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The wavelength of the first line of Lyma...

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.

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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`
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