Question

The total energy of an electron in the first excited state of the hydrogen atom is about $-3.4eV$ . The potential energy of the electron in this state is

• A. $-6.8eV$
• B. $-8.6eV$
• C. $6.8eV$
• D. $8.6eV$

Answer 1

Answer: (A)

The kinetic energy of an electron, $\left(KE\right)=\frac{KZ{e}^2}{2r}$
The potential energy of an electron, $(PE)=\frac{-KZ{e}^2}{r}\Rightarrow PE=-2KE$
In this calculation, the electric potential and potential energy are zero at infinity.
Total energy $=PE+KE=-2KE+KE=-KE$
$PE$ of an electron in this first excited state
$=-2KE=-2(3.4)=-6.8eV$