Photoelectric Effect by Lenard and it's Observations Questions in English

(A) The photoelectric current $(I_p)$ is directly proportional to the intensity $(I)$ of the incident light,i.e.,$I_p \propto I$.
Since the number of emitted photoelectrons is directly proportional to the intensity of the incident light,decreasing the intensity $(I)$ leads to a decrease in the number of emitted photoelectrons.
Therefore,the correct option is $A$.

(B) In the photoelectric effect,the emission of electrons from a metal surface depends on the energy of the incident photons.
According to Einstein's photoelectric equation,the energy of an incident photon is given by $E = h\nu$.
For the emission of electrons to occur,the incident frequency $\nu$ must be greater than or equal to the threshold frequency $\nu_0$ of the metal.
If $\nu < \nu_0$,no photoelectric emission occurs regardless of the intensity or amplitude of the light.
Therefore,a minimum threshold frequency is required for the photoelectric effect to take place.

(B) According to the experimental observations of the photoelectric effect,the photoelectric current $i$ is directly proportional to the intensity $I$ of the incident light,provided the frequency of the incident light is above the threshold frequency.
Mathematically,this is expressed as $i \propto I$.
This linear relationship,where the current is zero when the intensity is zero,is represented by a straight line passing through the origin.

(B) From the photoelectric equation,the stopping potential $V_s$ is given by $eV_s = h\nu - \phi = \frac{hc}{\lambda} - \phi$.
This implies that the stopping potential is inversely proportional to the wavelength $\lambda$ of the incident light.
Given that $|V_2| > |V_1|$,where $V_2$ and $V_1$ are the stopping potentials corresponding to wavelengths $\lambda_2$ and $\lambda_1$ respectively.
Since $V_s \propto \frac{1}{\lambda}$,a higher stopping potential corresponds to a smaller wavelength.
Therefore,$|V_2| > |V_1| \implies \lambda_2 < \lambda_1$ or $\lambda_1 > \lambda_2$.

(B) According to the experimental observations of the photoelectric effect:
$1$. The photoelectric current $(I_p)$ is directly proportional to the intensity $(I)$ of the incident light, provided the frequency is above the threshold frequency $(v_0)$. This is shown in graph $(2)$.
$2$. The maximum kinetic energy $(K_{max})$ of the emitted photoelectrons is given by Einstein's photoelectric equation: $K_{max} = hv - \Phi_0$, where $h$ is Planck's constant, $v$ is the frequency of incident light, and $\Phi_0$ is the work function of the metal.
$3$. Since $v = c / \lambda$, where $c$ is the speed of light and $\lambda$ is the wavelength, we have $K_{max} = (hc / \lambda) - \Phi_0$. This shows that $K_{max}$ depends on the frequency (or wavelength) of the incident light, not its intensity.
$4$. Option $B$ is correct because the saturation photoelectric current is directly proportional to the intensity of incident light for a fixed frequency above the threshold frequency.

(D) The saturation photoelectric current $I_P$ is directly proportional to the number of photoelectrons emitted per unit time,which in turn is proportional to the intensity of incident light.
For a point source,the intensity $I$ of light at a distance $d$ from the source follows the inverse square law,i.e.,$I \propto \frac{1}{d^2}$.
Since the saturation photoelectric current $I_P$ is proportional to the intensity of light falling on the metal plate,we have $I_P \propto I \propto \frac{1}{d^2}$.
This relationship $I_P \propto \frac{1}{d^2}$ represents a curve that decreases rapidly as the distance $d$ increases,which corresponds to curve $D$ in the given graph.

(A) According to the experimental observations of the photoelectric effect,the photoelectric current is directly proportional to the intensity of the incident radiation,provided the frequency of the radiation is above the threshold frequency.
Intensity represents the number of photons incident per unit area per unit time.
As intensity increases,the number of photons striking the metal surface per unit time increases.
Consequently,the number of photoelectrons emitted per unit time increases,which leads to an increase in the photoelectric current.
The kinetic energy of the emitted photoelectrons depends only on the frequency of the incident radiation and the work function of the metal,not on the intensity.

