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Q71P

Expert-verifiedFound in: Page 1218

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**An old model of a hydrogen atom has the charge **** of the proton uniformly distributed over a sphere of radius ${{\mathit{a}}}_{{\mathbf{0}}}$****, with the electron of charge -e**** and mass **** at its center. **

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

- The force on the electron is $F=\frac{{e}^{2}r}{4{\mathrm{\pi a}}_{0}^{3}}$.
- The angular frequency is given by $\omega =\sqrt{\frac{{e}^{2}}{4\pi m{a}_{0}^{3}}}$.

**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.**

Here we have, radius of sphere is ${a}_{0}$

Mass of sphere is *m*.

(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:

$E\left(4{\mathrm{\pi r}}^{2}\right)=\frac{e}{{\epsilon}_{0}}{\left(\frac{r}{{a}_{0}}\right)}^{3}\phantom{\rule{0ex}{0ex}}E=\frac{er}{4{\mathrm{\pi a}}_{0}^{3}}$

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 $F=-\frac{{e}^{2}r}{4{\mathrm{\pi a}}_{0}^{3}}$.

(b)

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

$F=ma\phantom{\rule{0ex}{0ex}}=m\frac{{d}^{2}r}{d{t}^{2}}$

Now, by step 3 we have,

$m\frac{{d}^{2}r}{d{t}^{2}}+\frac{{e}^{2}r}{4{\mathrm{\pi a}}_{0}^{3}}=0\phantom{\rule{0ex}{0ex}}\frac{{d}^{2}r}{d{t}^{2}}+\frac{{e}^{2}r}{4\pi m{a}_{0}^{3}}=0\phantom{\rule{0ex}{0ex}}\frac{{d}^{2}r}{d{t}^{2}}+{\omega}^{2}r=0$

So, we get

${\omega}^{2}=\frac{{e}^{2}}{4{\mathrm{\pi ma}}_{0}^{3}}\phantom{\rule{0ex}{0ex}}\omega =\sqrt{\frac{{e}^{2}}{4{\mathrm{\pi ma}}_{0}^{3}}}$

Hence, the angular frequency is given by $\omega =\sqrt{\frac{{e}^{2}}{4{\mathrm{\pi ma}}_{0}^{3}}}$.

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