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:

Question

Toppr Admin · Accepted Answer

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

An electron revolves around the nucleus of hydrogen atom in a circle of radius 5×10−11 m. The intensity of electric field at a point in the orbit of the electron is:9.216×10−8N/C5.76×1011N/C04N/C

We know hydrogen atom consist of one proton and one electronR=5×10−11mWe know attractive force =kq1(q2)r2=9×109×(1⋅6×10−19)2(5×10−11)2We know F=Eq∴E=Fq∴E=9×109×(1⋅6×10−19)5×10−11)2=5⋅76×1011N/C.

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:

$9.216 \times 10^{- 8} N / C$
$5.76 \times 10^{11} N / C$
$0$
$4 N / C$