Question

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

• A. $3$
• B. $4$
• C. $1$
• D. $2$

Answer 1

The correct option is D

$2$

Step 1: Given data

For the first line of the Lyman series, let the wavelength be ${\lambda }_1$

For the second line of the Balmer series, let the wavelength be ${\lambda }_2$

Given ${\lambda }_1={\lambda }_2$

Step 2: Formula used

Electron jumps from $n=2$ to $n=1$ to produce the first line of the Lyman series, therefore

$\frac{1}{{\lambda }_1}=R(\frac{1}{1^2}-\frac{1}{2^2})$ , Here R is rydberg constant $R=109677c{m}^{-1}$

Electron jumps from $n=4$ to $n=2$ to produce the second line of the Balmer series, therefore

$\frac{1}{{\lambda }_2}=R{Z}^2(\frac{1}{2^2}-\frac{1}{4^2})$ , Here R is rydberg constant $R=109677c{m}^{-1}$

Step 3: Calculating the atomic number of hydrogen-like ion

$$\begin{gathered} {\lambda }_1={\lambda }_2 \\ \frac{1}{{\lambda }_1}=\frac{1}{{\lambda }_2} \\ R(\frac{1}{1^2}-\frac{1}{2^2})=R{Z}^2(\frac{1}{2^2}-\frac{1}{4^2}) \\ R\frac{3}{4}=R{Z}^2\times \frac{3}{16} \\ {Z}^2=4 \\ Z=2 \end{gathered}$$

Hence, option D is the correct answer.