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

Explanation

The wavelength of the first line of lyman series for hydrogen atom is
`(1)/(lambda)=R[(1)/(1^(2))-(1)/(2^(2))]`
The wavelength of the second line of Balmer series for hydrogen like ion is
`(1)/(lambda.)=Z^(2)R[(1)/(2^(2))-(1)/(4^(2))]`
According to question, `lambda=lambda.`
`R[(1)/(1^(2))-(1)/(2^(2))]=Z^(2)R[(1)/(2^(2))-(1)/(4^(2))]`
`"or "(3)/(4)=(3Z^(2))/(16) or Z^(2)=4 or Z=2`