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. 4
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is C

2

Wavelength of the first line of lyman series for hydrogen atom is (Z = 1)

$\frac{1}{\lambda }=R[\frac{1}{(1{)}^2}-\frac{1}{(2{)}^2}]=\frac{3R}{4}$

$\lambda =\frac{4}{3R}$

Note: In such question, sometimes we have use term $\frac{1}{R}=911\stackrel{\circ }{A}$

Wavelength of the second line of Balmer series for [H – like atom] Z [let]

$\frac{1}{{\lambda }^{\prime }}{Z}^{2}R[\frac{1}{(2{)}^2}-\frac{1}{(4{)}^2}]=\frac{3R{Z}^2}{16}\Rightarrow {\lambda }^{\prime }\frac{16}{3R{Z}^2}{\lambda }^{\prime }=\lambda \frac{16}{3R{Z}^2}=\frac{4}{3R}\Rightarrow Z=2$