Question

Calculate the electrostatic potential energy of an electron-proton system of hydrogen atom. In the first Bohr orbit of hydrogen atom, the radius of the orbit is $5.3\times {10}^{-11}m$:

• A. $-4.35\times {10}^{-18}J$
• B. $-2.175\times {10}^{-18}J$
• C. $-4.35\times {10}^{-19}J$
• D. $-2.175\times {10}^{-19}J$

Answer 1

Answer: (A)

$-4.35\times {10}^{-18}J$

The electrostatic P.E. of proton electron pair is $-\frac{1}{4\pi {ϵ}_0}\frac{e^2}{r}$ where $r$ is the separation between the proton and electron and $e$ is electron's charge.
Since the charges are opposite in sign, P.E. is negative.
Given, $r=5.3\times {10}^{-11}m$
Hence, $P.E.=-(9\times {10}^9)\times \frac{(1.6\times {10}^{-19}{)}^2}{5.3\times {10}^{-11}}=-4.35\times {10}^{-18}J$