Question
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 = 0.1Å.
- R = 10 Å.
✓
Answer
For a point nucleus in H-atom,
Ground state: mvr = h,
$\frac{\text{mv}^2}{\text{r}_\text{B}}=-\frac{\text{e}^2}{\text{r}^2_\text{B}}.\frac{1}{4\pi\in_0}$
The electrostatic force of attraction between positively charged nucleus and negatively charged electrons (Coulombian force) provides necessary centripetal force of revolution.
$\frac{\text{mv}^2}{\text{r}_\text{B}}=-\frac{\text{e}_2}{\text{r}^2_\text{B}}.\frac{1}{4\pi\in_0}$
By Bohr's postulates in ground state, we have
mvr = h
$\therefore\ \text{m}\frac{\text{h}^2}{\text{m}^2\text{r}^2_\text{B}}=+\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)\frac{1}{\text{r}_\text{B}^2}$
$\therefore\ \frac{\text{h}^2}{\text{m}}.\frac{4\pi\in_0}{\text{e}^2}=\text{r}_\text{B}=0.51\mathring{\text{A}}$
Potential energy
$-\Big(\frac{\text{e}^2}{4\pi\text{r}_0}\Big).\frac{1}{\text{r}_\text{B}}=-27.2\text{ev}; \text{KE}=\frac{\text{mv}^2}{2}=\frac{1}{2}\text{m}.\frac{\text{h}^2}{\text{m}^2\text{r}^2_\text{B}}=\frac{\text{h}}{2\text{mr}^2_\text{B}}=+13.6\text{eV}$
For a spherical nucleus of radius R,
If R < rB, same result.
If R > > rB the electron moves inside the sphere with redius r'B (r'B= new Bohr radius).
Charge inside, $\text{r}_\text{B}^4=\text{e}\Big(\frac{\text{r}^3_\text{B}}{\text{R}^3}\Big)$
$\therefore\ \text{r}'_\text{B}=\frac{\text{h}^2}{\text{m}}\Big(\frac{4\pi\in_0}{\text{e}^2}\Big)\frac{\text{R}^3}{\text{r}_\text{B}'^3}$
$\text{r}_\text{B}'^4=(0.51\mathring{\text{A}}),\text{R}^3\ \ [\text{R}=10\mathring{\text{A}}]$
$=510(\mathring{\text{A}})^4$
$\therefore\ \text{r}'\text{B}\approx(510)^{\frac{1}{4}}\mathring{\text{A}}<\text{R}$
$\text{KE}=\frac{1}{2}\text{mv}^2=\frac{\text{m}}{2}.\frac{\text{h}}{\text{m}^2\text{r}'^2_\text{B}}=\frac{\text{h}}{2\text{m}}.\frac{1}{\text{r}'^2_\text{B}}$
$=\Big(\frac{\text{h}}{2\text{mr}^2_\text{B}}\Big)\Big(\frac{\text{r}^2_\text{B}}{\text{r}'^2_\text{B}}\Big)=(13.6\text{eV})\frac{(0.51)^2}{(510)^\frac{1}{2}}=\frac{3.54}{22.6}=0.16\text{eV}$
$\text{PE}=+\Big(\frac{\text{e}^2}{4\pi\in_0}\Big).\Big(\frac{\text{r}'^2-3\text{R}^2}{2\text{R}^3}\Big)=+\Big(\frac{\text{e}^2}{4\pi\in_0}.\frac{1}{\text{r}_\text{B}}\Big)\bigg(\frac{\text{r}_\text{B}(\text{r}'^2_\text{B}-3\text{R}^2)}{\text{R}^3}\bigg)$
$=+(27.2\text{eV})\bigg[\frac{0.51(\sqrt{510}-300)}{1000}\bigg]$
$=+(27.2\text{eV}),\frac{-141}{1000}=-3.83\text{eV}$
Ground state: mvr = h,
$\frac{\text{mv}^2}{\text{r}_\text{B}}=-\frac{\text{e}^2}{\text{r}^2_\text{B}}.\frac{1}{4\pi\in_0}$
The electrostatic force of attraction between positively charged nucleus and negatively charged electrons (Coulombian force) provides necessary centripetal force of revolution.
$\frac{\text{mv}^2}{\text{r}_\text{B}}=-\frac{\text{e}_2}{\text{r}^2_\text{B}}.\frac{1}{4\pi\in_0}$
By Bohr's postulates in ground state, we have
mvr = h
$\therefore\ \text{m}\frac{\text{h}^2}{\text{m}^2\text{r}^2_\text{B}}=+\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)\frac{1}{\text{r}_\text{B}^2}$
$\therefore\ \frac{\text{h}^2}{\text{m}}.\frac{4\pi\in_0}{\text{e}^2}=\text{r}_\text{B}=0.51\mathring{\text{A}}$
Potential energy
$-\Big(\frac{\text{e}^2}{4\pi\text{r}_0}\Big).\frac{1}{\text{r}_\text{B}}=-27.2\text{ev}; \text{KE}=\frac{\text{mv}^2}{2}=\frac{1}{2}\text{m}.\frac{\text{h}^2}{\text{m}^2\text{r}^2_\text{B}}=\frac{\text{h}}{2\text{mr}^2_\text{B}}=+13.6\text{eV}$
For a spherical nucleus of radius R,
If R < rB, same result.
If R > > rB the electron moves inside the sphere with redius r'B (r'B= new Bohr radius).
Charge inside, $\text{r}_\text{B}^4=\text{e}\Big(\frac{\text{r}^3_\text{B}}{\text{R}^3}\Big)$
$\therefore\ \text{r}'_\text{B}=\frac{\text{h}^2}{\text{m}}\Big(\frac{4\pi\in_0}{\text{e}^2}\Big)\frac{\text{R}^3}{\text{r}_\text{B}'^3}$
$\text{r}_\text{B}'^4=(0.51\mathring{\text{A}}),\text{R}^3\ \ [\text{R}=10\mathring{\text{A}}]$
$=510(\mathring{\text{A}})^4$
$\therefore\ \text{r}'\text{B}\approx(510)^{\frac{1}{4}}\mathring{\text{A}}<\text{R}$
$\text{KE}=\frac{1}{2}\text{mv}^2=\frac{\text{m}}{2}.\frac{\text{h}}{\text{m}^2\text{r}'^2_\text{B}}=\frac{\text{h}}{2\text{m}}.\frac{1}{\text{r}'^2_\text{B}}$
$=\Big(\frac{\text{h}}{2\text{mr}^2_\text{B}}\Big)\Big(\frac{\text{r}^2_\text{B}}{\text{r}'^2_\text{B}}\Big)=(13.6\text{eV})\frac{(0.51)^2}{(510)^\frac{1}{2}}=\frac{3.54}{22.6}=0.16\text{eV}$
$\text{PE}=+\Big(\frac{\text{e}^2}{4\pi\in_0}\Big).\Big(\frac{\text{r}'^2-3\text{R}^2}{2\text{R}^3}\Big)=+\Big(\frac{\text{e}^2}{4\pi\in_0}.\frac{1}{\text{r}_\text{B}}\Big)\bigg(\frac{\text{r}_\text{B}(\text{r}'^2_\text{B}-3\text{R}^2)}{\text{R}^3}\bigg)$
$=+(27.2\text{eV})\bigg[\frac{0.51(\sqrt{510}-300)}{1000}\bigg]$
$=+(27.2\text{eV}),\frac{-141}{1000}=-3.83\text{eV}$
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