Question

In the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius 5.3×10−11m with a speed of 2.2×106m/s. Find the direction of the electric field that the electron produces at the location of the nucleus (treated as a point).

Answer 1

Answer:

The magnitude of the electric field is 512171146711.7046 N/C.

The direction is toward the electron as electron has negative charge.

Explanation:

e = Charge of electron = $1.6\times 10^{-19}\ C$

r = Radius of circular orbit = $5.3\times 10^{-11}\ m/s$

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

The Electric field is given by

$E=\frac{e}{4\pi\epsilon_0 r^2}\\\Rightarrow B=\frac{1.6\times 10^{-19}}{4\pi\times 8.85\times 10^{-12}\times (5.3\times 10^{-11})^2}\\\Rightarrow E=512171146711.7046\ N/C$

The magnitude of the electric field is 512171146711.7046 N/C.

The direction is toward the electron as electron has negative charge.