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The total energy of an electron in the first excited state of H-atom is -3.4 eV, then its potential energy in this state is _____ eV.

Asked in GUJCET 2017 · Energy levels and ionisation

Answer: (3) -6.8

Step-by-step solution

Given: total energy E=-3.4 eV in n=2.

Idea: in a Bohr orbit K=-E and U=2E.

U=2×(-3.4)=-6.8 eV.

Check: K=3.4 eV, and K+U=3.4-6.8=-3.4 eV=E.

Why the other options are wrong

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