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

Answer

(1) `91 nm` (b) `23 nm`

Explanation

The smallest wavelength corresponds to transition from `n=1` to `n=oo`
(a) `DeltaE= 13.6 eV`
`rArr lambda=(1.24xx19=0^(4))/(13.6)Å=911 Å=91 nm`
(b) `DeltaE=13.6xx2^(2)ev`
`lambda=(124xx10^(4))/(13.6xx4)Å=228Å=23 nm`.