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

Answer: (D)

The wavelength of the first line of lyman series for hydrogen atom is
$\frac{1}{\lambda }=R\left[\frac{1}{1^2}-\frac{1}{2^2}\right]$
The wavelength of the second line of Balmer series for hydrogen like ion is
$\frac{1}{{\lambda }^{\prime }}=Z^{2}R\left[\frac{1}{2^2}-\frac{1}{4^2}\right]$
According to question $\lambda ={\lambda }^{\prime }$
$\Rightarrow R\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=Z^{2}R\left[\frac{1}{2^2}-\frac{1}{4^2}\right]$
or $\frac{3}{4}=\frac{3Z^2}{16}$
or $Z^2=4$
or $Z=2$