Energy of Li2+ atom according to Bohr's model

[Krushnaraj Pandya]

Homework Statement

If n=1 is taken to be the reference of the potential energy, what will be the kinetic and potential energy in second excited state of Li2+?

Homework Equations

E=-13.6z^2/n^2

The Attempt at a Solution

I know KE is independent of reference point, so from the formula above, KE=13.6 as KE=(-)E for a hydrogen like atom. Now, PE for n=1 is -27.2 eV but we consider it zero here, since it is a mcq question and only one option is positive I know the correct answer is 217.6 eV. This should be calculated by first calculating PE of n=3 with 0 at infinity and then noting the difference but I don't know the expression for PE in terms of Z, I'd appreciate some help-thank you

 

[Krushnaraj Pandya]

I just obtained one using kq1q2/r then replacing the values, I'll test whether that gives a correct value or not and return

 

[Krushnaraj Pandya]

Alright, it did. It was quite simple in fact. Thank you @Krushnaraj Pandya