Question

In the Auger process an 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 Auger electron). Assuming the nucleus to be massive, the kinetic energy (in $keV$) of an $n = 4$ Auger electron emitted by Chromium by absorbing the energy from a $n = 2$ to $n = 1$ transition is

• A. $4.6$
• B. $7.5$
• C. $5.38$
• D. $3.36$

Answer 1

Answer: (C)

As the nucleus is massive, recoil momentum of the atom can be ignored. We can assume that the entire energy of transition is transferred to the Auger 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 $n^{th}$ state is
$E_{n} = - \frac{RhcZ^{2}}{n^{2}} $ where $R$ is Rydberg constant.
In the transition from $n = 2$ to $n = 1$, energy released,
$\Delta E = -RhcZ^{2}\left(\frac{1}{4} -1\right)$
$= \frac{3}{4}RchZ^{2} $
The energy required to eject $n = 4$ electron
$=RhcZ^{2}\left(\frac{1}{4} \right)^{2}$
$ = \frac{RchZ^{2}}{16}$
$\therefore $ KE of Auger electron
$= \frac{3RhcZ^2}{4} - \frac {RhcZ^2}{16}$
$KE = RhcZ^{2}\left(\frac{3}{4}-\frac{1}{16}\right)$
$ = \frac{11}{16}RhcZ^{2}$
$ = \frac{11}{16}\left(13.6 eV\right)\times 24 \times24 $
$ = 5385.6 \,eV $
$ = 5.38\, keV $
$\left[\because Rhc = 13.6 \,eV\right]$