The energy of electron in the first excited state of Hydrogen atom is -3.4 eV. The potential energy of this electron is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → -6.8 eV
$E_L$, Energy in first excited state = $\frac{13.6eV}{2^2}=-3.42eV$ [given]
and,
in hydrogen atom, the potential energy is twice the total energy,
$U=2E$
$=2×-3.42eV=-6.8eV$