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)

2

Answer 2

The wavelength of the first line of Hyman series for hydrogen atom is

$\frac{1}{\lambda} = R\left\lbrack \frac{1}{1^{2}} - \frac{1}{2^{2}} \right\rbrack$

The wavelength of the second line of Balmer series for hydrogen like ion is

$\frac{1}{\lambda'} = Z^{2}R\left\lbrack \frac{1}{2^{2}} - \frac{1}{4^{2}} \right\rbrack$

According to the question $\lambda = \lambda'$

$\Rightarrow R\left\lbrack \frac{1}{1^{2}} - \frac{1}{2^{2}} \right\rbrack = Z^{2}R\left\lbrack \frac{1}{2^{2}} - \frac{1}{4^{2}} \right\rbrack$

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