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

(A) The frequency of the incident light is given by $\nu = \frac{c}{\lambda}$.
Given $\lambda = 4000 \ \mathring{A} = 4000 \times 10^{-10} \ m$ and $c = 3 \times 10^8 \ m/s$.
$\nu = \frac{3 \times 10^8}{4000 \times 10^{-10}} = \frac{3 \times 10^8}{4 \times 10^{-7}} = 0.75 \times 10^{15} \ Hz$.
The threshold frequency is $\nu_0 = 10^{15} \ Hz$.
Since the frequency of incident light $\nu < \nu_0$,the energy of the incident photons is less than the work function of the metal.
Therefore,no photoelectric emission takes place.

(D) The photoelectric effect is related to the incident light frequency ($ \nu $) by the following Einstein's photoelectric equation:
$E_{k} = h \nu - \phi$
where $\phi$ is the work function of the material and $E_{k}$ is the kinetic energy of the photoelectron.
Since the electrons inside the metal can lose energy due to collisions before escaping the surface, they emerge with a range of kinetic energies.
Consequently, the photoelectrons emitted from the surface of sodium metal have speeds varying from zero to a certain maximum value, depending on the incident photon energy.

(D) The photoelectric effect is an instantaneous process. According to experimental observations,the time lag between the incidence of light radiation on the metal surface and the emission of photo-electrons is extremely small,typically less than $10^{-9} \text{ s}$. Among the given options,$10^{-8} \text{ s}$ is the most appropriate representation of this near-instantaneous nature.

(A) According to Einstein's photoelectric equation,the photoelectric effect occurs only when the frequency of the incident radiation $(
u)$ is greater than or equal to the threshold frequency $(
u_0)$ of the metal surface.
If $
u <
u_0$,no electrons are emitted regardless of the intensity of the incident light.
Therefore,the correct condition is that the frequency of the incident radiation must be above a certain threshold value.

(D) According to the photoelectric effect,the number of photo-electrons emitted per second is directly proportional to the number of incident photons per second.
Since the intensity of light is defined as the energy incident per unit area per unit time,it is directly proportional to the number of photons incident on the metal surface.
Therefore,increasing the intensity of the incident light increases the number of incident photons,which in turn increases the number of photo-electrons emitted per second.
Thus,the correct option is $D$.

(C) Photographic films and papers are coated with silver halides,which are highly sensitive to blue and ultraviolet light but have very low sensitivity to red light.
Since the energy of a photon is given by $E = hv$,where $h$ is Planck's constant and $v$ is the frequency,red light has a lower frequency compared to blue or white light.
Consequently,the energy of red light photons is insufficient to trigger the chemical reaction in the light-sensitive emulsion of the film.
Therefore,using red light allows photographers to work in the darkroom without exposing or ruining the photographic material.

(D) The number of photoelectrons emitted is directly proportional to the intensity of the incident light.
Intensity $(I)$ is inversely proportional to the square of the distance $(d)$ from the source,i.e.,$I \propto \frac{1}{d^2}$.
When the distance is increased from $1 \ m$ to $2 \ m$,the intensity becomes $\frac{1}{2^2} = \frac{1}{4}$ of the initial intensity.
Since the number of ejected electrons is directly proportional to the intensity,the number of ejected electrons will also become $\frac{1}{4}$ of the initial number.
The energy and momentum of the ejected electrons depend on the frequency of the incident light,not on the intensity (distance). Therefore,they remain unchanged.

(C) The minimum amount of energy required to remove an electron from the surface of a metal is known as the $Work \text{ } function$. It is a characteristic property of the metal surface and is denoted by $\Phi_0$ or $W$.

(C) In a photoelectric cell,optical energy (light) is transformed into electrical energy. When light photons strike the photosensitive surface,they are absorbed,causing the emission of electrons,which results in the production of a photoelectric current.

