Question

An electron revolves around the nucleus of hydrogen atom in a circle of radius $5\times {10}^{-11}m$. The intensity of electric field at a point in the orbit of the electron is:

• A. $5.76\times {10}^{11}N/C$
• B. $9.216\times {10}^{-8}N/C$
• C. $0$
• D. $4N/C$

Answer 1

Answer: (A)

$5.76\times {10}^{11}N/C$

We know hydrogen atom consist of one proton and one electron
$R=5\times {10}^{-11}m$
We know attractive force $=\frac{k{q}_1\left(q_2\right)}{r^2}=\frac{9\times {10}^9\times (1\cdot 6\times {10}^{-19}{)}^2}{(5\times {10}^{-11}{)}^2}$
We know $F=Eq$
$\therefore E=\frac{F}{q}$
$\therefore E=\frac{9\times {10}^9\times (1\cdot 6\times {10}^{-19})}{5\times {10}^{-11}{)}^2}=5\cdot 76\times {10}^{11}N/C.$