Potential energy of electron present in #Li^(2+)# is?

  1. #(-e^2)/(2piepsilon_0r)#

  2. #-3/2e^2/(piepsilon_0r)#

  3. #-3/4e^2/(piepsilon_0r)#

  4. #-1/2e^2/(piepsilon_0r)#

1 Answer
Truong-Son N.
Nov 26, 2017

#"Li"^(2+)# is a hydrogenic atom, and so, it uses the same coulombic potential energy found in the hydrogen atom Hamiltonian, except with a different atomic number.

Here is the coulomb potential for a hydrogenic (one-electron) atom:

#hatV_("H-like atom") = -(Ze^2)/(4piepsilon_0vecr)#

where:

  • #Z# is the atomic number.
  • #e# is the elementary charge, #1.602 xx 10^(-19) "C/particle"#. The force of attraction for the nucleus with the electron is included in #hatV# already, since #overbrace(-e)^("electron") cdot overbrace(Ze)^"protons" = -Ze^2#.
  • #epsilon_0 = 8.854 xx 10^(-12) "F"cdot"m"^(-1)# is the vacuum permittivity.
  • #vecr# is the radial distance of the electron from the nucleus.

We assume under the Born-Oppenheimer approximation that the nucleus can be treated as nearly stationary, so that the net charge of it is #Zcdote#.

You know the atomic number of #"Li"# is #3#. So...

#color(blue)(hatV_("Li"^(2+)) = -(3e^2)/(4piepsilon_0vecr))#

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