Question

what is the PE of an electron present in N shell of Be3+ ion​

Answer 1

Atomic Number of Beryllium = Z= 4

Hence, Number of Electron in Be-atom (in neutral state) = 4

In case Be+3 ion, Number of Electron = 1 (electron in the K-shell)

In this case, Be+3 ion will behave as Hydrogen-like-entity. Also, electron in N-shell means that it is in excited state. Hence, we can apply Bohr’s theory in calculating energy of an electron in a particular orbit :—

Potential Energy of the electron in the N-shell of Be+3 ion = -27.2 eV/ion.

Answer 2

Answer:

-27.2ev

Explanation:

energy of an electron = - 13.6($Z^{2} /n^{2}$)

      where Z = atomic number

                  n= shell number = 4

  ∴ Energy of an electron present in N shell of Be3+ ion​ =  - 13.6($Z^{2} /n^{2}$)

                                                                                               = 13.6($4^{2}/4^{2}$)

                                                                                                =13.6

PE = 2 × Energy

     = 2 × 13.6

     = 27.2ev

The project code is #SPJ2