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)/(lamda)-W`
Given `h=4.14xx10^-15eV-s`
`=4.14xx10^-15xx1.6xx10^-19J-s`
`=6.624xx10^-34J-s`
`E=(hc)/(lamda)=(6.624xx10^-34xx3xx10^8)/(400xx10^-9)`
`=4.968xx10^-19J`
`=(4.968xx10^-19)/(1.6xx10^-19)=3.1eV`
`E_k=3.1eV-1.9eV`
`=1.2eV`
`e+HetoHe^++` Photon Energy of He atom in their fourth (n=5) excited state
`E_n=-(Z^2Rhc)/(n^2)=-(13.6Z^1)/(n^2)eV`
`=-((2)^2xx13.6)/((5)^2)=-2.176eV`
(for `He^+` ion Z=2)
From consevation of energy,
`1.2eV+0=-2.176eV+E_y`
Energy of photon during combination,
`E_y=1.2+2.176=3.376eV`
Energy of Helium ion,
`E_n=-(Z^2Rhc)/(n^2)=-(4xx13.6)/(n^2)`
`-(54.4)/(n^2)eV`,`n=1,2,3..`
`=-54.4eV,-13.6eV,-6.04eV,-3.4eV,-2.176eV,-1.51eV`
Difference of energies lying between 2 and 4 eV is
`-3.4+6.04=2.64eV`
` -2.176+6.04=3.86eV`
Energies of photons emitted are `2.64 eV` and `3.86eV.`