If the total energy of an electron in the first shell of an H-atom is -13.6 eV, then its potential energy in the 1st excited state would be:
Correct answer: C. - 6.8 eV
- A. +6.8 eV
- B. +20.4 eV
- C. - 6.8 eV
- D. +3.4 eV
Explanation
This is the following solution:The energy of an electron in its nth orbit = -13.6/n2. For first excited state n=2 T.E in first excited state = -13.6/4 = -3.4 eV P.E = - 2ke2/r T.E = -ke2/r P.E = 2 (T.E) PE in the first excited state = -6.8 eV.
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