A gas hydrogen like atoms can absorb radiations of `68eV`. Consequently, the atoms emit 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 radiation.
(`d`) Find the ionization energy and the respective wavelength for the gas atoms.

Explanation

(a) Since three raditions are emitted , therefore, the final excited stateof the gas is `n = 3`.
The initial state of the gas atoms is `N = 2` as all the wevelength are smaller and the energy will be higher.
(b) `13.6Z^(2) [(1)/(2^(2)) - (1)/(3^(2))] = 68`
or `13.6Z^(2) [(5)/(36)] = 68 implies Z = 6`
( c) The minimum wevelength corresponds to the transition
`n = 3 to n = 1`
`(1)/(lambda_(min)) = RZ^(2) [(1)/(1^(2)) - (1)/(3^(2))]`
or `lambda_(min) = (9)/(8RZ^(2)) = (9)/(8 (1.097 xx 10^(7)) (6)^(2)) = 25.32 Å`