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

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{max})$ of emitted photoelectrons is given by $K_{max} = h\nu - \Phi$,where $h$ is Planck's constant,$\nu$ is the frequency of incident radiation,and $\Phi$ is the work function of the metal.
Since the stopping potential $(V_{s})$ is related to the maximum kinetic energy by the relation $eV_{s} = K_{max}$,we have $eV_{s} = h\nu - \Phi$.
This equation shows that the stopping potential $(V_{s})$ depends only on the frequency $(\nu)$ of the incident radiation and the work function $(\Phi)$ of the metal surface.
Intensity of light affects the number of photons incident per unit area per unit time,which in turn affects the number of photoelectrons emitted (photoelectric current),but it does not affect the maximum kinetic energy of the individual photoelectrons.
Therefore,if the frequency $(\nu)$ is kept constant,the stopping potential $(V_{s})$ remains the same regardless of changes in the intensity of the incident light.

(D) . Interference is a phenomenon in which two waves of the same frequency superpose to give a resultant intensity different from the sum of their separate intensities. Thus,it cannot exhibit the particle nature of light.
$B$. Diffraction is a phenomenon in which light bends at the sharp edges of an obstacle or an aperture. Thus,it also cannot exhibit the particle nature of light.
$C$. Polarisation of light is a property due to which a light ray,after emerging through a crystal (like tourmaline),has vibrations in a plane perpendicular to its direction of propagation. Thus,it also cannot explain the particle nature of light.
$D$. The Photoelectric effect states that light travels in the form of bundles or packets of energy,called photons. This effect is explained on the basis of the quantum (particle) nature of light. Hence,it clearly explains the particle nature of light.
Therefore,option $D$ is correct.

(B) The number of photoelectrons emitted per second is directly proportional to the intensity of the incident light.
For a point source of light,the intensity $I$ follows the inverse square law: $I \propto \frac{1}{d^2}$,where $d$ is the distance from the source.
When the distance is doubled $(d' = 2d)$,the new intensity $I'$ becomes $I' = \frac{I}{2^2} = \frac{I}{4}$.
Since the number of photoelectrons emitted is directly proportional to the intensity,the number of photoelectrons emitted becomes one-fourth of the initial number.
The energy of each emitted electron depends on the frequency of the incident light,not on its intensity. Therefore,the energy of each emitted electron remains unchanged.

(A) According to Einstein's photoelectric equation,a single incident photon ejects a single electron.
When the intensity of light increases,the number of incident photons per unit area per unit time increases.
Since each photon ejects one electron,the number of emitted photoelectrons increases,which leads to an increase in the photo-current.
The maximum kinetic energy of emitted photoelectrons is given by $K_{\max} = h\nu - \Phi$,where $h\nu$ is the energy of the incident photon and $\Phi$ is the work function.
Since the energy of individual photons does not change with intensity,the maximum kinetic energy of the emitted photoelectrons remains independent of the intensity of incident light.

(D) The threshold frequency $\nu_0$ is related to the work function $\Phi$ by the equation $\Phi = h\nu_0$,where $h$ is Planck's constant.
Metals like Zinc,Magnesium,and Cadmium have high work functions,meaning their threshold frequencies lie in the ultraviolet $(UV)$ region of the electromagnetic spectrum.
Sodium is an alkali metal with a very low work function (approximately $2.3 \ eV$ to $2.7 \ eV$).
Because of this low work function,the threshold frequency of Sodium lies in the visible region,not the ultraviolet region.
Therefore,the correct option is $D$.

(A) In the photoelectric effect,the photoelectric current $I$ is directly proportional to the intensity of the incident light,provided the frequency $\nu$ of the incident light is greater than the threshold frequency $\nu_0$.
If the intensity of the incident light is kept constant while the frequency $\nu$ is increased,the number of photons incident per unit time remains constant (since $P = n h \nu$,if power $P$ is constant,$n$ decreases as $\nu$ increases). However,in practical experimental setups,the photoelectric current is independent of the frequency of light as long as $\nu > \nu_0$.
Therefore,the graph of photoelectric current $I$ versus frequency $\nu$ is a horizontal line for $\nu > \nu_0$ and zero for $\nu < \nu_0$. This corresponds to the graph shown in option $A$.

(B) In the photoelectric effect,the saturation photocurrent is directly proportional to the intensity of the incident radiation.
It is independent of the frequency of the incident radiation.
Since the intensity is doubled,the saturation photoelectric current will also be doubled.
The change in frequency does not affect the saturation current.
Therefore,the saturation photoelectric current becomes doubled.

(A) From the graph,we observe that the stopping potential $V_{0}$ is the same for curves $1$ and $2$. However,for curve $3$,the stopping potential is greater in magnitude than that for curves $1$ and $2$.
We know that the stopping potential is given by the relation: $e V_{0} = E_{\max} = h \gamma - \phi_{0}$,where $\phi_{0}$ is the work function of the metal.
Since $V_{0}$ is the same for curves $1$ and $2$,it implies that the incident frequencies are equal,i.e.,$\gamma_{1} = \gamma_{2}$.
For curve $3$,the stopping potential is higher,which means $\gamma_{3} > \gamma_{1} = \gamma_{2}$.
Regarding the saturation current,it depends on the intensity of the incident light. Curves $2$ and $3$ reach the same saturation current,meaning $I_{2} = I_{3}$,while curve $1$ has a lower saturation current,so $I_{1} < I_{2} = I_{3}$.
Therefore,comparing the options,$\gamma_{1} = \gamma_{2}$ and $I_{1} \neq I_{2}$ is the correct statement.

