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

(C) In the photoelectric effect experiment,the emission of photoelectrons from the cathode surface occurs only when incident radiation (such as ultraviolet light) has a frequency greater than the threshold frequency of the metal.
When the incident ultraviolet radiation is stopped,no more photons are striking the cathode surface.
Consequently,the emission of photoelectrons ceases immediately.
Since the photoelectric current is directly proportional to the number of photoelectrons emitted per unit time,the current drops to zero.

(N/A) The threshold frequency is defined as the minimum frequency of incident radiation (light) required to eject photoelectrons from the surface of a metal.
If the frequency of the incident light is less than the threshold frequency,no photoelectric emission occurs,regardless of the intensity of the light.
It is denoted by the symbol $\nu_0$ and is a characteristic property of the metal surface.

(B) The threshold frequency ($ \nu_0 $) is defined as the minimum frequency of incident radiation required to eject photoelectrons from a metal surface.
It is a characteristic property of the material of the photosensitive surface.
It depends on the work function ($ \Phi_0 $) of the metal, which is given by the relation $ \Phi_0 = h \nu_0 $, where $ h $ is Planck's constant.
Since the work function is specific to the nature of the material, the threshold frequency depends only on the material of the surface and is independent of the intensity or frequency of the incident light.

(C) The threshold frequency $( \nu_0)$ is the minimum frequency of incident radiation required to eject electrons from a metal surface.
For most metals, the work function $(\Phi = h \nu_0)$ is relatively high.
This requires incident photons to have high energy, which corresponds to high frequencies.
These frequencies typically fall in the ultraviolet $(UV)$ region of the electromagnetic spectrum.
Therefore, the correct option is $C$.

(A) The threshold frequency $( \nu_0)$ is the minimum frequency of incident radiation required to eject photoelectrons from a metal surface.
Alkali metals (such as Lithium, Sodium, Potassium, etc.) have very low work functions $( \Phi_0 = h \nu_0)$.
Because their work functions are low, the energy required to eject electrons is small, which corresponds to a lower threshold frequency.
This threshold frequency for alkali metals typically falls within the visible region of the electromagnetic spectrum.
Therefore, visible light is sufficient to cause the photoelectric effect in alkali metals.

(N/A) The experimental setup for the study of the photoelectric effect consists of an evacuated glass tube containing two metal plates: a photosensitive emitter plate $(C)$ and a collector plate $(A)$.
A quartz window $(W)$ is sealed on the tube, allowing ultraviolet radiation to pass through and irradiate the photosensitive plate $(C)$.
When monochromatic ultraviolet radiation of sufficiently low wavelength is incident on the cathode $(C)$, electrons are emitted from its surface.
These emitted electrons are attracted by the collector plate $(A)$ maintained at a positive potential relative to $C$, resulting in a flow of current known as the photoelectric current.
The potential difference between $A$ and $C$ can be varied using a potential divider arrangement, and its polarity can be reversed using a commutator.
The potential difference is measured by a voltmeter $(V)$, and the resulting photoelectric current, which is typically in the $\mu A$ range, is measured by a microammeter $(\mu A)$.
By changing the photosensitive material of plate $C$, the threshold frequency and work function can be studied.
The intensity and frequency of the incident radiation can be varied to study their effects on the photoelectric current.
Filters or colored glass of different frequencies can be placed in the path of the incident radiation to study the effect of frequency, and the distance between the light source and plate $C$ can be varied to study the effect of intensity.

(N/A) The collector $(A = \text{anode})$ is kept at a positive potential with respect to the emitter $(C = \text{cathode})$, so that the emitted electrons are attracted toward the collector $A$.
By keeping the frequency of the incident radiation and the potential difference between the collector and emitter constant, the intensity of the incident radiation is varied, and the resulting photoelectric current is measured.
Experimental observations show that the photoelectric current is directly proportional to the intensity of the incident radiation.
This indicates that the number of photoelectrons emitted per second is directly proportional to the intensity of the incident radiation, as shown in the graph.