(C) The intensity of light $I$ is inversely proportional to the square of the distance $d$ from the source,i.e.,$I \propto \frac{1}{d^2}$.
Since the number of photoelectrons emitted is directly proportional to the intensity of incident light,we have $N \propto I$.
Therefore,the ratio of the number of electrons is given by $\frac{N_2}{N_1} = \frac{I_2}{I_1} = \left( \frac{d_1}{d_2} \right)^2$.
Given $d_1 = 1 \ m$ and $d_2 = \frac{1}{2} \ m$,we get:
$\frac{N_2}{N_1} = \left( \frac{1}{1/2} \right)^2 = (2)^2 = 4$.
Thus,the number of emitted electrons increases by a factor of $4$.

(D) The photoelectric effect is the phenomenon where electrons are emitted from a metal surface when light of a suitable frequency is incident on it. The maximum kinetic energy of the emitted electrons depends only on the frequency of the incident light, not its intensity.
The Einstein photoelectric equation is given by: $K_{max} = h\nu - \Phi_0$
Where:
$K_{max}$ is the maximum kinetic energy of the emitted electrons.
$h\nu$ is the energy of the incident photon.
$\Phi_0 = h\nu_0$ is the work function of the metal.

(D) For the emission of photoelectrons from a surface, the wavelength of the incident light $\lambda$ must be < or equal to the threshold wavelength $\lambda_0$.
Given, $\lambda_0 = 5200\, \text{\AA}$.
Ultraviolet light has a wavelength range typically between $100\, \text{\AA}$ and $4000\, \text{\AA}$, which is < $5200\, \text{\AA}$.
Infrared light has a wavelength greater than $7000\, \text{\AA}$, which is greater than $5200\, \text{\AA}$.
Therefore, light from an ultraviolet lamp (regardless of power, as power only affects the number of photoelectrons, not the emission condition) will cause emission.
Thus, both $(A)$ and $(B)$ are correct.

(D) The threshold frequency is $f_0$. The initial frequency of incident light is $f = 1.5 f_0$.
Since $f > f_0$,photoelectric emission occurs.
Now,the frequency is halved: $f' = \frac{1.5 f_0}{2} = 0.75 f_0$.
For photoelectric emission to occur,the incident frequency must be greater than or equal to the threshold frequency $(f \ge f_0)$.
Since $0.75 f_0 < f_0$,the incident frequency is now below the threshold frequency.
Therefore,no photoelectric emission will take place,regardless of the increase in intensity.
Thus,the photoelectric current will be $0$.

(B) According to the experimental observations of the photoelectric effect,the emission of photoelectrons from a metal surface is an instantaneous process. The time lag between the incidence of a photon and the emission of a photoelectron is approximately $10^{-10} \ s$.

(A) The intensity of light $I$ at a distance $r$ from a point source is given by the inverse square law: $I \propto \frac{1}{r^2}$.
Since the photoelectric current $I_p$ is directly proportional to the intensity of incident light,we have $I_p \propto I$.
Therefore,$I_p \propto \frac{1}{r^2}$.
For distance $r_1$,the current is $I_1 \propto \frac{1}{r_1^2}$.
For distance $r_2$,the current is $I_2 \propto \frac{1}{r_2^2}$.
Taking the ratio of the two currents:
$\frac{I_1}{I_2} = \frac{1/r_1^2}{1/r_2^2} = \frac{r_2^2}{r_1^2}$.
Thus,the ratio $(I_1 : I_2)$ is $r_2^2 : r_1^2$.

(C) The photoelectric current is proportional to the number of photoelectrons emitted per unit time,which is equal to the number of incident photons per unit time (assuming one photon ejects one electron).
The number of photons incident per unit time is given by $\frac{dn}{dt} = \frac{P}{E}$,where $P$ is the power of the incident beam and $E = h\nu$ is the energy of a single photon.
Since power $P = I \times A$ (where $I$ is intensity and $A$ is area),we have $\frac{dn}{dt} = \frac{IA}{h\nu}$.
The photoelectric current $i$ is proportional to the rate of emission of electrons: $i \propto \frac{dn}{dt} = \frac{IA}{h\nu}$.
However,in the standard photoelectric effect,the number of photoelectrons emitted depends only on the number of incident photons (intensity) provided the frequency is above the threshold frequency. The frequency $\nu$ affects the kinetic energy of the electrons,not the number of electrons emitted per unit time for a fixed intensity $I$. Therefore,the photoelectric current is independent of the frequency $\nu$ and depends only on the intensity $I$ of the incident light.

