Photoelectrons are emitted when `400 nm` radiation is incident on a surface of work - function 1.9 eV. These photoelectrons pass through a region containing `alpha`-particles. A maximum energy electron combines with an `alpha` -particle to from 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 range, that are likely to be emitted during and after the combination. `[Take, h = 4.14xx10^(-15) eV -s]`

Explanation

`E=(1242)/(lambda(nm)) eV=(1242)/(400)=3.1 eV`
`K=phi+K_(max) implies 3.1=1.9+K_(max)`
`K_(max)=1.2 eV`
`e+._(2)He^(4) to He^(+) "photon"`
Energy of `He^(+)` in fourth excited state (n=5)
`E_(n)=-13.6 (Z^(2))/(n^(2)) eV`
` E_(5)=-13.6xx((2)^(2))/((5)^(2))=-2.17 eV`
During the formation one photon is emitted. By energy conservation :
`1.2+0=-2.176+E_("photon") implies E_("photon")=3.376 eV`
During excitation
`E_(n)=-13.6(Z^(2))/(n^(2))=-(13.6(2)^(2))/(n^(2))=-(54.4)/(n^(2))eV`
Total number of photon that can be emitted is
`=(5(5-1))/(2)=10`
We have to determine photons in the energy range 2 to 4 eV. These photons will be in transitions
`n=5 " to "n=3, n=4 " to " n=3`
Photon `(5 to3)=-2.176-(-60.4)=3.864 eV`
`(4 to 3)=-3.4-(-60.4)=2.64 eV`