By E=−Z^2RE/n2 where RE is the Rydberg energy As n increase, EPE becomes less -ve(i.e. more +ve) , indicating higher energy level<
EPE = 1/4πε( Qproton Qe-) /r, As r increase, EPE becomes less -ve(i.e. more +ve) , indicating higher energy level<
Thanks to everyone that helped !<
I beg to differ on the above explanation provided by the author @ De Day:
The highest energy acquired by an electron is at K shell , and slowly energy decreases as one moves to L,M,N ...shells.
the confirmation is the energy required to take out a K-shell electron is highest and in X-ray emission the high speed cathode electrons knock out K-shell electrons and it needs about 20-25 keV of energy .
Therefore I wish to add that energy levels which are closest to the nucleus is at the highest and the above contention by the author is not correct.
Moreover if a K-shell electron is knocked out and a vacancy is created then any transition from L.M....levels leads to emission lines of lowest wavelength and highest frequency X-rays characteristic lines .
This energy packet contains the difference of energy levels of the atom.
the magnitude of this energy also suggests that the E(K)-E(m)= h. frequency . is largest.
I think the confusion is that the bound states total energy is sum of its K.E. and P.E. and total energy has to be negative for bound states and it's highest as one moves closer to the nuclear charge +ze.
By just thinking about the potential energy, one has to consider that the nuclear charge field has done work on the electron to get it to a shell radius and this work done is highest if one goes closer.
the test is to supply energy to pull out a K-electron and the value of energy needed to extract will again be largeer than L, M,.. and other shell electrons.