shielding - 2s electron does not experience the Coulombic...

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Shielding The problem: Our naive application of what we learned with regard to the energy of an electron in a one-electron atom or ion does not work. For example, recall the appropriate energy expression E n = -R Z 2 /n 2 where R is the Rydberg Constant = 2.18 x 10 -18 J. If we apply this to Li, we predict that the energy of the 2s electron is 9 times the energy of the 1s electron in hydrogen. This is not observed experimentally. The problem is caused by the fact that the experimentally derived value of Z is not the theoretically expected value of 3, but actually 1.3. This decrease in effective atomic number, Z eff , is due to shielding. The situation is described in cartoon fashion (exaggerated form) below. We have used Bohr orbitals (not physically correct) and grouped the two 1s electrons into a single charge of -2 (also, not physically correct). The result is that the
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Unformatted text preview: 2s electron does not experience the Coulombic potential expected for a positive nuclear charge of 3. Instead, this crude cartoon predicts that the effective nuclear charge is 1, a value not too different from the experimental value. Our non-physical cartoon is qualitatively correct and the 1s electrons do shield the 2s electrons from the full nuclear charge. The experimental value is somewhat greater than one because our cartoon does not accurately reflect the physical situation. The 2s orbital has some small probability of penetrating close to the nucleus. This is evident when we study the radial probability distribution function for the 2s orbital shown here. Since the electron in the 2s orbital is not "always behind" the 1s, the value of Z eff calculated is actually an underestimate.-1-2 +3 2s 1s 2 r...
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This document was uploaded on 03/03/2012.

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