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The energy levels of a hypothetical one ...

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 level diagram.
(b) What is the first excitation potential
(c ) What wavelength (Å) can be emitted when these atoms in the ground state are bombarded by
electrons that have been accelerated through a potential difference of 16.2 V?
(d) If these atoms are in the ground state, can they absorb radiation having a wavelength of 2000 Å ?
(e) what is the photoelectric threshold wavelength of this atom?

Text Solution

Verified by Experts

a. `E_(1) = - (18.0)/(1^(2)) = - 18.0 eV, E_(2) = - (18.0)/(2^(2)) = - 4.5 e V`,
E_(3) = - (18.0)/(3^(2)) = - 2.0 eV, E_(4) = - (18.0)/(4^(2)) = - 1.125 e V`
The energy level diagram is showing in figure

b. The excitation potential of state `n = 2` is :
`18.0 - 4.5 = 13.5 V`
c. Energy of the electron accelerated by a potential difference of `16.2 eV`. With this energy the electron can excite the atom from `n = 3` as
`E_(4) - E_(1) = 1.125 - (- 18.0) = 16.875 e V gt 16.2 e V`
`E_(3) - E_(1) = 2.0 - (- 18.0) = 16.0 e V lt 16.2 e V`
`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 corresponding to `lambda = 2000 Å`
`E = (12375)/(2000) = 6.187.5 Å`
The minimum cxcitation energy is `13.5 e V (n = 1) to n = 2))`.
e. Threshold wevelength for photoemission to take place from such an atom is `lambda_(max) = (12375)/(18) = 687.5 Å`
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