The energy levels of a hypothetical one electron atom are shown in the figure. (a)Find the ionization potential of this atom.
(b) Find the short wavelength limit of the series terminating at n=2
(c ) Find the excitaion potential for the state n=3.
(d) find wave number of the photon emitted for the transition n=3 to n=1
n=`oo`__________________0eV
n=5 __________________ -0.80 eV
n=4 ______________ -1.45 eV
n=3__________________ -3.08 eV
n=2_______________ -5.30 eV
n=1 ____________________ -15.6 eV.

Explanation

(a) Ionization potential `=15.6 V`
(b) `lambda_(min)=(12400)/(5.3)=2340Å`
(c )`DeltaE_(31) = -3.08-(-15.6)=12.52eV`
Therefore, excitation potential for state `n=3` is `12.52` volt.
(d) `(1)/(lambda_(31))=(DeltaE_(31))/(12400)A^(-1)=(12.52)/(12400)A^(-1)`
(e ) (i) `E_(2)-E_(1)=10.3eV lt 6eV`.
(ii) `E_(2)-E_(1)=10.3 eV lt 11 eV`.
Hence electron can excite the atoms. So `K_(min)=(11-190.3)=0.7 eV`