Question

An old model of a hydrogen atom has the chargeof the proton uniformly distributed over a sphere of radius${\mathit{a}}_{\mathbf{0}}$, with the electron of charge -eand massat its center.

1. What would then be the force on the electron if it were displaced from the center by a distance$\mathit{r}\mathbf{\le }{\mathit{a}}_{\mathbf{0}}$?
2. What would be the angular frequency of oscillation of the electron about the center of the atom once the electron was released?

Short answer

1. The force on the electron is$F=\frac{e^{2}r}{4{\mathrm{\pi a}}_0^3}$.
2. The angular frequency is given by $\omega =\sqrt{\frac{e^2}{4\pi m{a}_0^3}}$.

Step-by-step solution

The force exerted on the electron

The force exerted on the electron if it is moved by a distance of $\mathit{r}\mathbf{\le }{\mathit{a}}_{\mathbf{0}}$is equal to its charge times the local electric field.

F = -eE (1)

Where, E is electric field and e is charge.

Identification of given data

Here we have, radius of sphere is$a_0$

Mass of sphere is m.

Finding the force on the electron if it were displaced from the center by a distance r≤a0

(a)

By Gauss law we have,

$\int \overrightarrow{E}.d\overrightarrow{a}=\frac{q}{{\epsilon }_0}$

Where, q is enclosed charge and ${\epsilon }_0$is permittivity.

We need to determine the electric field at r distance from the nucleus, therefore the enclosed charge is calculated as the product of the volumes of spheres with r and a multiplied by the proton charge, as follows:

$q=e{\left(\frac{r}{a_0}\right)}^3$

So, the electric field at distance of r is given by:

$$\begin{gathered} E\left(4{\mathrm{\pi r}}^2\right)=\frac{e}{{\epsilon }_0}{\left(\frac{r}{a_0}\right)}^3 \\ E=\frac{er}{4{\mathrm{\pi a}}_0^3} \end{gathered}$$

Now, by substituting the value of in equation (1) we get,

$F=-\frac{e^{2}r}{4{\mathrm{\pi a}}_0^3}$

Hence, the force on the electron is $F=-\frac{e^{2}r}{4{\mathrm{\pi a}}_0^3}$.

Finding the angular frequency of oscillation of the electron about the center of the atom once the electron was released

(b)

Now, from the Newton’s second law of motion we have,

$$\begin{gathered} F=ma \\ =m\frac{d^{2}r}{d{t}^2} \end{gathered}$$

Now, by step 3 we have,

$$\begin{gathered} m\frac{d^{2}r}{d{t}^2}+\frac{e^{2}r}{4{\mathrm{\pi a}}_0^3}=0 \\ \frac{d^{2}r}{d{t}^2}+\frac{e^{2}r}{4\pi m{a}_0^3}=0 \\ \frac{d^{2}r}{d{t}^2}+{\omega }^{2}r=0 \end{gathered}$$

So, we get

$$\begin{gathered} {\omega }^2=\frac{e^2}{4{\mathrm{\pi ma}}_0^3} \\ \omega =\sqrt{\frac{e^2}{4{\mathrm{\pi ma}}_0^3}} \end{gathered}$$

Hence, the angular frequency is given by $\omega =\sqrt{\frac{e^2}{4{\mathrm{\pi ma}}_0^3}}$.