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=1n=`oo`__________________0eV n=5 __________________ -0.80 eV n=4 ______________ -1.45 eVn=3__________________ -3.08 eV n=2_______________ -5.30 eVn=1 ____________________ -15.6 eV.

The energy levels of a hypothetical one electron atom are shown in the

1.	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=1n=`oo`__________________0eV n=5 __________________ -0.80 eV n=4 ______________ -1.45 eVn=3__________________ -3.08 eV n=2_______________ -5.30 eVn=1 ____________________ -15.6 eV.
Answer» (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`

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