Calculate the smaller wavelength of radiation that may be emitted by (a) hydrogen (b)`He^(+)`and (c ) `Li^(++)`

Answer

A, B, C

Explanation

Small wave length is emitted i.e. largest energy
`n_1 =1`, `n_2 = lambda`
(a) ` (1)/(lambda) = R((1)/(n_1^2)-(1)/(n_2^2))`
`lambda =(1)/(1.1 xx 10^7) = (1)/(1.1) xx 10^-7`
`=0.909 xx 10^-7`
`=90.9 xx 10^-4 = 91 nm`
(b) `(1)/(lambda) = Z^2R((1)/(n_1^2)-(1)/(n_2^2))`
`lambda = (91nm)/(4) = 23 nm`
(c) `1/lambda = Z^2R((1)/(n_1^2) - (1)/(n_2^2))`
implies ` lambda = (91 nm)/(Z^2) = 91/9 = 10nm`