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

From Einstein 's photoelectric equation, maximum kinetic energy of emitted electrons
`E_(K) = (hc)/(lambda) - W`
Now, `E = (hc)/(lambda) = (6.624 xx 10^(-34) xx 3 xx 10^(8))/(400 xx 10^(-9))`
`= 4.968 xx 10^(-19) J-s = (4.968 xx 10^(-19))/(1.6 xx 10^(-19)) = 3.1 e V`
`E_(K) = 3.1 e V - 1.9 e V = 1.2 e V`
`e + alpha to he^(+) + Photon
`Energy of He atom is their fourth excited state `(n = 5)` is
`E_(n) = - (Z^(2) Rhc)/(n^(2)) = - ((2) xx 13.6)/((5)^(2)) = - 2.176 e V`
From conservation of energy , `1.2 e V + 0 = - 2.176 e V + E_(lambda)` Energy of photon during combination ,
`E_(lambda) = 1.2 + 2.176 = 3.376 e V`
Energy of helium ion , `E_(n) = - (Z^(2) Rhc)/(n^(2)) = - (4 xx 13.6)/n^(2)`
` = - (54.4)/n^(2) e V, n = 1, 2, 3`...
`= - 54.4 e V, -13.6 e V, - 6.04 e V`,
`- 3.4 e V, -2.176 e V, - 1.51 e V`
Different of energies lying bbetween `2 and 4 eV` is
`- 3.4 + 6.04 = 2.64 e V`
`- 2.176 + 6.04 = 3.86 e V`
Energies of photons emitted are `2.64 e V and 3.86 e V`.