Question

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)`

Answer 1

`"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))`