(N/A) In the experiment of the photoelectric effect,the collector plate $A$ is kept at a positive potential relative to the emitter.
When the magnitude of the positive potential is increased,the photoelectric current also increases because more electrons are attracted to the collector.
For a particular positive voltage on plate $A$,all emitted electrons reach the collector,and the current reaches its maximum value.
If the collector voltage is increased further beyond this point,the photoelectric current does not increase. This maximum value of current is called the saturation current.
When the collector voltage is sequentially decreased and made negative,a repulsive force acts on the electrons.
As the collector voltage becomes more negative,the repulsive force increases,allowing only the most energetic electrons to reach the collector. Consequently,the collector current decreases rapidly.
The specific value of negative potential at which the photoelectric current becomes zero is known as the cut-off voltage or stopping potential,denoted by $V_{0}$.
Since the energy of electrons emitted from the metal surface varies,the stopping potential provides a measure of the maximum kinetic energy of the photoelectrons.
When the photoelectric current becomes zero,the maximum kinetic energy $(K_{\max})$ of the photoelectrons is equal to the work done by the retarding potential.
Therefore,$K_{\max} = e V_{0}$.

(N/A) The figure shows a graph of the variation of photoelectric current with collector plate potential for different frequencies of incident radiation at a constant intensity.
For different frequencies of incident radiation,the stopping potential value obtained is different,but the saturation current remains the same.
For frequencies $\nu_{3} > \nu_{2} > \nu_{1}$,the corresponding stopping potentials obtained are $V_{03} > V_{02} > V_{01}$.
From this,it can be concluded that when the frequency of incident radiation increases,its energy $[E = h\nu]$ also increases. Hence,the maximum kinetic energy of the emitted photoelectrons depends on the energy (frequency) of the incident radiation.
When the energy of the photoelectrons is higher,a greater retarding voltage (stopping potential) is required to stop them.

(N/A) $(i)$ For a given photosensitive substance, when the frequency of incident radiation is greater than the threshold frequency, the photoelectric current is directly proportional to the intensity of light.
$(ii)$ For a given photosensitive substance and frequency of incident radiation, the saturation current is proportional to the intensity of radiation, but the stopping potential is independent of the intensity.
$(iii)$ For a given photosensitive substance, if the frequency is less than the threshold frequency, photoelectric emission will not take place, regardless of the intensity of radiation.
$(iv)$ For a frequency greater than the threshold frequency $(v > v_{0})$, the stopping potential and maximum kinetic energy vary linearly with the frequency of the incident radiation, but they do not depend on the intensity of the radiation.
$(v)$ Photoelectric emission is an instantaneous process, occurring within a time interval of $10^{-9} \,s$ or less.

(N/A) In the photoelectric effect experiment,the frequency and intensity of incident radiation are controlled as follows:
$1$. Frequency: The frequency of the incident radiation is changed by using different light sources or by using filters that allow only specific wavelengths to pass through. Changing the frequency affects the maximum kinetic energy of the emitted photoelectrons.
$2$. Intensity: The intensity of the incident radiation is changed by varying the distance between the light source and the metal surface (using the inverse square law) or by using neutral density filters to reduce the number of photons per unit area per unit time. Changing the intensity affects the number of photoelectrons emitted per second (photoelectric current).

(A) According to the experimental observations of the photoelectric effect,the number of photoelectrons emitted per second is directly proportional to the intensity of the incident light,provided the frequency of the incident light is above the threshold frequency.
Increasing the intensity means increasing the number of incident photons per unit area per unit time,which results in a higher number of photoelectrons being ejected from the metal surface.

(N/A) $1$. Saturation Current: In the photoelectric effect,as the accelerating potential (anode potential) is increased,the photoelectric current increases. Eventually,a stage is reached where all emitted photoelectrons reach the collector plate. At this point,the current becomes constant and does not increase further with an increase in potential. This maximum constant current is called the saturation current.
$2$. Stopping Potential: For a given frequency of incident radiation,the minimum negative (retarding) potential applied to the anode with respect to the cathode that stops even the most energetic photoelectrons is called the stopping potential or cut-off potential $(V_0)$. At this potential,the photoelectric current becomes zero.
$3$. Cut-off Frequency (Threshold Frequency): For a given photosensitive material,there exists a minimum frequency of incident radiation below which no photoelectric emission takes place,no matter how high the intensity of the radiation is. This minimum frequency is called the threshold frequency or cut-off frequency $(
u_0)$.

