The potential energy of the orbital electron | vedclass

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(C) In a hydrogen-like atom,the potential energy $U$ and kinetic energy $K$ of an electron in an orbit of radius $r$ are related by the virial theorem

Vedclass · Accepted Answer

$\frac{E}{2}$

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(C) In a hydrogen-like atom,the potential energy $U$ and kinetic energy $K$ of an electron in an orbit of radius $r$ are related by the virial theorem.For a Coulomb potential $U = -\frac{kZe^2}{r}$,the electrostatic force provides the necessary centripetal force:$\frac{kZe^2}{r^2} = \frac{mv^2}{r}$Multiplying both sides by $\frac{r}{2}$,we get:$\frac{kZe^2}{2r} = \frac{1}{2} mv^2 = K$Since the potential energy $U = -\frac{kZe^2}{r}$,we can see that $K = -\frac{U}{2}$.Given that the potential energy $U = -E$,the kinetic energy is:$K = -\frac{(-E)}{2} = \frac{E}{2}$.

The potential energy of the orbital electron in the ground state of a hydrogen atom is $-E$. What is its kinetic energy?

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