The wavelength of the first line of Lyman series for hydrogen is idetical to that of the second line of Balmer series for some hydrogen like ion Y. Ca
The wavelength of the first line of Lyman series for hydrogen is idetical to that of the second line of Balmer series for some hydrogen like ion Y. Calculate energies of the first three levels of Y.
1 Answers
Wavelength of the first line of Lyman series for hydrogen atom (Z=1) will be given by the equation,
`(1)/(lambda)=R((1)/(1^(2))-(1)/(2^(2)))=(3R)/(4)`
The wavelength of second Balmer line for hydrogen like ion Y is
`(1)/(lambda_(2))=RZ^(2)((1)/(2^(2))-(1)/(4^(2)))=(3RZ^(2))/(16)`
Given that `" "lambda_(1)=lambda_(2)or (1)/(lambda_(1))=(1)/(lambda_(2))`
i.e.,`" " (3R)/(4)=(3ZR^(2))/(16)`
`:.` Z=2
i.e.n Y ion is `H^(+)`. The erergies of first four levels of Y are ,
`E_(1)=-(13.6)Z^(2)=-54.4 eV`
`E_(2)=(E_(1))/((2)^(2))=-13.6 eV`
and `" " E_(3)=(E_(1))/((3)^(2))=-6.04 eV`