An electron is orbiting a nucleus which has a charge of 19e, under the action of the coulomb force at a radius of 1.15 × 10-10 m

Question

rosijanka

3 months ago

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An electron is orbiting a nucleus which has a charge of 19e, under the action of the coulomb force at a radius of 1.15 × 10-10 m

. calculate the angular velocity of the electron, in radians per second.

Physics

2 answers:

Sav [3.1K] · Answer 1 · 2026-03-01T20:19:31+03:00

The angular velocity of an electron in its orbit around the nucleus, influenced by the Coulomb force, is $\boxed{5.63\times10^{16}\text{ rad/sec}}$ .

Explanation:

Both the electron and the nucleus carry charges of opposite signs, leading to an attractive electrostatic force between them.

Despite this attraction, the electron continues to move in a circular trajectory around the nucleus. Its circular motion induces a centripetal force that prevents it from being pulled directly into the nucleus.

The equation representing the equilibrium of forces on the electron during its orbital path is shown as:

$\boxed{F_c=F_e}$ ...... (1)

The centripetal force acting on the electron can be defined as:

$F_c=\dfrac{mv^2}{r}$

In this formula, $m$ represents the mass of the electron, $v$ denotes the speed of the electron, and $r$ is the radius of its circular path.

The electrostatic force exerted on the electron by the nucleus is expressed as:

$F_e=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r^2}$

Here, $q_1\&q_2$ represent the charges of the electron and nucleus, and $r$ refers to the separation distance between the two charges.

The angular velocity of a body can be expressed as:

$v=r\omega$

In this instance, $\omega$ signifies the angular velocity.

By substituting the values of centripetal force, electrostatic force, and angular velocity into equation (1), we obtain:

$\dfrac{m\omega^2r^2}{r}=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r^2}$

By plugging in the values into the previous expression, we get:

$\omega^2=\dfrac{(9\times10^9)\times19(1.6\times10^{-19})\times(1.6\times10^{-19})}{(9.1\times10^{-31})\times(1.15\times10^{-10})^3}\\\omega^2=\dfrac{4.377\times10^{-27}}{1.38\times10^{-60}}\\\omega^2=3.17\times10^{33}\\\omega=5.63\times10^{16}\text{ rad/sec}$

Thus, the angular velocity of the electron in its orbit around the nucleus, guided by the Coulomb force, is $\boxed{5.63\times10^{16}\text{ rad/sec}}$ .

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Answer Details:

Grade: College

Subject: Physics

Chapter: Electrostatics

Keywords:

electron, orbiting, nucleus, coulomb force, radius of circular, attraction, centripetal force, angular velocity, rad/sec, charge, particles.