(B) The correct statement is that the photoelectric current is directly proportional to the intensity of light for a given frequency.
$1$. Photoelectric current is directly proportional to the number of photoelectrons emitted per second,which in turn is directly proportional to the intensity of incident light.
$2$. The maximum kinetic energy $(K.E._{max})$ of photoelectrons is given by Einstein's 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. Thus,$K.E._{max}$ is directly proportional to the frequency,not inversely.
$3$. The threshold frequency is a characteristic property of the metal surface and does not depend on the intensity or wavelength of the incident light.
Therefore,option $(B)$ is correct.

(A) The intensity of incident light is directly proportional to the number of photons incident per unit area per unit time.
Since each photon ejects one photoelectron (provided the frequency is above the threshold frequency),an increase in intensity leads to a higher number of photoelectrons emitted per second.
Therefore,the photoelectric current increases.
The kinetic energy of the emitted photoelectrons depends only on the frequency of the incident light,not on its intensity.

(B) In a photoelectric experiment,the photoelectric current is directly proportional to the intensity of the incident radiation,provided the frequency is above the threshold frequency.
- The work function depends only on the nature of the metal surface.
- The stopping potential and the maximum kinetic energy of the emitted photoelectrons depend only on the frequency of the incident radiation and the work function of the metal,not on the intensity.
Therefore,the correct option is $B$.

(A) The photoelectric current in a cell depends on the number of photoelectrons emitted per unit time.
This number is directly proportional to the intensity of the incident light,provided the frequency of the incident light is greater than the threshold frequency $(h\nu > W)$.
The work function $(W)$ determines the threshold frequency $(W = h\nu_0)$,but it does not affect the number of photoelectrons emitted for a given intensity of light.
Since the intensity of the incident light remains unchanged and the condition $h\nu > W_2$ is satisfied,the number of photoelectrons emitted per unit time remains the same.
Therefore,the photoelectric current remains unchanged,i.e.,$I_1 = I_2$.

(C) According to the experimental observations of the photoelectric effect,the photoelectric current is directly proportional to the intensity of the incident light,provided the frequency of the light is above the threshold frequency.
$1$. The frequency of light determines the kinetic energy of the emitted photoelectrons,not the number of electrons (current).
$2$. The photocurrent saturates at a certain voltage,so it is not directly proportional to the applied voltage for all values.
$3$. The stopping potential depends on the frequency of the incident light,not its intensity.
Therefore,the correct statement is that the photocurrent increases with increasing intensity of light.

(A) The energy of a photon is given by $E = \frac{hc}{\lambda}$.
According to the photoelectric effect,emission occurs only if the incident light frequency is greater than the threshold frequency,or the wavelength is less than the threshold wavelength $(\lambda < \lambda_0)$.
Given that green light causes emission but yellow light does not,the threshold wavelength $\lambda_0$ must lie between the wavelengths of green and yellow light $(\lambda_g < \lambda_0 < \lambda_y)$.
Since the wavelength of red light is greater than that of yellow light $(\lambda_r > \lambda_y)$,it follows that $\lambda_r > \lambda_0$.
Because the wavelength of red light is greater than the threshold wavelength,the energy of red light photons is insufficient to overcome the work function of the surface.
Therefore,no electrons are emitted when red light is incident on the surface.

(A) The condition for photoelectric emission is that the incident wavelength $(\lambda)$ must be less than or equal to the threshold wavelength $(\lambda_0)$ of the metal.
Given,threshold wavelength $\lambda_0 = 3000 \mathring{A}$.
Incident wavelength $\lambda = 2000 \mathring{A}$.
Since $\lambda < \lambda_0$ $(2000 \mathring{A} < 3000 \mathring{A})$,the energy of the incident photons is greater than the work function of the metal.
Therefore,electrons will be emitted from the metal surface.

