$E=-\frac{d V}{d r}=-\frac{d}{d r} \frac{q e^{-\alpha r}}{4 \pi \varepsilon_{0} r}$
$=\frac{-q}{4 \pi \varepsilon_{0}} \frac{d}{d r}\left(\frac{e^{-\alpha r}}{r}\right)$
$=\frac{-q}{4 \pi \varepsilon_{0}}\left(\frac{-\alpha r e^{-\alpha \alpha_{r}}-e^{-\alpha r}}{r^{2}}\right)$
$=\frac{q}{4 \pi \varepsilon_{0}} \cdot e^{-\alpha r} \cdot\left(\frac{\alpha r+1}{r^{2}}\right)$
Elcctric field at $r=\frac{1}{1}$ is
Electric field at $r=\frac{1}{\alpha}$ is
$E\left(r=\frac{1}{\alpha}\right)=\frac{q}{4 \pi \varepsilon_{0}} \cdot e^{-\alpha \times \frac{1}{\alpha}}\left(\frac{\alpha \times \frac{1}{\alpha}+1}{1 / \alpha^{2}}\right)$
$=\frac{(q / e)}{4 \pi \varepsilon_{0}} \cdot 2 \alpha^{2}$
Flux through a sphere of radius $\frac{1}{\alpha}$ is
$\phi=\int E \cdot d A$
For spherical distribution,
$E. d A =E d A \cos 0^{\circ}=E d A$
and $E=$ uniform
So, we have
$\therefore \quad \phi=E \int d A$
$=\frac{ q / e}{4 \pi \varepsilon_{0}} \cdot 2 \alpha^{2} \cdot 4 \pi\left(\frac{1}{\alpha}\right)^{2}$
or $\quad \phi=\frac{2 q}{\varepsilon_{0} e}$
From Gauss' law, we have $\phi=\frac{q_{\text {enclosed }}}{\varepsilon_{0}}$
So, charge enclosed in given sphere is $\frac{2 q}{e}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
(Energy released per fission $= 200\,MeV, 1\,eV = 1.6 \times {10^{ - 19}}J)$
Assertion $A$ : The photoelectric effect does not take place, if the energy of the incident radiation is less than the work function of a metal.
Reason $R$ : Kinetic energy of the photoelectrons is zero, if the energy of the incident radiation is equal to the work function of a metal.
In the light of the above statements, choose the most appropriate answer from the options given below