A gas of hydrogen like atoms can absorb radiation of 68 eV. Consequently, the atom emits radiations of only three different wavelengths. All the wavelengths are equal or smaller than that of the absorbed photon.
(a) Determine the initial state of the gas atoms.
(b) Identify the gas atoms. (c ) find the minimum wavelength of the emitted radiations.
(d) Find the ionization energy and the respective wavelength for the gas atoms.

Explanation

(a) `(n(n-1))/(2)=3`
`:. n=3`
i.e., after excitation atom jumps to second excited state.
hence `n_(f)=3`. So `n_(1)` can be `1` or `2`
If `n_(1)=1` then energy emitted is either equal to, greater than or less than the energy absorbed.
Hence the emitted wavelength is either equal to, less than or greater than the absorbed wavelength. Hence `n!=2`.
If `n_(1)=2`, then `E_(e )geE_(a)` Hence `lambda_(e ) lelambda_(0)`
(b) `E_(3)-E_(2)=68ev`
`:. (13.6)(Z^(2))((1)/(4)-(1)/(9))=68`
`:. Z=6`
(c ) `lambda_(min)=(12400)/(E_(3)-E_(1))=(12400)/((13.6)(6)^(2)(1-(1)/(9)))=(12400)/(435.2)=28.49`
(d) Ionization energy `= (13.6)(6)^(2)=489.6eV`
`lambda=(12400)/(489.6)=25.33A`