(D) The intensity of light $I$ from a point source follows the inverse square law,$I \propto \frac{1}{d^2}$.
The number of photoelectrons emitted per unit time,$\frac{dn}{dt}$,is directly proportional to the intensity of incident light,which in turn is proportional to the number of photons incident on the surface.
Since $I \propto \frac{1}{d^2}$,the number of photoelectrons emitted per unit time is $\frac{dn}{dt} \propto \frac{1}{d^2}$.
When the distance is increased from $d_1 = 1 \ m$ to $d_2 = 2 \ m$,the ratio of the number of emitted electrons is:
$\frac{(\frac{dn}{dt})_2}{(\frac{dn}{dt})_1} = \frac{d_1^2}{d_2^2} = \frac{1^2}{2^2} = \frac{1}{4}$.
Thus,the number of emitted electrons becomes one-fourth of the initial number. The kinetic energy of the emitted electrons depends on the frequency of the incident light,not on the intensity,so the energy remains unchanged.

(C) photoelectric cell is a device that converts light energy into electrical energy. When light photons of sufficient energy (greater than the work function of the metal) strike the cathode,electrons are emitted due to the photoelectric effect. These electrons flow through the circuit,creating an electric current. Therefore,the energy conversion is from light to electrical energy.

(A) The photoelectric effect occurs when the frequency of incident radiation $v$ is greater than the threshold frequency $v_0$ of the metal.
Ultraviolet $(UV)$ rays have a higher frequency than visible light and infrared rays.
Infrared rays have a lower frequency than $UV$ rays,meaning $v_{IR} < v_{UV}$.
Since the photoelectric effect is triggered by $UV$ rays,the threshold frequency $v_0$ must be less than or equal to $v_{UV}$.
Because $v_{IR} < v_{UV}$,it is possible that $v_{IR} < v_0$,in which case the photoelectric effect will not occur for infrared rays.
$X$-rays and $\gamma$-rays have frequencies much higher than $UV$ rays,so they will definitely cause the photoelectric effect.

(A) In the photoelectric effect,the number of photoelectrons emitted per second is directly proportional to the intensity of the incident light.
Intensity $(I)$ of light from a point source follows the inverse square law,$I \propto \frac{1}{r^2}$,where $r$ is the distance from the source.
When the distance is increased from $r_1 = 1 \ m$ to $r_2 = 2 \ m$,the intensity becomes $I_2 = I_1 \times (\frac{r_1}{r_2})^2 = I_1 \times (\frac{1}{2})^2 = \frac{I_1}{4}$.
Since the number of emitted photoelectrons is proportional to the intensity,the number of emitted electrons will become one-fourth of the initial number.

(A) The photoelectric current depends only on the intensity of the incident radiation and the number of photons striking the surface per unit time,provided the frequency of the incident light is above the threshold frequency.
The work function $W$ determines the threshold frequency $(W = h
u_0)$,but it does not affect the number of photoelectrons emitted per unit time if the intensity remains constant.
Since $hf > W_2$ (and consequently $hf > W_1$),photoemission occurs in both cases.
Therefore,the photocurrent remains the same,i.e.,$I_1 = I_2$.

(C) The intensity $I$ of light from a point source follows the inverse square law,$I \propto \frac{1}{d^2}$,where $d$ is the distance from the source.
Since the number of photoelectrons emitted is directly proportional to the intensity of incident light,we have $N \propto I \propto \frac{1}{d^2}$.
Let $N_1$ be the number of photons at distance $d_1 = 1 \ m$ and $N_2$ be the number of photons at distance $d_2 = 2 \ m$.
Then,$\frac{N_1}{N_2} = \left( \frac{d_2}{d_1} \right)^2$.
Substituting the values: $\frac{N_1}{N_2} = \left( \frac{2}{1} \right)^2 = 4$.
Therefore,$N_1 = 4N_2$,which means the number of photons emitted at $1 \ m$ is $4$ times the number of photons emitted at $2 \ m$.

