In the Angular process as atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom. (this is called an Augar elecrons ) . Assuming the nuclesu to be massive , calculate the kinetic energy of an `n=4` Augar electron emitted by chromium by absorbing the energy from a `n=2` to `n=1` transition .A. 4.6B. 7.5C. 5.38D. 3.36

In the Angular process as atom makes a transition to a lower state wit

1.	In the Angular process as atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom. (this is called an Augar elecrons ) . Assuming the nuclesu to be massive , calculate the kinetic energy of an `n=4` Augar electron emitted by chromium by absorbing the energy from a `n=2` to `n=1` transition .A. 4.6B. 7.5C. 5.38D. 3.36
Answer» Correct Answer - C As the uncleus is massive, recoil momentum of the atom can be ignore. We can assume that the entire energy of tranition is transferred to the augar electron. As there is a single valence electron in chromium `(Z=24)` the energy states may be thought of as given by bohr model. the energy of the `n^th` stae is `E_n=-(RchZ^2)/(n^2)` when R is hydberg constant. In the transition form `n=2` to `n=1`, energy released , `DeltaE=-RhcZ^2((1)/(4)-1)=(3)/(4)RhcZ^2` The energy rquired to eject ` n=4` electron `=RhcZ^2((1)/4)^2=(RhcZ^2)/(16)` `therefore` KE of augar electron `=(3RhcZ^2)/(4)-(RhcZ^2)/(16)` `KE=RhcZ^2((3)/(4)-(1)/(16))=(11)/(16)RhcZ^2` `(11)/(16)(13.6eV)xx24xx24=5385.6eV=5.38keV` `[because Rhc=13.6eV]`