Question

A) If a proton had a radius $R$ and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a $H-atom$ when $R=0.1\stackrel{o}{A}$.

B) If a proton had a radius $R$ and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a $H-atom$ when $R=10\stackrel{o}{A}$.

Answer 1

A) $\text{Find the expression for radius}.$

Let the ground state radius of normal hydrogen atom be $r$
From Bohr equation for angular momentum,

$mvr=\frac{nh}{2\pi }\Rightarrow v=\frac{nh}{2\pi mr}$ …$\left(i\right)$

Centripetal force on electron is balanced by electrostatic force,

$\frac{m{v}^2}{r}=\frac{k{e}^2}{r^2}$

Now putting the equation $\left(i\right)$ in the above equation we get,

$\Rightarrow r=\frac{n^2{h}^2}{k{e}^{2}4{\pi }^{2}m}$ …$\left(ii\right)$

$\text{Find the kinetic energy and potential energy of the electron}.$

On putting the values of charge and mass of electron and $n=1$,

we get, $r=0.51\stackrel{o}{A}$

Since the $r>R$, nucleus is treated as point charge since electron will be outside the proton,

$\therefore K.E.=\frac{1}{2}m{v}^2$

$=\frac{m}{2}\times {\left(\frac{n\times h}{2\times \pi \times m\times r}\right)}^2$

$=\frac{n^2{h}^2}{8{\pi }^{2}m{r}^2}=13.7eV$ …$\left(iii\right)$

$P.E.=-\frac{e^2}{4\pi {ϵ}_{0}r}=-27.2eV$

$\text{Find the total energy of the electron}.$

On putting the values,

we get, total energy $=K.E+P.E.$

$=(13.7–27.2)eV$

$=-13.5eV$

$\text{Final Answer}:-13.5eV$


B) $\text{Find the expression for radius}.$

Let the ground state radius of normal hydrogen atom be $r$
From Bohr equation for angular momentum we have,

$mvr=\frac{nh}{2\pi }\Rightarrow v=\frac{nh}{2\pi mr}$ …$\left(i\right)$

And from force balance equation,

$\frac{m{v}^2}{r}=\frac{k{e}^2}{r^2}$

Now putting the equation $\left(i\right)$ in the above equation we get,

$\Rightarrow r=\frac{n^2{h}^2}{k{e}^{2}4{\pi }^{2}m}$ …$\left(ii\right)$

$\text{Find the radius of the electron}.$

On putting the values of charge and mass of electron and $n=1$,

we get, $r=0.53\stackrel{o}{A}$

It is clear that radius $r<R$, nucleus is treated as the solid sphere with electron orbiting inside it due to which there will be a new orbital radius of the electron at ground state $r^{\prime }$.

As charge is uniformly distributed, charge enclosed by the electron radius in the nucleus

$e^{\prime }=\frac{e\left(\frac{4}{3}\pi r^{\prime 3}\right)}{\frac{4}{3}\pi R^3}=\frac{e{r}^{\prime 3}}{R^3}$

Force acting on the electron will be equal to the centripetal force,

$\frac{m{v}^2}{r^{\prime }}=\frac{ke{e}^{\prime }}{r^{\prime 2}}$

$\Rightarrow r^{\prime }=\frac{ke{e}^{\prime }}{m{v}^2}=\frac{k{e}^2{r}^{\prime 3}}{m{v}^2{R}^3}$

Replacing $r$ with $r^{\prime }$ and substituting the value of $v$, we get

$r^{\prime }=\frac{n^2{h}^2}{ke{e}^{\prime }4{\pi }^{2}m}$

$r^{\prime }=\frac{n^2{h}^2{R}^3}{k{e}^2{r}^{\prime 3}4{\pi }^{2}m}$

$r^{\prime 4}=\left(\frac{n^2{h}^2}{k{e}^{2}4{\pi }^{2}m}\right)R^3=0.51\stackrel{o}{A}\times R^3$

Putting the value of $R$, we get,

$r^{\prime 4}=0.53\stackrel{o}{A}\times (10\stackrel{o}{A}{)}^4$

$\Rightarrow r^{\prime }=(510{)}^{1/4}\stackrel{o}{A}=4.79\stackrel{o}{A}$

$\text{Find the total energy of the electron}.$

$K.E.=\frac{m{v}^2}{2}$

$=\frac{m}{2}{\left(\frac{nh}{2\pi m{r}^{\prime }}\right)}^2$

$=\frac{n^2{h}^2}{8{\pi }^{2}m{r}^{\prime 2}}$

$=\left(\frac{h^2}{8{\pi }^{2}m{r}^2}\right){\left(\frac{r}{r^{\prime }}\right)}^2$

$K.E.=13.7eV\times {\left(\frac{0.53}{4.79}\right)}^2=0.167eV$

Using the formula for $P.E.$ at point inside the solid charged sphere,

$P.E.=\frac{k\times e^2\times (r^{\prime 2}–3R^2)}{2R^3}$

$=\left(\frac{k\times e^2}{r}\right)\times \left(\frac{r\times (r^{\prime 2}-3R^2)}{2R^3}\right)$

Where $r^{\prime }$ is the point inside the sphere and $R$ is the radius of the sphere

$\therefore$ On putting the values,

$P.E.=27.2eV\times 0.53\stackrel{o}{A}\times \frac{({4.79}^2-3\times {10}^2){\stackrel{o}{A}}^2}{10\times {\stackrel{o}{A}}^3}$

$P.E.=-3.84eV$

$\therefore$ Total energy at ground state $=(0.167-3.84)eV=-3.673eV$

$\text{Final Answer}:-3.673eV$