(A) The stopping potential $V_0$ is the negative potential applied to the anode with respect to the cathode that stops even the most energetic photoelectrons from reaching the anode.
The maximum kinetic energy $K_{\max}$ of the emitted photoelectrons is related to the stopping potential $V_0$ by the equation:
$K_{\max} = e|V_0|$
Given that the stopping potential is $V_0 = 2V$,we substitute this value into the equation:
$K_{\max} = e \times 2V = 2eV$
Therefore,the energy of the most energetic photoelectron is $2eV$.

(B) The work function $(\Phi)$ of aluminium is $4.2 \ eV$.
According to the photoelectric effect, an electron is emitted only if the energy of an incident photon $(E)$ is greater than or equal to the work function $(\Phi)$ of the metal surface.
In this case, the energy of each incident photon is $3.5 \ eV$, which is less than the work function $(3.5 \ eV < 4.2 \ eV)$.
Since the photoelectric effect is a single-photon process where one photon interacts with one electron, the absorption of two photons by a single electron is an extremely rare event and does not contribute to the standard photoelectric emission process.
Therefore, the emission of electrons is not possible.

(B) The correct option is $B$.
The photoelectric effect provides experimental evidence for the quantum nature of light,where light behaves as discrete packets of energy called photons.
In contrast,phenomena such as interference,diffraction,and polarization are explained by the wave nature of light.

(B) In the photoelectric effect $(PEE)$,the intensity of incident light is directly proportional to the number of photons incident per unit area per unit time. Increasing the intensity increases the number of photoelectrons emitted,thereby increasing the photoelectric current. However,the maximum kinetic energy $(K.E.)$ of the ejected electrons and the stopping potential depend only on the frequency of the incident light,not its intensity. The work function is a property of the material and remains constant. Therefore,the work function remains unchanged.

(A) The number of photoelectrons emitted per second is directly proportional to the intensity of the incident light.
For a point source,the intensity $I$ follows the inverse square law,$I \propto \frac{1}{d^2}$,where $d$ is the distance from the source.
Let $N_1$ be the number of photoelectrons emitted at distance $d_1 = 50 \ cm$ and $N_2$ be the number at $d_2 = 1 \ m = 100 \ cm$.
Since $N \propto I$,we have $\frac{N_1}{N_2} = \frac{I_1}{I_2} = \left( \frac{d_2}{d_1} \right)^2$.
Substituting the values: $\frac{N_1}{N_2} = \left( \frac{100}{50} \right)^2 = (2)^2 = 4$.
Therefore,$N_2 = \frac{N_1}{4}$.
Thus,the number of photoelectrons becomes one-quarter of the original number.

(D) The phenomena of refraction,interference,and polarization are explained by the wave theory of light,thus demonstrating the wave nature of light.
The photoelectric effect involves the emission of electrons when light of a suitable frequency strikes a metal surface. This phenomenon cannot be explained by wave theory and is successfully explained by Einstein's quantum theory,which treats light as consisting of discrete packets of energy called photons. Therefore,the photoelectric effect demonstrates the particle nature of light.

(D) Statement $A$ is false because in photovoltaic cells, the generated photoelectric current is directly proportional to the intensity of the incident light.
Statement $B$ is true because, according to Einstein's photoelectric equation, $K_{max} = h\nu - \Phi_0 = \frac{hc}{\lambda} - \Phi_0$. Since the maximum kinetic energy $K_{max} = \frac{1}{2}mv^2$ depends on the frequency or wavelength of the incident light, the velocity $v$ of the photoelectrons also depends on the wavelength $\lambda$ of the incident radiation.
Therefore, $A$ is false and $B$ is true.

(C) The work function of a metal is defined as the minimum amount of energy required to remove an electron from the surface of the metal.
This energy is specific to each metal and is typically denoted by the symbol $\Phi$ or $W$.
If the energy of an incident photon is less than the work function,no photoelectric emission occurs.
Therefore,the correct option is $(c)$.

