The energy levels of a hypothetical one electron atom are given by `E_n = -(18.0)/(n^2) eV`
where n = 1,2,3,…. (a) Compute the four lowest energy levels and construct the energy levels diagram. (b) What is the first excitation potential (c ) What wavelength (Å) can be emitted when these atoms in the ground state are bombarged by electrons that have been accelerated through a potential difference of 16.2 V? (e) what is the photoelectric threshold wavelength of this atom?

Answer

A, C, D

Explanation

(a) `E_1 = (-18.0)/((1))^2 = 18.0 eV`
`E_2 = (18.0)/((2))^2 = -4.5 eV`
`E_3 = (-18.0)/((3))^2 =-2.0 eV`
`E_4 = (-18.0)/((4))^2 =-1.125 eV`
The energy level diagram is shown in figure
`_________E_4 = -1.125 eV`
`_________E_3 = - 2.0 eV`
`_________E_2 =-4.5 eV`
`________ E_1 =- 18.-0 eV`
(b) `E_2 = 13.5 eV`
`:. First excitation potential is 13.5 V
(c ) Energy of the electron accelerated by a potential difference of 16.2 V is 16.2 eV. With this energy the electron can excite the atom from n =1 to n = 3 as.
`E_4 - E_1 = -1.125 - (-18.0) = 16.875 eVgt 16.2 eV`
and `E_3 - E_1 = - 2.0 - (-18.0)=16.0 eV lt16.2 0 eVlt 16.2 eV`
Now, `lambda_32 = (12375)/(E_3 - E_2 = (12375)/(-2.0 - ((-4.5))`
`=4950 Å`
lambda_31 = (12375)/(E_3 - E_2) = (12375)/(16) = 773 Å`
and `lambda_21 = (12375)/(E_2 - E_1) = (12375)/((-4.5 - (-18.0)`
`= 917 Å`
(d) No, the energy correspondingt to `lambda = 2000 Å is`
E = (12375)/(2000) = 6.1875 eV`
The minimum excitation energy is 13.5 eV (n=1 to n=2)
. (e) Threshold wavelength for photoemission to take place from such an atom is `lambda_(min) = (12375)/(18)`
`=687.5 Å`