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

Options

A. `3`
B. `4`
C. `1`
D. `2`

Answer

D

Explanation

For hydrogen atom, the first line of Lyman series is
`1/lambda=R[1/(1^(2))-1/(2^(2))]=(3R)/4........(i)`
For hydrogen like ion, the second line of Balmer series is
`1/(lambda')=RZ^(2)[1/(2^(2))-1/(4^(2))]=RZ^(2)xx3/16`
According to question, `lambda=lambda'`
`:. (3R)/4=RZ^(2)xx3/16 or Z^(2) =4 or Z=2`