(D) The photoelectric current is directly proportional to the intensity of the incident light,provided the incident frequency is greater than the threshold frequency.
Since the saturation current depends only on the number of photoelectrons emitted per second,which is directly proportional to the intensity of the incident light,doubling the intensity will double the saturation current.
The frequency of the incident light affects the kinetic energy of the emitted photoelectrons,not the number of photoelectrons (saturation current),as long as it remains above the threshold frequency.
Therefore,when the intensity is doubled,the photoelectric saturation current doubles.

(A) The classical electromagnetic theory of light treats light as a continuous wave. According to this theory,the energy of the wave depends on its intensity (amplitude). However,experimental observations of the photoelectric effect show that the emission of electrons depends on the frequency of the incident light,not its intensity. Furthermore,the emission is instantaneous,which contradicts the classical prediction of a time lag for energy absorption. Therefore,the electromagnetic theory of light failed to explain the photoelectric effect.

(C) For a fixed frequency of incident light,the saturation photo-current is directly proportional to the intensity of the incident light.
From the given graph,the saturation photo-current for light of intensity $I_{2}$ is higher than that for light of intensity $I_{1}$.
Therefore,we can conclude that $I_{1} < I_{2}$.

(B) The work function $\phi_0$ of a metal is defined by the threshold wavelength $\lambda_0$ as $\phi_0 = \frac{hc}{\lambda_0}$.
Given the threshold wavelength $\lambda_0 = 5420 Å$.
Using the relation $\phi_0 (\text{in eV}) = \frac{12400}{\lambda_0 (\text{in Å})}$.
Substituting the value: $\phi_0 = \frac{12400}{5420} eV$.
$\phi_0 \approx 2.29 eV$.
Thus,the work function of sodium is $2.29 eV$.

(C) photocell is a device that operates based on the photoelectric effect. In this process,when light (photons) of suitable frequency falls on a photosensitive surface,electrons are emitted. This flow of electrons constitutes an electric current. Therefore,photocells convert light energy into electrical energy.

(B) The photoelectric effect occurs when light of a sufficiently high frequency (or wavelength shorter than the threshold wavelength) strikes a metal surface.
According to Einstein's photoelectric equation, $KE_{max} = h\nu - \Phi$, where $\Phi$ is the work function.
$1$. Thermionic emission occurs at high temperatures, not photoelectric emission.
$2$. Photoemission occurs if the incident wavelength $\lambda$ is less than the threshold wavelength $\lambda_0$ (critical value).
$3$. The $KE$ of photoelectrons depends on the frequency of incident radiation, not the amplitude.
$4$. The photoelectric current is proportional to the intensity of incident radiation, not the frequency.
Therefore, statement $2$ is correct.

(A) $1$. The stopping potential $V_0$ is determined by the frequency of the incident radiation $(eV_0 = hf - \phi)$. Curves $a$ and $b$ intersect the $V$-axis at the same point,meaning they have the same stopping potential. Therefore,$f_a = f_b$.
$2$. The saturation current is proportional to the intensity of the incident radiation. Curves $a$ and $b$ have different saturation current levels,which implies $I_a \neq I_b$.
$3$. Thus,the correct relationship is $f_a = f_b$ and $I_a \neq I_b$.

(A) The wave theory of light treats light as a continuous electromagnetic wave. While this model successfully explains phenomena like interference,diffraction,and polarization,it fails to explain the photoelectric effect. In the photoelectric effect,the emission of electrons depends on the frequency of incident light rather than its intensity,which contradicts the classical wave theory. This phenomenon is explained by the particle nature of light,where light consists of discrete packets of energy called photons.

(D) In the photoelectric effect,the stopping potential $(V_0)$ depends only on the material (work function) and the frequency of the incident radiation. For a fixed frequency,the stopping potential is a characteristic of the material.
Looking at the graph,curves $1$ and $2$ intersect the voltage axis at the same point,meaning they have the same stopping potential. Therefore,curves $1$ and $2$ represent the same material.
Similarly,curves $3$ and $4$ intersect the voltage axis at a different,common point,meaning they represent another material with a different work function.
Thus,the pairs that represent the same material are $(1, 2)$ and $(3, 4)$.

(D) In photovoltaic cells, the photoelectric current produced is directly proportional to the intensity of the incident light. Therefore, statement $A$ is false.
According to Einstein's photoelectric equation, $K_{max} = h\nu - \Phi = \frac{hc}{\lambda} - \Phi$. Since the maximum kinetic energy $K_{max}$ of the photoelectrons depends on the wavelength $\lambda$ of the incident radiation, the velocity of the photoelectrons also depends on the wavelength. Therefore, statement $B$ is true.

(B) 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.
When the distance $d$ is doubled,the intensity $I$ becomes $\frac{1}{4}$ of its original value.
The photoelectric current is directly proportional to the intensity of incident light.
Since the intensity decreases,the number of photons hitting the surface per unit time decreases,resulting in a decrease in the photoelectric current.
Stopping potential and maximum kinetic energy depend on the frequency of the incident light,not on its intensity.
Therefore,the correct result is that the photoelectric current will decrease.

(A) The intensity $I$ of light at a distance $r$ from a point source is given by $I = \frac{P}{4\pi r^2}$,so $I \propto \frac{1}{r^2}$.
When the distance is changed from $d$ to $\frac{d}{2}$,the new intensity $I'$ becomes $I' = \frac{1}{(d/2)^2} = 4 \times \frac{1}{d^2} = 4I$.
Since the number of photoelectrons emitted per second is directly proportional to the intensity of incident light,the emission rate becomes $4$ times the original value.
According to Einstein's photoelectric equation,$KE_{max} = h\nu - \phi$,where $\nu$ is the frequency of incident light and $\phi$ is the work function.
Since the frequency $\nu$ remains unchanged,the maximum kinetic energy $(KE_{max})$ and the stopping potential $(V_s)$ remain unchanged,as $eV_s = KE_{max}$.