91Ó°ÊÓ

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

$\mathbf{∫}\stackrel{\mathbf{→}}{\mathbf{E}}\mathbf{.}\mathit{d}\stackrel{\mathbf{→}}{\mathbf{a}}\mathbf{=}\frac{\mathbf{q}}{{\mathbf{Î\mu }}_{\mathbf{0}}}$ (1)

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

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

The value of $a_{0}0.53×{10}^{-10}m$.

The distance from the center of the atom is r .

(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

$∫p\left(r\right)dv=-e_0$ (2)

Here we have,

$$\begin{gathered} V=\frac{4}{3}{\mathrm{Ï€°ù}}^3 \\ \mathrm{dV}=4{\mathrm{Ï€°ù}}^2\mathrm{dr} \end{gathered}$$

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

$$\begin{gathered} ∫A{e}^{\left(-2r/a_0\right)}\left(4{\mathrm{Ï€°ù}}^2\mathrm{dr}\right)=-e_0 \\ A4\mathrm{Ï€}∫{\mathrm{r}}^2{\mathrm{e}}^{\left(-2\mathrm{r}/{\mathrm{a}}_0\right)}\mathrm{dr}=-{\mathrm{e}}_0 \end{gathered}$$

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,

$$\begin{gathered} A4\mathrm{Ï€}∫{\left({\mathrm{a}}_0\mathrm{u}/2\right)}^2{\mathrm{e}}^{-\mathrm{u}}\left({\mathrm{a}}_0\mathrm{du}/2\right)=-{\mathrm{e}}_0 \\ \frac{{\mathrm{AÏ€²¹}}_0^3}{2}∫{\mathrm{u}}^2{\mathrm{e}}^{-\mathrm{u}}\mathrm{du}=-{\mathrm{e}}_0 \end{gathered}$$

Now, we know that

$∫u^n{e}^{-u}du=n!$

Therefore, $∫u^2{e}^{-u}du=2!$

So, we get

$$\begin{gathered} \frac{{\mathrm{AÏ€²¹}}_0^3}{3}.2!=-e_0 \\ A=\frac{-e_0}{{\mathrm{Ï€²¹}}_0^3} \end{gathered}$$

Hence, the value of A is $A=\frac{-e_0}{{\mathrm{Ï€²¹}}_0^3}$.

(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,

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

By substituting the values in above equation we get,

$$\begin{gathered} q=e_0+A4\mathrm{Ï€}{∫}_0^1{\left({\mathrm{a}}_0\mathrm{u}/2\right)}^2{\mathrm{e}}^{-\mathrm{u}}\left({\mathrm{a}}_0\mathrm{du}/2\right) \\ \mathrm{q}={\mathrm{e}}_0\frac{{\mathrm{AÏ€²¹}}_0^3}{2}{∫}_0^1{\mathrm{u}}^2{\mathrm{e}}^{-\mathrm{u}}\mathrm{du} \end{gathered}$$

Now, we know that

$∫u^n{e}^{-u}du=n!$

Therefore, $∫u^2{e}^{-u}du=2!$

So, we get

$$\begin{gathered} q=e_0+\frac{A{\mathrm{Ï€²¹}}_0^3}{2}\left(1-\frac{5}{e^2}\right) \\ q=e_0-e_0\left(1-\frac{5}{e^2}\right) \\ q=\frac{5e_0}{e^2} \end{gathered}$$

From equation (1) we have,

$∫\stackrel{→}{E}.d\stackrel{→}{a}=\frac{q}{{\mathrm{Î\mu }}_0}$

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

$$\begin{gathered} E\left(4{\mathrm{Ï€²¹}}_0^2\right)=\frac{5e_0}{{\mathrm{Î\mu }}_0{e}^2} \\ E=\frac{5e_0{e}^{-2}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{a}}_0^2} \end{gathered}$$

Hence, the magnitude is$E=\frac{5e_0{e}^{-2}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{a}}_0^2}$ .

(c)

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

Hence, the direction is outward.