(N/A) Light is an electromagnetic wave consisting of oscillating electric and magnetic fields. Phenomena like interference,diffraction,and polarization are explained well by the wave theory.
According to the wave theory,when light is incident on a metal surface,free electrons absorb radiant energy continuously. As the intensity of incident radiation increases,the amplitude of the electric and magnetic fields increases,leading to more energy absorption by the electrons.
Consequently,the wave theory predicts that the maximum kinetic energy of emitted electrons should increase with the intensity of light. Furthermore,it suggests that a sufficiently intense beam of any frequency should be able to eject electrons,implying that no threshold frequency should exist.
These predictions contradict experimental results. Experiments show that maximum kinetic energy is independent of intensity and that a threshold frequency is required for emission.
Additionally,the wave theory suggests that energy absorption occurs continuously over the wavefront. Since energy is spread over a large number of electrons,the energy absorbed per electron is very small. Calculations estimate that it would take hours for an electron to accumulate enough energy to escape. However,experiments show that electron emission is instantaneous,occurring within $10^{-9} \,s$.
Thus,the wave theory fails to explain the fundamental characteristics of the photoelectric effect.

(B) The statement is $False$.
According to the photoelectric effect, the number of photoelectrons emitted per second is directly proportional to the intensity of the incident light, not its frequency.
The frequency of the incident light determines the kinetic energy of the emitted photoelectrons, provided the frequency is above the threshold frequency $(\nu > \nu_0)$.
Increasing the frequency increases the energy of individual photons, but it does not increase the number of photoelectrons emitted.

(N/A) According to Einstein's photoelectric equation,the intensity of incident radiation is defined as the energy incident per unit area per unit time.
In the particle nature of light,intensity is directly proportional to the number of photons incident on the metal surface per unit time.
Since each photon interacts with a single electron,an increase in the number of incident photons (i.e.,an increase in intensity) leads to an increase in the number of photoelectrons emitted from the surface,provided the frequency of incident light is greater than the threshold frequency $(v > v_{0})$.
Therefore,the photoelectric current,which is the rate of flow of these photoelectrons,is directly proportional to the intensity of the incident radiation.

(A) The intensity of radiation (light) is defined as the amount of energy incident per unit area per unit time,perpendicular to the direction of propagation of the wave.
Mathematically,it is given by $I = \frac{E}{A \cdot t}$,where $E$ is the energy,$A$ is the area,and $t$ is the time.
In the context of the photoelectric effect,intensity is directly proportional to the number of photons incident on the surface per unit time.

(C) The particle nature of light,represented by photons,explains phenomena where light interacts with matter as discrete packets of energy.
$1$. Interference,diffraction,and polarization are wave phenomena that require the wave nature of light to be explained.
$2$. The photoelectric effect involves the emission of electrons when light hits a metal surface,which can only be explained by considering light as a stream of particles (photons) with energy $E = h
u$.
Therefore,the photoelectric effect is the correct phenomenon explained by the particle nature of light.

(B) No. When a photon interacts with an electron,the electron absorbs the energy of the photon. However,for the electron to be emitted from the metal surface,it must overcome the work function (binding energy) of the metal. If the energy absorbed by the electron is less than the work function,the electron will not be emitted; it will instead lose its energy through collisions with other atoms within the metal and eventually settle back into a lower energy state. Therefore,only those electrons that absorb energy greater than their binding energy are emitted as photoelectrons.

(A) The threshold frequency is denoted by $\nu_0$ and the work function is $\phi_0 = h\nu_0$.
In the first case,the incident frequency is $\nu_1 = 1.5\nu_0$. Since $\nu_1 > \nu_0$,photoelectric emission occurs.
In the second case,the frequency is halved,so the new frequency is $\nu_2 = \frac{1.5\nu_0}{2} = 0.75\nu_0$.
Since the new incident frequency $\nu_2$ is less than the threshold frequency $\nu_0$ $(\nu_2 < \nu_0)$,no photoelectric emission can take place regardless of the intensity of the light.
Therefore,the photoelectric current will be $0$.

(D) $\rightarrow$ Increasing the intensity of incident light corresponds to an increase in the number of photons incident per unit area per unit time.
$\rightarrow$ The kinetic energy $(K.E.)$ of the ejected photoelectrons is determined by the Einstein photoelectric equation: $K.E._{max} = h\nu - \phi$,where $h$ is Planck's constant,$\nu$ is the frequency of incident light,and $\phi$ is the work function of the metal.
$\rightarrow$ Since the kinetic energy depends only on the frequency of the incident light and not on its intensity,increasing the intensity does not change the $K.E.$ of the ejected electrons.

