In the Auger process an atom makes a transition to a lower state … - Vidyadip

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, calculate the kinetic energy of an n = 4 Auger electron emitted by Chromium by absorbing the energy from a n = 2 to n = 1 transition.

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Answer

As the nucleus is massive, recoil momentum of the atom may be neglected and the entire energy of the transition may be considered transferred to the Auger electron. As there is a single valence electron in Cr, the energy states may be thought of as given by the Bohr model.

The energy of the nth state $\text{En}=-\text{Z}^2\text{R}\frac{1}{\text{n}^2}$ where R is the Rydberg constant and Z = 24.

The energy released in a transition form 2 to 1 is $\Delta\text{t}=\text{Z}^2\text{R}\Big(1-\frac{1}{4}\Big)=\frac{3}{4}\text{Z}^2\text{R}$.

The energy required to eject a n = 4 electron is $\text{E}_4=\text{Z}^2\text{R}\frac{1}{16}$.

Thus, the kinetic energy of the Auger electron is $\text{KE}=\text{Z}^2\text{R}\Big(\frac{3}{4}-\frac{1}{16}\Big)=\frac{1}{16}\text{Z}^2\text{R}$

$=\frac{11}{16}\times24\times24\times13.6\text{eV}=5385.6\text{eV}.$

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