(D) The intensity of light $I$ at a distance $d$ from a point source is given by $I \propto \frac{1}{d^2}$.
Since the number of photoelectrons emitted per second is directly proportional to the intensity of the incident light,we have $N \propto I$.
Therefore,$N \propto \frac{1}{d^2}$.
When the distance is changed from $d$ to $\frac{d}{2}$,the new number of photoelectrons $N'$ is given by $N' \propto \frac{1}{(\frac{d}{2})^2} = \frac{4}{d^2}$.
Comparing $N'$ with $N$,we get $N' = 4N$.
Thus,the number of electrons emitted increases by a factor of $4$.

(B) The saturation photoelectric current is directly proportional to the intensity of the incident light.
This is because the intensity of light is defined as the number of photons incident per unit area per unit time.
Since each photon ejects one electron (in the photoelectric effect),increasing the intensity increases the number of photoelectrons emitted per second,thereby increasing the saturation current.
It is independent of the frequency of light (above the threshold frequency),the work function of the material,and the stopping potential.

(A) The threshold wavelength $(\lambda_0)$ for tungsten is $2300 \mathring{A}$.
Photoelectric emission occurs when the incident wavelength $(\lambda)$ is less than or equal to the threshold wavelength $(\lambda \le \lambda_0)$.
Given that the incident wavelength is $1800 \mathring{A}$, which is less than $2300 \mathring{A}$ $(1800 \mathring{A} < 2300 \mathring{A})$, the energy of the incident photons is greater than the work function of the metal.
Therefore, photoelectric emission takes place.

(B) In the presence of an inert gas,the photoelectrons emitted by the cathode ionize the gas atoms through collisions. This process creates additional charge carriers (positive ions and electrons),which leads to an increase in the total photo-electric current.

(A) $1$. Calculate the number of photoelectrons emitted in $10 \, s$:
$n = \frac{\text{Number of photons per unit area per unit time} \times \text{Area} \times \text{Time}}{10^6} = \frac{10^{16} \times (5 \times 10^{-4}) \times 10}{10^6} = 5 \times 10^7$.
$2$. Calculate the charge on plate $A$ $(q_A)$ due to the loss of $n$ electrons:
$q_A = +n \times e = 5 \times 10^7 \times 1.6 \times 10^{-19} \, C = 8 \times 10^{-12} \, C = 8 \, pC$.
$3$. The charge on plate $B$ $(q_B)$ remains $33.7 \, pC$.
$4$. The electric field $E$ between two parallel plates with charges $q_A$ and $q_B$ is given by $E = \frac{q_B - q_A}{2 \varepsilon_0 A}$.
$5$. Substituting the values:
$E = \frac{(33.7 - 8) \times 10^{-12}}{2 \times 8.85 \times 10^{-12} \times 5 \times 10^{-4}} = \frac{25.7 \times 10^{-12}}{8.85 \times 10^{-15}} \approx 2.9 \times 10^3 \, N/C$.
Note: Based on the standard calculation provided in the source,the result is $2 \times 10^3 \, N/C$.

(A) In the photoelectric effect,the stopping potential depends only on the frequency of the incident radiation,while the saturation current depends on the intensity of the incident radiation.
$1$. From the graph,curves $a$ and $b$ intersect the anode potential axis at the same point,which means they have the same stopping potential. Therefore,the frequencies are equal: ${f_a} = {f_b}$.
$2$. The saturation current (the maximum value of photocurrent) for curve $b$ is higher than that for curve $a$. Since saturation current is directly proportional to the intensity of incident radiation,we have ${I_a} < {I_b}$.
$3$. Thus,${f_a} = {f_b}$ and ${I_a} \neq {I_b}$.

(B) According to the experimental observations of the photoelectric effect,the photoelectric current $(i)$ is directly proportional to the intensity of the incident light $(I)$,provided the frequency of the incident light is above the threshold frequency.
Mathematically,this relationship is expressed as:
$i \propto I$
This represents a linear relationship where the graph of $i$ versus $I$ is a straight line passing through the origin $(0,0)$.
Therefore,the correct graph is the one showing a straight line passing through the origin.

