Solveeit Logo
Login

Question

Physics Question on Atoms

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=12mv2K = \frac{1}{2}mv^{2} =e28πε0r[∵v2=e24πε0mr]= \frac{e^{2}}{8\pi \varepsilon_{0}r} \left[\because v^{2} = \frac{e^{2}}{4\pi\varepsilon_{0} mr}\right] Electrostatic potential energy, U=−e24πε0rU =\frac{-e^{2}}{4 \pi \varepsilon_{0} r} The total energy EE of the electron in a hydrogen atom is E=K+U,EE = K + U, E =e28πε0r+(−e24πε0r)= \frac{e^{2}}{8\pi\varepsilon_{0} r} + \left(\frac{-e^{2}}{4\pi\varepsilon_{0} r}\right) =−e28πε0r = - \frac{e^{2}}{8\pi \varepsilon_{0} r} here negative sign shows that electron is bound to the nucleus.