(C) The photoelectric current is directly proportional to the intensity of the incident light (illumination).
According to the photoelectric effect,when light of a suitable frequency falls on a metal surface,electrons are emitted.
The number of photoelectrons emitted per second is directly proportional to the intensity of the incident radiation.
Therefore,a change in the intensity of illumination results in a corresponding change in the photoelectric current.

(B) According to the photoelectric effect,the number of photoelectrons emitted per unit time is directly proportional to the number of incident photons per unit time.
Since the intensity of light is defined as the energy incident per unit area per unit time,and for a given frequency $\nu$,the energy of each photon is $E = h\nu$ (a constant),the intensity is directly proportional to the number of incident photons.
Therefore,the number of photoelectrons emitted is directly proportional to the intensity of the incident light,provided the frequency $\nu$ is greater than the threshold frequency $\nu_0$.

(B) The number of photoelectrons ejected is directly proportional to the intensity of the incident light. Therefore, if the intensity is doubled from $I$ to $2I$, the number of photoelectrons emitted will also double, becoming $2N$.
Maximum kinetic energy $(K)$ of the photoelectrons is determined by Einstein's photoelectric equation: $K = h\nu - \Phi$, where $h$ is Planck's constant, $\nu$ is the frequency of incident radiation, and $\Phi$ is the work function of the metal.
Since the frequency $\nu$ remains unchanged, the maximum kinetic energy $K$ remains constant regardless of the change in intensity.
Thus, the new values are $2N$ and $K$.

(B) The number of photoelectrons emitted per second is given by $\frac{n}{\Delta t} = \frac{I}{e}$.
Given $I = 0.16 \, \mu A = 0.16 \times 10^{-6} \, A$ and $e = 1.6 \times 10^{-19} \, C$.
$\frac{n}{\Delta t} = \frac{0.16 \times 10^{-6}}{1.6 \times 10^{-19}} = 10^{12} \, \text{photons/s}$.
The number of incident photons per second is given by $\frac{N}{\Delta t} = \frac{W \lambda}{hc}$.
Given $W = 10^{-3} \, W$,$\lambda = 5000 \, \mathring{A} = 5 \times 10^{-7} \, m$,$h = 6.63 \times 10^{-34} \, J \cdot s$,and $c = 3 \times 10^8 \, m/s$.
$\frac{N}{\Delta t} = \frac{10^{-3} \times 5 \times 10^{-7}}{6.63 \times 10^{-34} \times 3 \times 10^8} \approx \frac{5 \times 10^{-10}}{19.89 \times 10^{-26}} \approx 0.251 \times 10^{16} = 2.51 \times 10^{15} \, \text{photons/s}$.
Percentage efficiency = $\frac{n/\Delta t}{N/\Delta t} \times 100 = \frac{10^{12}}{2.51 \times 10^{15}} \times 100 \approx 0.04 \%$.

(D) The intensity of light from a point source at a distance $r$ is given by $I_{light} \propto \frac{1}{r^2}$.
Since the saturation photocurrent is directly proportional to the intensity of the incident light, we have $I \propto I_{light}$.
Therefore, the saturation photocurrent $I$ is inversely proportional to the square of the distance $r$, i.e., $I \propto \frac{1}{r^2}$.
This relationship is represented by a curve that decreases rapidly as $r$ increases, which corresponds to the shape shown in option $D$.

(B) According to the photoelectric effect, the number of photoelectrons emitted per second is directly proportional to the intensity of the incident light, provided the frequency is above the threshold frequency $(v > v_0)$.
Increasing the intensity means increasing the number of photons incident per unit area per unit time. Since each photon interacts with one electron, the number of emitted photoelectrons increases.
The maximum kinetic energy of the photoelectrons depends only on the frequency of the incident light $(K_{max} = hv - \Phi_0)$. Since the frequency is kept constant, the kinetic energy of each emitted photoelectron remains unchanged.
Therefore, increasing the intensity only increases the number of emitted electrons.

