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 radiations are emitted, therefore, the final excited state of the gas is `n=3`.
The initial state of the gas atoms is
`n=2` as all the wavelengths 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)]=68rArrZ=6`
(`c`) The minimum wavelength 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(1.097xx10^(7))(6)^(2))=28.5Å`
(`d`) The ionization energy of the gas atoms is
`E=13.6Z^(2)=(13.6)(6)^(2)=489.6eV`
`lambda=(1)/(RZ^(2))=(1)/((1.097xx10^(7))(6)^(2))=25.32Å`