(A) The saturation current in a photoelectric effect experiment is directly proportional to the number of photoelectrons emitted per unit time.
This number depends on the intensity of the incident light and the quantum efficiency of the material.
Since the intensity of the incident light and the frequency are kept constant,and the condition $h
u > w_2$ is satisfied,photoemission occurs in both cases.
The work function affects the maximum kinetic energy of the emitted electrons,but it does not affect the number of photoelectrons emitted per unit time for a given intensity of incident light.
Therefore,the saturation current remains the same.
Thus,$I_1 = I_2$.

(D) The photoelectric effect is an instantaneous process,so it will not be delayed.
Since the frequency of the incident radiation remains the same,the maximum kinetic energy of the emitted photoelectrons remains unchanged.
The saturation current is directly proportional to the intensity of the incident radiation.
Therefore,if the intensity of the radiation is decreased,the number of photoelectrons emitted per unit time decreases,which leads to a decrease in the saturation current.
Thus,the correct option is $D$.

(C) According to the photoelectric effect, when light of a suitable frequency falls on a metal surface, electrons are emitted. However, these electrons do not all emerge with the same kinetic energy. This is because electrons inside the metal occupy different energy levels. The work function $(\Phi)$ is defined as the minimum energy required to remove an electron from the highest occupied energy level (Fermi level) of the metal. Electrons located at deeper energy levels require more energy than the work function to escape the surface. Therefore, when a photon of energy $(h\nu)$ is absorbed, the kinetic energy of the emitted electron is given by $K_{max} = h\nu - \Phi$ for electrons at the surface, while electrons from deeper levels emerge with less kinetic energy. Thus, option (C) is the most appropriate explanation.

$(A)$ Incorrect. The stopping potential $(V_0)$ depends on both the frequency of incident light ($\nu$) and the work function ($\Phi$) of the metal, given by $eV_0 = h\nu - \Phi$.
$(B)$ Correct. The saturation current is directly proportional to the intensity of the incident light because intensity determines the number of photons, and thus the number of photoelectrons emitted per second.
$(C)$ Incorrect. The maximum kinetic energy $(K_{max})$ of a photoelectron depends only on the frequency of the incident light and the work function of the metal; it is independent of the intensity of light.
$(D)$ Incorrect. The photoelectric effect cannot be explained by the wave theory of light; it requires Einstein's particle theory (photon model) to explain the instantaneous emission and frequency dependence.

(B) The condition for photoelectric emission is that the energy of the incident photon $(E)$ must be greater than or equal to the work function $(\Phi_0)$ of the metal surface.
Given incident energy $E = 2.20\,eV$.
Work functions are:
For $Cs$: $\Phi_0 = 2.14\,eV$
For $K$: $\Phi_0 = 2.30\,eV$
For $Na$: $\Phi_0 = 2.75\,eV$
Comparing $E$ with $\Phi_0$:
For $Cs$: $2.20\,eV > 2.14\,eV$ (Emission occurs)
For $K$: $2.20\,eV < 2.30\,eV$ (No emission)
For $Na$: $2.20\,eV < 2.75\,eV$ (No emission)
Therefore, only $Cs$ will emit photoelectrons.

(D) The photoelectric current depends on the number of photons incident on the photoelectric cell per unit time.
Since the lens is used to focus light from an extended source onto the cell,the total amount of light energy (and thus the number of photons) collected by the lens depends on its aperture (diameter).
Given that the diameter of the new lens is the same as the original lens,the amount of light energy intercepted by the lens remains unchanged.
Therefore,the number of photons incident on the photoelectric cell remains the same.
As a result,the photoelectric current $I$ remains unchanged.

(C) Let the threshold frequency be $f_0$. The initial frequency of incident light is $f_i = 1.5 f_0$.
When the frequency is halved,the new frequency becomes $f' = \frac{1.5 f_0}{2} = 0.75 f_0$.
According to the photoelectric effect,emission only occurs if the incident frequency $f$ is greater than or equal to the threshold frequency $f_0$ $(f \ge f_0)$.
Since $0.75 f_0 < f_0$,the incident light does not have enough energy to eject electrons from the surface,regardless of the intensity.
Therefore,the number of photoelectrons emitted will be zero.

