An electron is revolving in a Bohr orbit of a doubly ionized lithium atom ( Li²⁺ ). If its angular momentum is h , what is the total energy of the electron in this orbit?

Step-by-step solution

The angular momentum of an electron in the n^ th Bohr orbit is given by L = nh 2 . Given L = h , we have: nh 2 = h n = 2 The total energy of an electron in a hydrogen-like ion is given by E = -13.6 Z^2 n^2 eV . For a doubly ionized lithium atom ( Li²⁺ ), the atomic number Z = 3 . Substituting Z = 3 and n = 2 , we get: E = -13.6 3^2 2^2 E = -13.6 9 4 = -30.6 eV Failing to account for the atomic number Z leads to the incorrect value of -3.4 eV . Answer: -30.6 eV