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 a-particles. A maximum energy electron combines with an a -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

A

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 and `lambda`) Therefore, maximum kinetic energy of phototelectrons `K_(max) = E - W = (3.1 - 1.9) = 1.2 eV`
Now the situatin is as shown in figure. Energy of electron in 4th excited state of `He^+ (n = 5) will be`
`E_5 =- 13.6 (Z)/(n^2) eV`
`rArr 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 energy 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 `Delta E_(5rarr4)=E_5-E_4 = 1.2 eVlt2eV`
`Delta E_(5rarr3)=E_5-E_3 = 3.84eV`
`Delta E_(5rarr2)=E_5-E_2 = 11.4 eV gt4eV`
`Delta E_(4rarr3)=E_4-E_3 = 2.64eV`
`Delta E_(4rarr2)=E_4-E_2 = 10.2 eV gt 4eV`
Hence, the energy of emitted photons in the range of 2 eV and 4 eV are 3.4 eV during combination and 3.84 eV and 2.64 after combination.