(C) The photoelectric effect depends on the intensity and frequency of incident light.
Since the incident lights have the same wavelength,their frequencies are identical. Therefore,the maximum kinetic energy of the emitted photoelectrons is the same,which means the stopping potential $(V_0)$ remains the same for both intensities.
Since the intensity $I_2 > I_1$,the number of photoelectrons emitted per unit time is greater for $I_2$ than for $I_1$. Consequently,the saturation current for $I_2$ will be greater than that for $I_1$.
Comparing this with the given figures,Figure $C$ correctly shows that both curves start from the same stopping potential $(-V_0)$ on the negative $V$-axis,and the saturation current for $I_2$ is higher than for $I_1$.
Thus,the correct figure is $C$.

(B) The wave nature of light successfully explains phenomena such as reflection,refraction,interference,diffraction,and polarization.
However,the photoelectric effect involves the emission of electrons from a metal surface when light of a suitable frequency strikes it. This phenomenon cannot be explained by the wave theory of light because the wave theory suggests that the energy of light depends on intensity,whereas the photoelectric effect depends on the frequency of the incident light.
Therefore,the photoelectric effect provides evidence for the particle nature of light (photons).

(B) Given power $P = 200 \ W$ and work function $\phi = 6.25 \ eV$.
Since the frequency is just above the threshold frequency,the initial kinetic energy of photoelectrons is zero.
Energy of one photon $E_1 = h\nu = \phi = 6.25 \ eV = 6.25 \times 1.6 \times 10^{-19} \ J = 10^{-18} \ J$.
Number of photons incident per second $N = P / E_1 = 200 / 10^{-18} = 2 \times 10^{20} \ s^{-1}$.
Since efficiency is $100 \%$,the number of electrons emitted per second is also $N = 2 \times 10^{20} \ s^{-1}$.
When these electrons are accelerated by a potential $V = 500 \ V$,their final kinetic energy $K = eV = 1.6 \times 10^{-19} \times 500 = 8 \times 10^{-17} \ J$.
The momentum of each electron upon impact is $p = \sqrt{2m_eK} = \sqrt{2 \times 9 \times 10^{-31} \times 8 \times 10^{-17}} = \sqrt{144 \times 10^{-48}} = 12 \times 10^{-24} \ kg \ m/s$.
The force exerted on the anode is $F = N \times p = (2 \times 10^{20}) \times (12 \times 10^{-24}) = 24 \times 10^{-4} \ N$.
Comparing with $F = n \times 10^{-4} \ N$,we get $n = 24$.

(B) The stopping potential $V_S$ is given by the Einstein's photoelectric equation: $V_S = \frac{h\nu - \phi}{e}$,where $h$ is Planck's constant,$\nu$ is the frequency of incident light,$\phi$ is the work function,and $e$ is the charge of an electron.
From this equation,it is clear that the stopping potential depends only on the frequency of the incident light and the material of the surface (work function). It does not depend on the intensity of the incident light.
Therefore,Assertion $A$ is false.
The intensity of light is defined as the energy incident per unit area per unit time,which is proportional to the number of photons incident per unit time. Increasing the intensity increases the number of photons,which in turn increases the number of photoelectrons emitted per second,provided the frequency is above the threshold frequency. Thus,Reason $R$ is true.

(A) $1$. The photoelectric current $(I)$ is directly proportional to the intensity of incident light,provided the frequency of light is above the threshold frequency. Therefore,the graph $A$ is correct,showing a linear relationship between photoelectric current and intensity.
$2$. The photoelectric current does not depend on the frequency of incident light (as long as it is above the threshold frequency). Once the threshold frequency is reached,the current remains constant with respect to frequency. Therefore,graphs $C$ and $D$ are incorrect.

(D) 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 surface.
The kinetic energy of photoelectrons depends only on the frequency of the incident radiation $(\nu)$ and the work function of the material $(\Phi)$.
Intensity of incident radiation affects the number of photoelectrons emitted per unit time (photoelectric current),but it does not affect the kinetic energy of individual photoelectrons.
Therefore,both Statement-$I$ and Statement-$II$ are incorrect.

(B) The Assertion $(A)$ is correct. When monochromatic light is incident on a metal surface, photoelectrons are emitted with a range of kinetic energies from $0$ to $K_{max} = h\nu - \Phi$. This spread occurs because electrons originating from deeper layers of the metal lose some of their kinetic energy due to collisions with other atoms before escaping the surface.
The Reason $(R)$ is incorrect. The work function $\Phi$ is a property of the metal surface itself, defined as the minimum energy required to remove an electron from the surface. It does not vary with the depth of the electron inside the metal; rather, the energy loss during the escape process causes the observed spread in kinetic energy.

