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{3 Z^{2}}{16} $
or $ Z^{2}=4$
or $Z=2$