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]`

Answer

During combination `3.4 eV`. After combination `3.84` eV, `2.64 eV`

Explanation

Given work function `W = 1.9 eV`
Wavelength of incident light, `lambda = 400 nm`
`:.` Energy of incident light, `E = (hc)/(lambda) = 3.1 eV`
(Substituting the values of h, c anf `lambda`)
therefore, maximum kinetic energy of photoelectron `K_("max")=E-W=(3.1-1.9)=1.2 eV`
Now the situation is as shown below:

Energy of electron in `4^(th)` excited state of `He^(+) (n=5)` will be `E_(5)= - 13.6 Z^(2)/n^(2) eV`
`E_(5)= -(13.6) ((2)^(2))/((5)^(2))=2.2 eV`
Therefore, energy released during the combination
`=1.2-(-2.2)=3.4 eV`
Similarly energies in other states of `He^(+)` will be `E_(4)=-13.6 ((2)^(2))/((4)^(2))= -3.4 eV`
`E_(3)= -13.6 ((2)^(2))/((3)^(2))= -6.04 eV`
`E_(2)= -13.6 ((2)^(2))/2^(2)= -13.6 eV`
The possible transitions are
`DeltaE_(5 rarr 4)=E_(5)-E_(4)=1.3 eV lt 2 eV`
`DeltaE_(5 rarr 3)=E_(5)-E_(3)=3.84 eV`
`DeltaE_(5 rarr 2)=E_(5)-E_(2)=11.5 eV gt 4 eV`
`DeltaE_(4 rarr 3)=E_(4)-E_(3)=2.64 eV`
`DeltaE_(4 rarr 2)=E_(4)-E_(2)=10.2 eV gt 4 eV`
Hence, the energy of emitted photons in the range of `2 eV` and `4 eV` are
`3.3 eV` during combination and `3.84 eV` and `2.64` after combination.