Question

An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The coulomb force f ur between the two is

Answer 1

Answer:

Coulomb's force of attraction between electron and nucleus is given as

$F = \frac{e^2}{4\pi \epsilon_0 r^2}$

Explanation:

As we know that charge of the nucleus of the hydrogen atom is equal to the charge of proton

And the force on the circulating electron is given as coulomb force on it

here we know that the orbital radius of the electron is given as r

now we have

$F = \frac{kq_1q_2}{r^2}$

here we know that

$q_1 = q_2 = e$

now we have

$F = \frac{ke e}{r^2}$

$F = \frac{e^2}{4\pi \epsilon_0 r^2}$

#Learn

Topic : Coulomb's law

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