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Question: In a hydrogen atom the total energy of electron is...

In a hydrogen atom the total energy of electron is

A

e24πε0r\frac{e^{2}}{4\pi\varepsilon_{0}r}

B

−e24πε0r\frac{- e^{2}}{4\pi\varepsilon_{0}r}

C

−e28πε0r\frac{- e^{2}}{8\pi\varepsilon_{0}r}

D

e28πε0r\frac{e^{2}}{8\pi\varepsilon_{0}r}

Answer

−e28πε0r\frac{- e^{2}}{8\pi\varepsilon_{0}r}

Explanation

Solution

The kinetic energy of the electron in hydrogen atom are

K=12mv2=e28πε0r[∵v2=e24πε0mr]K = \frac{1}{2}mv^{2} = \frac{e^{2}}{8\pi\varepsilon_{0}r}\left\lbrack \because v^{2} = \frac{e^{2}}{4\pi\varepsilon_{0}mr} \right\rbrack

Electrostatic potential energy

U=−e24πε0rU = \frac{- e^{2}}{4\pi\varepsilon_{0}r}

The total energy E of the electron in a hydrogen atom is

E=K+UE = K + U

E=e28πε0r+(−e24πε0r)=−e28πε0rE = \frac{e^{2}}{8\pi\varepsilon_{0}r} + \left( \frac{- e^{2}}{4\pi\varepsilon_{0}r} \right) = - \frac{e^{2}}{8\pi\varepsilon_{0}r}

Here negative sign shows that electron is bound to the nucleus