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
For an electron in a Bohr orbit the kinetic energy is \( K = \frac{e^{2}}{8\pi\varepsilon_0 r} \) and the potential energy is \( U = -\frac{e^{2}}{4\pi\varepsilon_0 r} \), so \( U = -2K \). The total energy is \( E = K + U = -K \).
\( E = -3.4\,\mathrm{eV} \Rightarrow K = +3.4\,\mathrm{eV} \)
\( U = 2E = 2\times(-3.4\,\mathrm{eV}) = -6.8\,\mathrm{eV} \)