(C) In the photoelectric effect,the saturation photoelectric current is directly proportional to the intensity of the incident light,provided the frequency of the incident light is greater than the threshold frequency $(ν > ν_0)$.
Since the intensity of the incident light is doubled,the number of photons incident per unit time also doubles.
Consequently,the number of photoelectrons emitted per unit time doubles,leading to a doubling of the saturation photoelectric current.
Therefore,the correct option is $C$.

(D) According to Einstein's photoelectric equation,$K_{max} = h\nu - \Phi_0$,where $K_{max}$ is the maximum kinetic energy,$h\nu$ is the energy of the incident photon,and $\Phi_0$ is the work function of the metal.
$1$. The wave theory of light fails to explain the photoelectric effect because it cannot account for the instantaneous emission of electrons or the existence of a threshold frequency.
$2$. $K_{max}$ depends on the frequency of incident light $(
u)$,not its intensity.
$3$. The stopping potential $(V_s)$ is given by $eV_s = K_{max} = h\nu - \Phi_0$. Thus,it depends on both the frequency of incident light and the work function of the metal.
$4$. The saturation current is directly proportional to the number of photoelectrons emitted per second,which is directly proportional to the intensity of the incident light. Therefore,as intensity increases,the saturation current increases.

(A) $1$. Frequency $(f)$: The stopping potential $(V_0)$ is the potential at which the photocurrent becomes zero. It is given by $eV_0 = hf - \phi$,where $\phi$ is the work function. $A$ more negative stopping potential indicates a higher frequency of incident radiation.
From the graph,the stopping potentials are $V_{0a} = V_{0b} < V_{0c} < V_{0d}$.
Therefore,$f_a = f_b < f_c < f_d$.
$2$. Intensity $(I)$: The saturation current is directly proportional to the intensity of the incident radiation. From the graph,the saturation currents for curves $c$ and $d$ are equal,while $a$ and $b$ have lower saturation currents.
Thus,$I_c = I_d$ and $I_a < I_b < I_c = I_d$.
Comparing the given options,option $A$ is the most consistent with the relationships derived,specifically $f_b = f_a$ (from $f_a = f_b$) and $I_c = I_d$.

(C) Photoelectric emission occurs only when the energy of the incident photon $(E)$ is greater than or equal to the work function $(\Phi_0)$ of the metal surface.
Given incident energy $E = 2.41 \ eV$.
Work functions are:
For $Cs$: $\Phi_0 = 2.14 \ eV$
For $K$: $\Phi_0 = 2.30 \ eV$
For $Na$: $\Phi_0 = 2.75 \ eV$
Comparing $E$ with $\Phi_0$:
For $Cs$: $2.41 \ eV > 2.14 \ eV$ (Emission occurs)
For $K$: $2.41 \ eV > 2.30 \ eV$ (Emission occurs)
For $Na$: $2.41 \ eV < 2.75 \ eV$ (No emission)
Therefore, both $Cs$ and $K$ surfaces will emit photoelectrons.

(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 $(I)$ is defined as the energy incident per unit area per unit time,and for a given frequency,the energy of each photon is constant $(E = hv)$,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 $(I)$,provided the frequency $v$ is greater than the threshold frequency $v_0$.

(C) The photocurrent is independent of the frequency of the incident photon,provided the frequency is above the threshold frequency.
However,when the intensity of the incident light increases,the number of photons incident per unit area per unit time increases.
This leads to an increase in the number of photoelectrons emitted from the metal surface,thereby increasing the photocurrent.

(C) In the photoelectric effect,the stopping potential depends only on the frequency of the incident radiation and the work function of the metal surface. Since all four curves intersect the potential axis at the same point (the same stopping potential),the frequency of all four radiations must be the same. Therefore,$f_a = f_b = f_c = f_d$.
The saturation photocurrent is directly proportional to the intensity of the incident radiation. From the graph,the saturation current values are $I_a < I_b < I_c < I_d$. Therefore,the intensities follow the same order: $I_a < I_b < I_c < I_d$.

