# 22 2 n forntheenergyiszero

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Unformatted text preview: ⎛ n ⎞ m⎜ ⎟ ⎝ mr ⎠ Ze2 = r r n 2 2 Ze2 n22 = ⇒r= 2 mr r Zme2 2 a0 = (Bohr radius) 2 me Ioniza,on energy (I)  •  When the distance between the nucelus and  the electron is very large, the electron is  dissociated from the proton(s), and the  interac,on energy between the electron and  the proton(s) must be zero  e  p  1 2 Ze2 E = mv − 2 r using the centrifugal/electrostatic force relation mv 2 Ze2 Ze2 2 = 2 → mv = r r r Ze2 E=− →0 r→∞ 2r Ioniza,on energy (II)  •  Using the expression for the distance r we can  write the energy as a func,on of the quantum  number n  n22 r= Zme2 1 Ze2 E=− 2r 1 Z 2 e4 m E=− n = 1, 2, 3, ... 22 2 n •  For n‐>∞ the energy is zero  •  The lowest energy (most stable is when n=1).  The dissocia,on energy is therefore  ⎛ 1 Z 2 e4 m ⎞ 1 Z 2 e4 m Ionization energy=En = ∞ − En =1 = 0 − ⎜ − 2 ⎟ = 2 2 ⎝2 ⎠ Spectrum of hydrogen atom  •  A transi,on between any orbit‐states ni and nf  of the electron in a hydrogen‐like atoms has  an energy diﬀerence  1 Z 2 e4 m ⎛ 1 1⎞ ΔE = E f − Ei = ⎜ n2 −...
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## This document was uploaded on 03/04/2014 for the course CH 354L at University of Texas at Austin.

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