The ground state energy of hydrogen atom is – 13.6 eV. What are the kinetic and potential energies of the electron in this state?

Solution

Here,

Ground state Energy, E = – 13.6 eV

Kinetic Energy,

E_{k} = \frac{1}{4 {πε}_{0}} . \frac{e^{2}}{2 r}

and,
Potential Energy,

E_{p} = - \frac{1}{4 {πε}_{0}} \frac{e^{2}}{r}

[∵ E_{p} = \frac{- {kq}_{1} q_{2}}{r} a n d [E_{k} = 2 E_{p}]]

Therefore, total energy,

E = E_{k} + E_{p}

= \frac{1}{4 {πε}_{0}} . \frac{e^{2}}{2 r} - \frac{1}{4 {πε}_{0}} . \frac{e^{2}}{r}

E = - \frac{1}{2} (\frac{1}{4 {πε}_{0}} . \frac{e^{2}}{r})

\Rightarrow

- 13.6 = - \frac{1}{2} (\frac{1}{4 {πε}_{0}} . \frac{e^{2}}{r})

∴ \frac{1}{4 {πε}_{0}} \frac{e^{2}}{r} = 27.2

∴

Kinetic energy,

E_{k} = \frac{1}{4 {πε}_{0}} . \frac{e^{2}}{2 r} = \frac{27.2}{2} eV = 13.6 eV

Potential energy,

E_{p} = - \frac{1}{4 {πε}_{0}} . \frac{e^{2}}{r} = - 27.2 eV .

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