Given, $\mathrm{P}=20 \mathrm{~W}, \lambda=5000 Å=5000 \times 10^{-10} \mathrm{~m}$
$$
\mathrm{d}=2 \mathrm{~m}, \phi_0=2 \mathrm{eV}, \mathrm{r}=1.5 \mathrm{~A}=1.5 \times 10^{-10} \mathrm{~m}
$$
(i) Let number of photon emitted by bulb per second then
$$
\begin{aligned}
&\mathrm{P}=\mathrm{n}^{\prime} \mathrm{h}\left(\frac{\mathrm{c}}{\lambda}\right) \\
&\mathrm{n}^{\prime}=\frac{\mathrm{p}}{\mathrm{hv} / \lambda}=\frac{\mathrm{p} \lambda}{\mathrm{hc}} \\
&=\frac{20 \times\left(5000 \times 10^{-10}\right)}{\left(6.62 \times 10^{-34}\right) \times\left(3 \times 10^8\right)} \\
&\Rightarrow \quad 5 \times 10^{19} \mathrm{~s}^{-1}(\text { no of photons per sec) } \\
&\begin{array}{l}
\text { Number of photons emitted by bulb per second } \mathrm{n}^{\prime}=5 \\
\times 10^{19}
\end{array} \\
&\text { (ii) Energy of the incident photon } \mathrm{E}=\mathrm{hv} \\
&=\frac{\mathrm{hc}}{\lambda}=\frac{\left(6.62 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{5000 \times 10^{-10} \mathrm{j}}
\end{aligned}
$$

$$
\begin{aligned}
&=\frac{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^8\right)}{5000 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV} \\
&=2.48 \mathrm{eV}
\end{aligned}
$$
As the energy is greater than $2 \mathrm{eV}$, i.e. the work function of metal surfae hence photoelectric emission takes place.
$$
\text { (iii) Consider the figure }
$$


Let $\Delta t$ be the time spent in getting the energy $\phi$ (work function of metal).
Energy received by atomic disk in $\Delta \mathrm{t}$ time is
$$
\mathrm{E}=\mathrm{P} \times \mathrm{A} \cdot \Delta \mathrm{t}=\mathrm{P} \times \pi \mathrm{r}^2 . \Delta \mathrm{t}
$$
energy transfered by bulb in full solid angle
$4 \pi \mathrm{d}^2$ to atoms $=4 \pi \mathrm{d}^2 \%$
So, $\mathrm{P} \pi \mathrm{r}^2 \Delta \mathrm{t}=4 \pi \mathrm{d}^2 \%$
$$
\begin{aligned}
& \frac{\mathrm{P}}{4 \pi \mathrm{d}^2} \times \pi \mathrm{r}^2 \Delta \mathrm{t}=\phi_0 \\
\Rightarrow \quad & \Delta \mathrm{t}=\frac{4 \phi_0 \mathrm{~d}^2}{\operatorname{Pr}^2}=\frac{4 \times\left(2 \times 1.6 \times 10^{-19}\right) \times 2^2}{20 \times\left(1.5 \times 10^{-10}\right)^2} \approx 11.4 \mathrm{sec}
\end{aligned}
$$
(iv) Number of photons received by one atomic disc in time $\Delta \mathrm{t}$ is
$$
\begin{aligned}
\mathrm{N} &=\frac{\mathrm{n}^{\prime} \times \pi \mathrm{r}^2}{4 \pi \mathrm{d}^2} \times \Delta \mathrm{t} \\
\Rightarrow \quad &=\frac{\mathrm{n}^{\prime} \mathrm{r}^2 \Delta \mathrm{t}}{4 \mathrm{~d}^2} \\
&=\frac{\left(5 \times 10^{19}\right) \times\left(1.5 \times 10^{-10}\right)^2 \times 11.4}{4 \times(2)^2} \approx 1
\end{aligned}
$$
(1 photon per atom)
(v) As time of emission of electrons is $11.4 \mathrm{sec}$
So, the photoelectric emission is not instantaneous in the problem. It takes about $11.4 \mathrm{sec}$ in photoelectric emission, there is an collision between incident photon and free electron of the metal surface which lasts for very very short interval of time $\left(\approx 10^{-9} \mathrm{~S}\right)$, so we say photoelectric emission is instantaneous.