in different series to find the atomic number.

Answer 1

The Correct Option is C

To determine the atomic number Z of an H-like ion where the wavelength matches between two different transitions, we analyze the energy transitions in hydrogen-like atoms.

1. Lyman Series Transition (Hydrogen Atom):
For the Lyman series (transition to n=1):

$ \frac{hc}{\lambda} = R_h c \left(1 - \frac{1}{n_2^2}\right) $
For the first line (n=2 → n=1):

$ \frac{hc}{\lambda} = R_h c \left(1 - \frac{1}{4}\right) = \frac{3}{4} R_h c $

2. Balmer Series Transition (H-like Ion):
For the second line of Balmer series (n=4 → n=2) in an H-like ion:

$ \frac{hc}{\lambda} = Z^2 R_h c \left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3}{16} Z^2 R_h c $

3. Equating the Transitions:
Since both transitions produce the same wavelength:

$ \frac{3}{4} R_h c = \frac{3}{16} Z^2 R_h c $
Canceling common terms:

$ \frac{3}{4} = \frac{3}{16} Z^2 $

4. Solving for Z:
$ Z^2 = \frac{3/4}{3/16} = 4 $
$ Z = \sqrt{4} = 2 $

Final Answer:
The atomic number of the H-like ion is $2$.

Answer 2

To find the atomic number \( Z \) of the H-like ion, we use the Rydberg formula for the wavelength of emitted light during electron transitions in hydrogen-like atoms:

For hydrogen, the Rydberg formula for the Lyman series transition (from \( n = 2 \) to \( n = 1 \)) is:

\[\frac{1}{\lambda_1} = R_H \left(1 - \frac{1}{4}\right) = \frac{3R_H}{4}\]

For the H-like ion, the formula for the Balmer series (from \( n = 4 \) to \( n = 2 \)) is:

\[\frac{1}{\lambda_2} = RZ^2 \left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3RZ^2}{16}\]

We are given that the wavelength of the first line of the Lyman series for hydrogen is equal to the second line of the Balmer series for the ion. Thus:

\[\frac{3R_H}{4} = \frac{3RZ^2}{16}\]

Simplifying and solving for \( Z \):

\[R_H = \frac{RZ^2}{4} \Rightarrow Z^2 = 4 \Rightarrow Z = 2\]

Therefore, the atomic number \( Z \) of the H-like ion is 2.