photoelectrons are emited when `40 nm ` radiation is incident on a surface of work function `1.9 eV` These photoelectron pass tjhrough a ragain cotaining a - particle A maximum energy electron conbines 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 typing is the `2` to `4 eV` rage, that are likrly to be emitted during and after the combiution `[Take h = 4014 xx 10^(-15) eV s]`

Answer

A, B, C, D

Explanation

The energy of the incident photon is
`E_(1) = (hc)/(lambda) = ((4.14 xx 10^(-15)eVs)(3 xx 10^(8)m//s))/((400 xx 10^(-9) m)) = 3.1 eV`
The +miximum kinetic energy of the emitted electron is
`E_(max) = E_(1)- W = 3.1eV - 19 eV = 1.2eV`
it is given that,
(Emitted electron of maximum energy) ` + _(2)He^(2+) rarr He^(+)`
in 4th excited state
+ photon
The fourth excited state implies that the electron enters in the `n = 5` state
in this state its energy is
`E_(5) = ((13.6eV)Z^(2))/(n^(2)) = ((13.6 eV)(2)^(2))/(5^(2))`
` = - 2.18 eV`
The energy of the emitted photon in tha above combination reaction is
`E = E_(max) + (- E_(5)) = 1.2 eV + 2.18 eV = 3.4 eV`