1.

If the total energy of an electron in the first shell of H atom = 0.0 eV tgen its potential energy in the ffirst excited statewould be

Answer»

The energy of an electron in its nth orbit = -13.6/n2.
T.E in first excited state = -13.6/4 = -3.4 eV
PE in the first excited state = -6.8 eV.
(as PE:TE = 2:1)

For this to be zero, we must add +6.8.

So, +6.8 



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