Question

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.

Answer 1

The angular velocity of the electron as it orbits the nucleus due to the coulomb force is $\boxed{5.63\times10^{16}\text{ rad/sec}}$.

Explanation:

The electron and the nucleus both have a particular amount of charge with opposite sign. Since the charges are opposite in nature, there will be an electrostatic force of attraction acting between the charged particles.

The electron does not get attracted towards the nucleus. It keep on orbiting the nucleus i a circular path. The circular motion of the electron exerts a centripetal force that acts on the electron and keep it from being attracted towards the nucleus.

Write the expression for the balanced forces acting on the electron during its motion around the nucleus.

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

The centripetal force acting on the electron is given as:

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

Here, $m$ is the mass of electron, $v$ is the velocity of electron during motion and $r$ is the radius of the path in which electron rotates.

The electrostatic force acting on the electron due to the nucleus is:

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

Here, $q_1\&q_2$ are the charge of electron and nucleus and $r$ is the distance between the two charges.

The angular velocity of a body can be represented as:

$v=r\omega$

Here, $\omega$ is the angular velocity.

Substitute the values of the centripetal force, electrostatic force and angular velocity in equation (1).

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

Substitute the values in above expression.

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

Therefore, the angular velocity of the electron as it orbits the nucleus due to 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.

Answer 2

Answer : $\omega = 13.41 \times10^{16}\ rad/sec$

Explanation :

Given that,

Charge   $q = 19\times1.6\times 10^{-19} c$

Radius $r = 1.15\times 10^{-10}\ m$

We know that,

Centripetal force = Coulomb force

$\dfrac{mv^{2}}{r} = \dfrac{1}{4\pi\epsilon_{0}}\times \dfrac{q_{1}q_{2}}{r^{2}}$ ......(I)

Now, angular velocity

$\omega = \dfrac{v}{r}$

$v = \omega\ r$

Now, put the value of linear velocity in equation (I)

$\dfrac{m\omega^{2}r^{2}}{r} = \dfrac{1}{4\pi\epsilon_{0}}\times \dfrac{q_{1}q_{2}}{r^{2}}$

$\omega^{2} = \dfrac{9\times10^{9}\times19\times1.6\times10^{-19}\times1.6\times10^{-19}}{1.6\times10^{-31}\times(1.15\times10^{-10})^{3}}$

$\omega ^{2} = 179.89\times 10^{32}\ rad/sec$

$\omega = 13.41\times 10^{16}\ rad/sec$

Hence, this is the required solution.