Question

A singly-ionized lithium ion Li2+ de-excites from a n= 4 state to a n= 2 state.

What is the energy of photon emitted during this process?

Answer 1

you go through the derivation of the energy of the electron, you observe that the term Z^2 appears in the value, but is ignored for hydrogen because z=1.
And that means the energy of the electron in lithium is simply 9 of that in hydrogen, because z^2 is 9.
I'll tell you why it's z^2.
We multiply the whole thing by z once owing to the fact that more the charge of the nucleus, the close the electron is to the nucleus (which naturally means there is more attraction and the negative energy is more).
It appears the second time simply because of Coulomb's law.
So the value of the energy of the electron in the first orbit is around 122.4 eV
And the general value is 122.4eV\n^2.

En = -122.4 evV/ n^2

E2 = -122.4 eV / 2^2

E2 = -30.6 eV

E4 = - 122.4/ 4^2

E4 = - 7.65 eV

Energy of photon = -7.65 -(-30.6)

Energy of photon = 22.65 eV

Thank You ?