# Photoelectrons are emitted when 40 nm radiation is incident on a surface of work function 1.9 eV. These photoelectrons pass through a region containing α – particles. A maximum energy electron combines with an α-particle to form a He+ ion, emitting a single photon in this process. He+ ions thus formed are in their fourth excited state. Find the energies in eV of the photons, lying in the 2 to 4 eV rang, that are likely to be emitted during and after the combination [Take h = 4.14 x 1015 eV.s.]

Navjyot Kalra
10 years ago
Hello Student,
The energy of the incident photon is
1 = hc / λ = (4.14 x 10-15 eVs) (3 x 108 m/s) / (400 x 10-9 m) = 3.1 eV
The maximum kinetic energy of the electrons is Emax = E1 – W = 3.1 eV – 1.9 eV = 1.2 eV
It is given that,
The fourth excited state implies that the electron enter I the n = 5 state.
In this state its energy is
E = - (13.6 eV)Z2 / n2 = - (13.6eV)(2)2 / 52
= - 2.18 eV
This energy of the emitted photon in the above combination reaction is
E =Emax­ + (- E5) = 1.2 eV + 2.18 eV = 2.4 eV
Note : After the recombination reaction, the electron may undergo transition from a higher level to a lower level thereby emitting photons.
The energies in the electronic levels of He+ are
E4 = (- 13.6 eV)(22) / 42 = - 3.4 eV
E3 = ( - 13. 6eV)(22) / 32 = -6.04 eV
E2 = (-13.6eV) (22) / 22 = - 13.6 eV
The possible transitions are
n = 5 ⟶ n = 4
∆E = E5 – E4 = [ - 2.18 – ( - 3.4 )] eV = 1.28 eV
n = 5 ⟶ n = 3
∆E = E5 – E3 = [ - 2.18 – ( -6.04 )] eV = 3.84 eV
n = 5 ⟶ n = 2
∆E = E5 – E2 = [ - 2.18 – ( - 13.6 )] eV = 11.4 eV
n = 4 ⟶ n = 3
∆E = E4 – E3 = [ - 3.4 – ( - 6.04 )] eV = 2.64 eV
Thanks
Navjot Kalra