If potential energy of an electron in a hydrogen atom in first excited state is taken to be zero, kinetic energy (in eV) of an electron in ground state will be

Asked by kumarisakshi0209 | 19th Mar, 2019, 09:11: PM

Expert Answer:

energy of electron in its nth orbit = -13.6/nˆ2
TE in first excited state = -13.6/4 = -3.4 eV
PE in the first excited state = -6.8 eV (-PE = 2 X -TE)
For this to be zero, We must add 6.8 eV. So, add this 6.8 eV to all energy values to get the new values because shifting of zero or reference simply shifts all energy by the same value.
KE in ground state = 13.6 eV
So, in new system = 13.6 + 6.8 = 20.4eV

Answered by Ankit K | 19th Mar, 2019, 10:45: PM

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