The energy level excitation potential of a hydrogen -like atom is shown in Figure

a Find the value of `Z`
b If initially the atom is in the ground state ,then
(i) determine its excitation potential, and
m (ii) determine its ionization potential.
c Can it absorbe a photom of `42 eV`?
d Can it absorbe a photom of `56 eV`?
e Calculate the radius of its first Bohr orbit.
f Calculate the kinectic and potential energy of an electron in the first orbit.

Explanation

We know that `E_(n) = - 13.6 (Z^(2))/(n^(2) e V)`
Here for `n = 1. E_(1) = - 54.4 e V`, therefore
`-54.4 = - 13.6 (Z^(2))/(l^(2))`
`:. Z = sqrt((54.4)/(13.6)) = 2`
b. (i). The first excitation energy is required to excite the electron from `n = 1` to ` n = 2` .Thus ,
`Delta E_(12) = E_(2) - E_(1) = - 13.6 - (54.4) - 40.8 e V`
Therefore, excitation potential is `40.8 V`
(ii). The ionization energy is required to eject the electron from `n = 1` to `n = oo`, thus , `Delta E_(1 oo) = E_(oo) - E_(1) = - 0 - (54.4) - 54.4 e V`
Therefore, the ionization potential is `54.4 V`
c. No , by absorbing a photon of `42 e V`, the energy of electron will become
`E = - 54.4 + 42 = - 12.4 e V`
This energy level axists between `n = 1` and `n = 2` shells which is not possible. Hence , an electron in the first orbit cannot absorbe it.
d. Yes. By obsorving a a photon of `56 e V` , the electron comes out of the atom where the energy is not quantized. Hence a photon of `56 e V` can be absorbed.
e. We know that
`r_(n) = 0.53 (n^(2))/(Z) (Å)`
Hence , `n = 1 :Z = 2`, therefore, `r = 0.53 ((1)^(2))/(2) = 0.265(Å)`
f. Since `K = - E` and `U = 2 E`, therefore .
`K = - 54.4 e V = 8.7 xx 10^(18)j` and
`U = - 108.8 e V = 1.74 xx 10^(17)j`