Question

A hydrogen atom can be considered as having a central point- like proton of positive charge eand an electron of negative charge -ethat is distributed about the proton according to the volume charge density$\mathit{\rho }\mathbf{=}\mathit{A}\mathit{e}\mathit{x}\mathit{p}\left(-2r/a_0\right)$. Hereis a constant,${\mathit{a}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{53}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathit{m}$, andris the distance from the center of the atom.

(a) Using the fact that the hydrogen is electrically neutral, find A. the

(b) Then find magnitude

(c) Then find direction of the atom’s electric field at${\mathit{a}}_{\mathbf{0}}$.

Short answer

(a) The value of A is $A=\frac{-e}{{\mathrm{\pi a}}_0^3}$.

(b) The magnitude is $q=\frac{5e_0}{e^2}$.

(c) The direction of the atom’s electric field is outward.

Step-by-step solution

Gauss law

According to this lawthe charge enclosed divided by the permittivity determines the total electric flux out of a closed surface. It is defined by,

$\mathbf{\int }\overrightarrow{\mathbf{E}}\mathbf{.}\mathit{d}\overrightarrow{\mathbf{a}}\mathbf{=}\frac{\mathbf{q}}{{\mathbf{\epsilon }}_{\mathbf{0}}}$ (1)

Where, q is enclosed charge and ${\mathit{\epsilon }}_{\mathbf{0}}$ is permittivity.

Identification of given data

Here we have, volume charge density$\rho =Aexp\left(-2r/a_0\right)$

The value of $a_{0}0.53\times {10}^{-10}m$.

The distance from the center of the atom is r .

Finding the value of A .

(a)

Suppose, the hydrogen is electrically neutral.

So, the volume integral over the charge density gives the total charge (say$-e_0$)

Therefore, we get

$\int p\left(r\right)dv=-e_0$ (2)

Here we have,

$$\begin{gathered} V=\frac{4}{3}{\mathrm{\pi r}}^3 \\ \mathrm{dV}=4{\mathrm{\pi r}}^2\mathrm{dr} \end{gathered}$$

Now, substitute all numerical values in equation (1) we get,

$$\begin{gathered} \int A{e}^{\left(-2r/a_0\right)}\left(4{\mathrm{\pi r}}^2\mathrm{dr}\right)=-e_0 \\ A4\mathrm{\pi }\int {\mathrm{r}}^2{\mathrm{e}}^{\left(-2\mathrm{r}/{\mathrm{a}}_0\right)}\mathrm{dr}=-{\mathrm{e}}_0 \end{gathered}$$

Now, let

$$\begin{gathered} 2r/a_0=u \\ 2dr/a_0=du \\ dr=a_{0}du/2 \end{gathered}$$

By substituting the values in above equation we get,

$$\begin{gathered} A4\mathrm{\pi }\int {\left({\mathrm{a}}_0\mathrm{u}/2\right)}^2{\mathrm{e}}^{-\mathrm{u}}\left({\mathrm{a}}_0\mathrm{du}/2\right)=-{\mathrm{e}}_0 \\ \frac{{\mathrm{A\pi a}}_0^3}{2}\int {\mathrm{u}}^2{\mathrm{e}}^{-\mathrm{u}}\mathrm{du}=-{\mathrm{e}}_0 \end{gathered}$$

Now, we know that

$\int u^n{e}^{-u}du=n!$

Therefore, $\int u^2{e}^{-u}du=2!$

So, we get

$$\begin{gathered} \frac{{\mathrm{A\pi a}}_0^3}{3}.2!=-e_0 \\ A=\frac{-e_0}{{\mathrm{\pi a}}_0^3} \end{gathered}$$

Hence, the value of A is $A=\frac{-e_0}{{\mathrm{\pi a}}_0^3}$.

finding the magnitude

(b)

Here we have enclosed charge by a Gaussian sphere of radius$r=a_0$including the proton charge$+e_0$ at the center is given by,

$$\begin{gathered} q=e_{0}A4\mathrm{\pi }{\int }_0^{{\mathrm{a}}_0}{\mathrm{r}}^2{\mathrm{e}}^{\left(-2\mathrm{r}/{\mathrm{a}}_0\right)}\mathrm{dr} \\ 2\mathrm{r}/{\mathrm{a}}_0=\mathrm{u} \\ 2\mathrm{dr}/{\mathrm{a}}_0=\mathrm{du} \\ \mathrm{dr}={\mathrm{a}}_0\mathrm{du}/2 \end{gathered}$$

By substituting the values in above equation we get,

$$\begin{gathered} q=e_0+A4\mathrm{\pi }{\int }_0^1{\left({\mathrm{a}}_0\mathrm{u}/2\right)}^2{\mathrm{e}}^{-\mathrm{u}}\left({\mathrm{a}}_0\mathrm{du}/2\right) \\ \mathrm{q}={\mathrm{e}}_0\frac{{\mathrm{A\pi a}}_0^3}{2}{\int }_0^1{\mathrm{u}}^2{\mathrm{e}}^{-\mathrm{u}}\mathrm{du} \end{gathered}$$

Now, we know that

$\int u^n{e}^{-u}du=n!$

Therefore, $\int u^2{e}^{-u}du=2!$

So, we get

$$\begin{gathered} q=e_0+\frac{A{\mathrm{\pi a}}_0^3}{2}\left(1-\frac{5}{e^2}\right) \\ q=e_0-e_0\left(1-\frac{5}{e^2}\right) \\ q=\frac{5e_0}{e^2} \end{gathered}$$

From equation (1) we have,

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

Now, from step 3 and from $q=\frac{5e_0}{e^2}$we obtained that

$$\begin{gathered} E\left(4{\mathrm{\pi a}}_0^2\right)=\frac{5e_0}{{\epsilon }_0{e}^2} \\ E=\frac{5e_0{e}^{-2}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{a}}_0^2} \end{gathered}$$

Hence, the magnitude is$E=\frac{5e_0{e}^{-2}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{a}}_0^2}$ .

Finding the direction of the atom’s electric field at a0 .

(c)

Now, here net charge enclose is given by $q=\frac{5e_0}{e^2}$is positive.

Hence, the direction is outward.