(C) In the photoelectric effect,the photoelectric current $(I)$ is directly proportional to the number of photoelectrons emitted per unit time.
This number of emitted photoelectrons is directly proportional to the number of incident photons per unit time (intensity of light).
Since the intensity of the light is kept constant,the number of incident photons per unit time remains unchanged.
Although the frequency of the light is doubled,which increases the kinetic energy of the emitted photoelectrons,it does not change the number of photons incident on the surface.
Therefore,the number of photoelectrons emitted per unit time remains constant,and the photoelectric current remains the same.

(A) The power of incident light $P = 1 \ mW = 10^{-3} \ W$. The wavelength $\lambda = 4560 \ \mathring{A} = 4560 \times 10^{-10} \ m$.
First,calculate the number of photons incident per second $(n_p)$:
$n_p = \frac{P}{E_{photon}} = \frac{P \lambda}{hc} = \frac{10^{-3} \times 4560 \times 10^{-10}}{6.62 \times 10^{-34} \times 3 \times 10^8} \approx 2.295 \times 10^{15} \text{ photons/s}$.
The number of photoelectrons emitted per second $(n_e)$ is given by the quantum efficiency $\eta = 0.5\% = 0.005$:
$n_e = \eta \times n_p = 0.005 \times 2.295 \times 10^{15} \approx 1.1475 \times 10^{13} \text{ electrons/s}$.
The photoelectric current $I_p$ is given by $I_p = n_e \times e$,where $e = 1.6 \times 10^{-19} \ C$:
$I_p = 1.1475 \times 10^{13} \times 1.6 \times 10^{-19} \approx 1.836 \times 10^{-6} \ A$.
Rounding to the provided option,the correct value is $1.856 \times 10^{-6} \ A$.

(D) $1$. The saturation current is directly proportional to the intensity of the incident light. Since the intensity is reduced, the saturation current must decrease, which means the new curve must lie below the original broken-line curve.
$2$. The stopping potential is directly proportional to the frequency of the incident light $(eV_s = h\nu - \phi)$. Since the frequency is increased, the magnitude of the stopping potential increases (it becomes more negative). This means the curve must intersect the negative potential axis at a point further to the left than the original curve.
$3$. Comparing the options, curve $D$ has a lower saturation current and a larger magnitude of stopping potential compared to the original broken-line curve. Therefore, curve $D$ represents the new situation.

(B) At $t = 0$,the position of the photosensitive material is $x_m(0) = -9\,m$ and the position of the source is $x_s(0) = +7\,m$. The distance between them is $r_1 = |7 - (-9)| = 16\,m$.
At $t = 3\,s$,the positions are given by $x = x_0 + vt$.
For the material: $x_m(3) = -9 + (8 \times 3) = -9 + 24 = 15\,m$.
For the source: $x_s(3) = 7 + (4 \times 3) = 7 + 12 = 19\,m$.
The distance between them is $r_2 = |19 - 15| = 4\,m$.
The intensity $I$ of light from a point source follows the inverse square law,$I \propto 1/r^2$.
Therefore,the ratio of intensities is $\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2} = \left(\frac{4}{16}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$.

