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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» Solution :According to formula `E_(n)=-13.6(Z^(2))/(n^(2))` `:.E_(2)=-13.6xx((24)^(2))/((2)^(2)) ( :.` From chromium atom Z=24 here n=2) `:. E_(2)=-1958.4eV` Now `E_(1)=-13.6xx((25)^(2))/((1)^(2))` `:.E_(1)=-7833.6eV` `rArr E_(2) gt E_(1)` `rArr` Energy obtained, `DeltaE=E_(2)-E_(1)` `:.DeltaE=-1958.4-(-7833.6)` `=7833.6-1958.4` `DeltaE=5875.2eV ....(1)` Now, TOTAL energy of electron in n = 4 state in chromium atom, `E_(4)=-13.6eVxx((24)^(2))/((4)^(2))` `:.E_(4)=-489.6eV` `rArr` When above electron is given energy AE = 5875.2 eV, it will firstly USE energy 489.6 eV for its emission and the REMAINING energy will be possessed by it in the form of kinetic energy. Its value would be `K=5875.2-489.6` `:.K=5385.6eV` |
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