Question

In a hydrogen atom, an electron with a charge of (-)1.6E-19 C orbits a proton with the same but positive charge at a distance 61 picometers away (that's trillionths of a meter or 10-12 m). What electric force F, in N, binds the two particles together

Answer 1

This question involves the concept of Colomb's Law and electrostatic force.

The electrostatic force will be "6.2 x 10⁻⁸ N".

COLOMB'S LAW:

According to Colomb's Law, every charge exerts an electrostatic force on the other charge, which is directly proportional to the product of the magnitudes of both the charges and inversely proportional to the square of the distance between them.

$F = \frac{kq_1q_2}{r^}$

where,

• F = electrostatic force = ?

• k = Colomb's constant = 9 x 10⁹ N.m²/C²

• q₁ = magnitude of first charge = 1.6 x 10⁻¹⁹ C

• q₂ = magnitude of second charge = 1.6 x 10⁻¹⁹ C

• r = distance between charges = 61 x 10⁻¹² m

Therefore,

$F = \frac{(9\ x\ 10^9\ N.m^2/C^2)(1.6\ x\ 10^{-19}\ C)(1.6\ x\ 10^{-19}\ C)}{(61\ x\ 10^{-12}\ m)^2}$

F = 6.2 x 10⁻⁸ N

Learn more about Colomb's Law here:

brainly.com/question/9774180