Question

If we are viewing the atom in such a way that the electron's orbit is in the plane of the paper with the electron moving clockwise. Find the magnitude of the electric field that the electron produces at the location of the nucleus (treated as a point). Express your answer in newtons per coulomb.

Answer 1

The magnitude of the electric field that the electron produces at the location of the nucleus is 5.13 x 10¹¹ N/C.

Electric field strength of the electron

The magnitude of the electric field strength of the electron is the force experienced in the field per unit charge.

$E = \frac{F}{Q} = \frac{kQ}{r^2}$

where;

• Q is the charge of the electron
• r is the distance of the electron from the nucleus

The magnitude of the electric field strength is calculated as follows;

$E = \frac{(9\times 10^9) \times (1.6 \times 10^{-19})}{(5.3 \times 10^{-11})^2} \\\\E = 5.13 \times 10^{11} \ N/C$

Thus, the magnitude of the electric field that the electron produces at the location of the nucleus is 5.13 x 10¹¹ N/C.

Complete question is below:

In the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius 5.3 x 10^-11 m with a speed of 2.2 x 10^6 m/s.

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