(C) The radiation pressure $P$ exerted by a beam with intensity $I$ is given by $P = \frac{R}{c} I + \frac{A}{c} I$,where $R$ is the reflection coefficient and $A$ is the absorption coefficient. Here,$R = 0.25$ and $A = 0.75$. Since the reflected light exerts $2I/c$ and absorbed light exerts $I/c$,we have:
$P = 0.25 \left(\frac{2I}{c}\right) + 0.75 \left(\frac{I}{c}\right) = \frac{0.5I + 0.75I}{c} = \frac{1.25I}{c}$.
Given $P = 5 \times 10^{-7}\,N/m^2$,we find intensity $I = \frac{P \times c}{1.25} = \frac{5 \times 10^{-7} \times 3 \times 10^8}{1.25} = 120\,W/m^2$.
The number of photons incident per second $n_p = \frac{I \times A_{area}}{E_{photon}} = \frac{I \times A_{area} \times \lambda}{hc}$.
$n_p = \frac{120 \times (5 \times 10^{-4}) \times 400 \times 10^{-9}}{6.63 \times 10^{-34} \times 3 \times 10^8} \approx 1.206 \times 10^{17} \approx 1.2 \times 10^{17}\,s^{-1}$.
Given $0.1\%$ of photons produce electrons,$n_e = 10^{-3} \times 1.2 \times 10^{17} = 1.2 \times 10^{14}\,s^{-1}$.
Saturation current $i_s = n_e \times e = 1.2 \times 10^{14} \times 1.6 \times 10^{-19} = 1.92 \times 10^{-5}\,A = 19.2\,\mu A$.

(D) The $Photoelectric \text{ } effect$ does not support the wave nature of light.
Interference, diffraction, and polarization are phenomena that can be explained by the wave theory of light.
However, the $Photoelectric \text{ } effect$ can only be explained by considering light as a stream of particles, known as photons, which demonstrates the particle nature of light.

(A) The intensity of light $I$ at a distance $r$ from a point source is given by $I = \frac{P_0}{4 \pi r^2}$, where $P_0$ is the power of the source.
Since the number of photoelectrons liberated per second is directly proportional to the number of photons incident on the surface, and the number of photons is proportional to the intensity of light, we have $N \propto I$.
Therefore, $N \propto \frac{1}{r^2}$.
Given $r_1 = 0.5\; m$ and $r_2 = 1.0\; m$, the ratio of the number of photoelectrons is:
$\frac{N_2}{N_1} = \frac{r_1^2}{r_2^2} = \left( \frac{0.5}{1.0} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
Thus, the number of photoelectrons liberated is reduced by a factor of $4$.

(B) In the photoelectric effect,the photoelectric current $(i)$ is directly proportional to the intensity of incident light $(I)$ for a fixed frequency of incident radiation.
Given that $I_1 > I_2$,the saturation current for intensity $I_1$ will be greater than that for intensity $I_2$,i.e.,$i_1 > i_2$.
However,the stopping potential $(V_0)$ depends only on the frequency of the incident light and the nature of the material of the photosensitive surface,not on the intensity of the light.
Therefore,both curves must intersect the voltage axis at the same stopping potential $(V_0)$.
Comparing this with the given options,the figure that shows different saturation currents but the same stopping potential is represented by the curve where both lines start from the same point $-V_0$ on the voltage axis.

(A) When a high energy $X-$ray beam falls on a metallic ball,photoelectrons are emitted from the surface of the ball due to the photoelectric effect.
As the ball loses electrons,it acquires a net positive charge.
Since the ball is placed in a uniform electric field,it experiences an electrostatic force $F = qE$.
Because the charge $q$ is positive,the direction of the force is the same as the direction of the electric field.
Therefore,the ball will be deflected in the direction of the electric field until it reaches a new equilibrium position.

(B) The condition for photoelectric emission is that the wavelength of the incident radiation $(\lambda)$ must be less than or equal to the threshold wavelength $(\lambda_0)$.
Given: $\lambda_0 = 5200\,\mathring{A}$.
We know that the wavelength range for $UV$ (Ultraviolet) light is typically $100\,\mathring{A}$ to $4000\,\mathring{A}$,and the wavelength range for $IR$ (Infrared) light is greater than $7000\,\mathring{A}$.
Since the wavelength of $UV$ light is less than $5200\,\mathring{A}$ and the wavelength of $IR$ light is greater than $5200\,\mathring{A}$,only the $UV$ lamp can cause photoelectric emission.
Therefore,the correct option is the $50\,W$ $UV$ lamp.

