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 transiton n=3 to n=1 `prop _______________ 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`.

Answer

A, B, C, D

Explanation

(a) lonization energy = 15.6 eV
`:. lonization potential = 15.6 V`
(b) `E_(prop) - E_2 = 5.3 eV`
`:. Lambda = (12375)/(5.3) = 2335 Å`
(c ) `E_3 - E_1 = 12.52 eV`
`:. Excitation potential is 12.52 V.`
(d) `E_3 - E_1 = 12.52 eV`
`:. lambda = (12375)/(12.52) Å`
`= 988.41 Å`
`:. Wave number = (1)/(lambda) = (1)/(988.41xx10^(-10) m^(-1)`
`=1.01xx10^7 m^(-1)`