Question

calculate the energy of Li^2+atom for 2nd exited state ?
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Answer 1

Answer:

108eV

Explanation:

The formula for the energy of the electron in n^th orbit is En = -13.6Z^2/n^2 eV.

So, for the first orbit (n=1), E1 = -13.6*(3)^2/(1)^2 = -122.4 eV [Here, Z = 3]

For second orbit, n = 2, E2 = -13.6*(3)^2/(3)^2 = -13.6 [Here, Z = 3]

The second excitation energy of Li^2+ is the energy when an electron jumps from the first orbit (n = 1) to the third orbit (n = 3), = E2 – E1 = −13.6− (−122.4) =108 eV