(B, D) The stopping potential (cut-off voltage) depends only on the frequency (or wavelength) of the incident light and is independent of the intensity or the distance of the source from the photoelectric cell. Therefore,the stopping potential remains $0.6\, V$.
The saturation current is directly proportional to the intensity of light. Since the source is a point source,the intensity $I$ follows the inverse square law: $I \propto \frac{1}{d^2}$.
Given the initial distance $d_1 = 0.2\, m$ and initial current $I_1 = 18.0\, mA$,and the new distance $d_2 = 0.6\, m$,the new current $I_2$ is calculated as:
$I_2 = I_1 \times \left(\frac{d_1}{d_2}\right)^2 = 18.0 \times \left(\frac{0.2}{0.6}\right)^2 = 18.0 \times \left(\frac{1}{3}\right)^2 = 18.0 \times \frac{1}{9} = 2.0\, mA$.
Thus,both the stopping potential remains $0.6\, V$ and the saturation current becomes $2.0\, mA$.

(B) In the photoelectric effect,the emission of electrons from a metal surface occurs only when the incident light has a frequency greater than or equal to a specific minimum value. This minimum frequency is known as the threshold frequency $(\nu_0)$. If the frequency of the incident light is less than the threshold frequency,no photoelectric emission occurs,regardless of the intensity of the light.

(D) The kinetic energy of emitted photoelectrons depends on the frequency of the incident photon, not the intensity. Intensity only determines the number of photoelectrons emitted per unit time.
According to Einstein's photoelectric equation, $K_{max} = h\nu - \Phi_0$, where $\nu$ is the frequency of incident light and $\Phi_0$ is the work function.
Furthermore, the ejection of electrons from a metallic surface is only possible if the frequency of the incident photon is greater than or equal to the threshold frequency $(\nu \ge \nu_0)$.
Therefore, both the Assertion and the Reason are incorrect.

(A) Photoelectrons emitted from a metal surface exhibit a range of kinetic energies because electrons within the metal occupy different energy levels within a continuous band structure. When a photon strikes the metal,it transfers its energy to an electron. The energy required to remove an electron from the surface is the work function. Electrons located deeper within the metal require additional energy to reach the surface before they can be emitted,effectively increasing the energy barrier for them. Consequently,electrons emitted from different depths have different kinetic energies. Thus,both the Assertion and the Reason are correct,and the Reason provides a valid explanation for the Assertion.

(D) The photoelectric saturation current depends only on the intensity of the incident light,which corresponds to the number of photons incident per unit time.
It is independent of the frequency of the incident light.
Increasing the frequency of incident light increases the kinetic energy of the emitted photoelectrons,but it does not change the number of photoelectrons emitted per unit time.
Therefore,both the Assertion and the Reason are incorrect.

(B) According to Einstein's photoelectric equation, $K_{max} = h\nu - \Phi$, where $h\nu$ is the energy of the incident photon and $\Phi$ is the work function of the metal.
While the maximum kinetic energy is fixed for a given frequency, photoelectrons are emitted from different depths within the metal surface.
Electrons emitted from deeper layers lose some energy due to collisions before escaping the surface, resulting in a range of kinetic energies from $0$ to $K_{max}$.
Thus, the Assertion is correct.
The work function $\Phi$ is indeed a characteristic property of the metal, making the Reason correct.
However, the Reason does not explain why there is a spread in kinetic energies; the spread is due to the varying depths of emission and subsequent energy loss.
Therefore, both are correct, but the Reason is not the correct explanation of the Assertion.

(B) In the photoelectric effect,the stopping potential depends on the frequency of the incident radiation,while the saturation current depends on the intensity of the incident radiation.
From the graph,curves $(a)$ and $(b)$ have the same stopping potential (the point where the curve intersects the negative x-axis),which implies they have the same frequency.
However,they have different saturation currents (the plateau value of the photo current),which implies they have different intensities.
Therefore,curves $(a)$ and $(b)$ represent incident radiations of the same frequency but different intensities.

