Fqqrkqr==bg40222πεwhere qis the charge on either of them and ris the distance between them. We solve for the charge: qrFk×××⋅=×−−−50 1037 10899 1032 10109919....mNNm CC.22ch(b) Let Nbe the number of electrons missing from each ion. Then,
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This note was uploaded on 09/06/2009 for the course PHYSICS 14358 taught by Professor Bolton during the Spring '09 term at Kansas State University.