If the series limit of Lyman series for Hydrogen atom is equal to the series limit Balmer series for a hydrogen like atom, then atomic number of this hydrogen-like atom will be

Options

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

Answer

B

Explanation

By using `(1)/(lambda)RZ^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
For hydrogen atom `(1)/((lambda_("min"))_(H))=R[(1)/(n_(1)^(2))-(1)/(oo)\]=R`
`(1)/((lambda_("min"))_(H))=Ror (lambda_("min"))_(H)=(1)/(R)`
For another atom,
`(1)/((lambda_("min"))_("atom"))=RZ^(2)((1)/(2^(2))-(1^(1))/(oo))=(RZ^(2))/(4)`
`rArr (lambda_("min"))_("atom")=(4)/(RZ^(2))`
From Eqs. (i) and (ii), we get , `(1)/(4RZ^(2))rArrZ=2`