(N/A) Intensity of incident light,$I = 10^{-5} \; W m^{-2}$.
Surface area of a sodium photocell,$A = 2 \; cm^2 = 2 \times 10^{-4} \; m^2$.
Incident power of the light,$P = I \times A = 10^{-5} \times 2 \times 10^{-4} = 2 \times 10^{-9} \; W$.
Work function of the metal,$\phi_0 = 2 \; eV = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \; J$.
Number of layers of sodium that absorb the incident energy,$n = 5$.
The effective atomic area of a sodium atom,$A_e$,is approximately $10^{-20} \; m^2$.
The number of conduction electrons in $n$ layers is $n' = n \times \frac{A}{A_e} = 5 \times \frac{2 \times 10^{-4}}{10^{-20}} = 10^{17}$.
The incident power is uniformly absorbed by all the electrons continuously. The energy absorbed per second per electron is $E = \frac{P}{n'} = \frac{2 \times 10^{-9}}{10^{17}} = 2 \times 10^{-26} \; J/s$.
Time required for photoelectric emission,$t = \frac{\phi_0}{E} = \frac{3.2 \times 10^{-19}}{2 \times 10^{-26}} = 1.6 \times 10^7 \; s \approx 0.507 \; \text{years}$.
The time required for photoelectric emission is nearly half a year,which is not practical. Hence,the wave picture is in disagreement with the experimental observation of instantaneous emission.

(N/A) The work function of a metal is defined as the minimum amount of energy required by an electron to just escape from the metal surface.
It is a characteristic property of the metal and depends on the nature of the metal and its surface conditions.
The $SI$ unit of work function is the Joule $(J)$,but in atomic and nuclear physics,it is commonly expressed in electron-volt $(eV)$,where $1 \ eV = 1.602 \times 10^{-19} \ J$.

(N/A) When a clean metal surface is irradiated with electromagnetic radiation of sufficiently high frequency,electrons are emitted from the metal surface. These emitted electrons are known as photoelectrons. This phenomenon is called the photoelectric effect.

(N/A) The photoelectric effect is the phenomenon in which electrically charged particles (electrons) are released from or within a material when it absorbs electromagnetic radiation,such as light.
When light of a sufficiently high frequency (above the threshold frequency) falls on a metal surface,it causes the emission of electrons from the surface.
These emitted electrons are known as photoelectrons.
The process is governed by the equation: $K_{max} = h\nu - \Phi$,where $K_{max}$ is the maximum kinetic energy of the emitted photoelectrons,$h$ is Planck's constant,$\nu$ is the frequency of incident light,and $\Phi$ is the work function of the metal.

(N/A) The photoelectric effect was first observed by the scientist $Hertz$.
$Hertz$ conducted an experimental investigation into the production of electromagnetic waves using spark discharges.
$Hertz$ observed that high-voltage sparks across a detector loop were enhanced when the emitter plate was illuminated by ultraviolet light from an arc lamp.
When light is incident on a metal surface,some electrons near the surface absorb enough energy from the incident radiation to overcome the attraction of positive ions within the metal.
After gaining sufficient energy from the incident light,these electrons escape from the surface of the metal into the surrounding space.

(N/A) Between $1886$ and $1902$,Hallwach and Lenard studied the phenomenon of photoelectric emission.
Lenard's experiment:
As shown in the figure,two metal electrodes are kept at opposite ends inside an evacuated glass tube. One electrode is called the cathode $(C)$ (emitter),which is a photosensitive surface,and the second electrode is called the collector $(A)$. $A$ potential difference is applied between $C$ and $A$.
$A$ quartz window is provided in this tube through which ultraviolet rays are incident on the cathode $(C)$.
When ultraviolet radiation is incident on the cathode $(C)$,photoelectrons are emitted from the surface.
The anode $(A)$ is kept at a positive voltage with respect to the cathode $(C)$,hence it attracts the emitted electrons,resulting in a flow of electrons. Consequently,a current flows in the circuit.

(N/A) The minimum frequency of incident radiation required to eject electrons from a metal surface is called the threshold frequency. It is denoted by $\nu_0$.
The value of the threshold frequency depends solely on the nature of the metal (the work function of the material).
For most metals,such as zinc,cadmium,and magnesium,the threshold frequencies lie in the ultraviolet region,which corresponds to short wavelengths.
For alkali metals,such as lithium,sodium,potassium,caesium,and rubidium,the threshold frequencies lie in the visible region.