(D) The threshold frequency of the material is $\nu_0$. The initial frequency of incident light is $\nu_1 = 1.5 \nu_0$. Since $\nu_1 > \nu_0$,photoelectric emission occurs.
When the frequency is halved,the new frequency becomes $\nu_2 = \frac{1.5 \nu_0}{2} = 0.75 \nu_0$.
For photoelectric emission to occur,the incident frequency must be greater than or equal to the threshold frequency $(\nu \ge \nu_0)$.
Since $0.75 \nu_0 < \nu_0$,the new frequency is below the threshold frequency.
Therefore,no photoelectric emission will take place regardless of the intensity of the incident light.
Thus,the photocurrent becomes zero.

(C) The threshold frequency of the metal is given as $f_0 = 15 \times 10^{14} \, Hz$.
The frequency of the incident light with wavelength $\lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, m$ is calculated as:
$f = \frac{c}{\lambda} = \frac{3 \times 10^8 \, m/s}{6000 \times 10^{-10} \, m} = \frac{3 \times 10^8}{6 \times 10^{-7}} \, Hz = 0.5 \times 10^{15} \, Hz = 5 \times 10^{14} \, Hz$.
Since the frequency of the incident light $(f = 5 \times 10^{14} \, Hz)$ is less than the threshold frequency $(f_0 = 15 \times 10^{14} \, Hz)$, the energy of the incident photons is insufficient to overcome the work function of the metal.
Therefore, no photoelectric emission takes place.

(C) Initial frequency is $v = 2v_0$,where $v_0$ is the threshold frequency.
When the incident frequency is made $\left(\frac{1}{3}\right)^{rd}$,the new frequency becomes $v' = \frac{1}{3} \times 2v_0 = \frac{2}{3}v_0$.
Since $v' < v_0$,the incident light frequency is now less than the threshold frequency required for photoelectric emission.
According to the laws of the photoelectric effect,if the incident frequency is less than the threshold frequency,no photoelectrons are emitted regardless of the intensity of the light.
Therefore,the photoelectric current will be zero.

(D) The stopping potential $(V_0)$ is determined by the maximum kinetic energy of the emitted photoelectrons, which is given by Einstein's photoelectric equation: $K_{max} = h\nu - \Phi = eV_0$.
Here, $h$ is Planck's constant, $\nu$ is the frequency of incident radiation, $\Phi$ is the work function of the surface, and $e$ is the charge of an electron.
Since the stopping potential depends only on the frequency of the incident radiation and the nature of the material (work function), it is independent of the intensity of the incident radiation.
Therefore, increasing the intensity of the incident radiation does not change the stopping potential.

(C) In the photoelectric effect,the photoelectric current is directly proportional to the intensity of the incident light,provided the frequency is above the threshold frequency. Since the frequency and accelerating potential are kept constant,increasing the intensity of the incident light increases the number of photons incident per unit time,which in turn increases the number of photoelectrons emitted per unit time. Therefore,the photoelectric current increases.

(A) In the photoelectric effect,the photocurrent is directly proportional to the number of photoelectrons emitted per unit time.
Since each incident photon ejects at most one photoelectron,the number of photoelectrons emitted depends solely on the number of incident photons per unit time,which is determined by the intensity of the incident light.
The frequency of the incident light determines the kinetic energy of the emitted photoelectrons,provided the frequency is above the threshold frequency,but it does not affect the magnitude of the photocurrent.

(B) The stopping potential $(V_0)$ is determined by the maximum kinetic energy of the emitted photoelectrons, which depends on the frequency of the incident radiation, not its intensity.
According to Einstein's photoelectric equation: $K_{max} = h\nu - \Phi_0 = eV_0$.
Since the intensity of light only affects the number of photons incident per unit area per unit time (and thus the number of photoelectrons emitted), it does not change the maximum kinetic energy of the individual electrons.
Therefore, the stopping potential remains unchanged when the intensity is increased.

(D) The threshold frequency is given as $\nu_{0} = 15 \times 10^{14} \,Hz$.
The threshold wavelength $\lambda_{0}$ is calculated as $\lambda_{0} = \frac{c}{\nu_{0}} = \frac{3 \times 10^{8}}{15 \times 10^{14}} = 0.2 \times 10^{-6} \,m = 2000 \text{ Å}$.
The wavelength of incident light is $\lambda = 6000 \text{ Å}$.
For photoelectric emission to occur, the incident wavelength must be less than or equal to the threshold wavelength $(\lambda \leq \lambda_{0})$.
Since $\lambda = 6000 \text{ Å} > \lambda_{0} = 2000 \text{ Å}$, the energy of the incident photons is less than the work function of the metal.
Therefore, photoelectrons will not be emitted.