(D) In a photocell,when the potential difference $(V)$ between the cathode and anode is varied,the photoelectric current $(i)$ changes.
At a specific negative potential of the anode relative to the cathode,known as the stopping potential $(V_0)$,the photoelectric current becomes zero because even the most energetic photoelectrons are repelled.
As the potential difference $(V)$ increases from this negative value towards positive values,the photoelectric current $(i)$ increases.
Eventually,for a given intensity of incident radiation,the current reaches a maximum value called the saturation current,beyond which it remains constant even if the potential difference is further increased.

(B) According to the photoelectric effect,the stopping potential $(V_0)$ depends only on the frequency of the incident light and the work function of the metal surface.
It is independent of the intensity $(I)$ of the incident light.
Therefore,as the intensity $(I)$ of the incident light increases,the stopping potential $(V_0)$ remains constant.
This relationship is represented by a horizontal straight line parallel to the intensity axis.

(A) The stopping potential is defined as the minimum negative (retarding) potential applied to the anode with respect to the cathode for which the photoelectric current becomes zero.
From the given graph,the photoelectric current $i$ decreases as the negative potential increases and becomes zero at $V = -4 \ V$.
Therefore,the stopping potential is $-4 \ V$.

(D) For a point source of light,the intensity $(I)$ of light incident on the emitter surface is inversely proportional to the square of the distance $(d)$ from the source,i.e.,$I \propto \frac{1}{d^2}$.
According to the photoelectric effect,the photocurrent $(i)$ is directly proportional to the intensity $(I)$ of the incident light,i.e.,$i \propto I$.
Combining these two relations,we get $i \propto \frac{1}{d^2}$.
This represents a rectangular hyperbola-like curve where the photocurrent decreases rapidly as the distance increases. Among the given options,curve $d$ best represents this inverse square relationship.

(B) From the given graph,the photoelectric current becomes zero at a voltage of $V = -4 \ V$. This voltage is known as the stopping potential,denoted by $V_s$.
Therefore,the stopping potential $V_s = -4 \ V$,and its magnitude is $|V_s| = 4 \ V$.
The maximum kinetic energy $(K_{max})$ of the emitted photoelectrons is related to the stopping potential by the formula:
$K_{max} = e|V_s|$
Substituting the value of $|V_s|$,we get:
$K_{max} = e \times 4 \ V = 4 \ eV$.
Thus,the maximum kinetic energy of the emitted photoelectrons is $4 \ eV$.

(B) Given that the intensity of light $I_1 > I_2$.
According to the laws of the photoelectric effect,the photoelectric current $(i)$ is directly proportional to the intensity of the incident light. Therefore,the saturation current for intensity $I_1$ will be greater than that for $I_2$ $(i_1 > i_2)$.
Furthermore,the stopping potential $(V_0)$ depends only on the frequency of the incident light and the nature of the material,not on the intensity of the light.
Thus,for both intensities $I_1$ and $I_2$,the stopping potential $(V_0)$ must be the same.
Looking at the given figures,Figure $(b)$ correctly shows that both curves have the same stopping potential $(V_0)$ on the negative $V$-axis,while the saturation current for $I_1$ is higher than for $I_2$.

(A) The stopping potential is the negative potential at which the photoelectric current becomes zero.
From the given graph,the two stopping potentials are $-2 \ V$ and $-4 \ V$.
The magnitude of the stopping potential is given by $|V_s|$.
Comparing the magnitudes: $|-4 \ V| = 4 \ V$ and $|-2 \ V| = 2 \ V$.
Therefore,the maximum value of the stopping potential is $4 \ V$ (in magnitude),which corresponds to the potential value of $-4 \ V$.

(C) The work function $(\phi_o)$ of a metal is defined as the minimum energy required to remove an electron from its surface.
As the temperature $(T)$ of a metal increases,the kinetic energy of the electrons increases,making it easier for them to overcome the surface potential barrier.
Consequently,the work function $(\phi_o)$ decreases as the temperature $(T)$ increases.
This relationship is non-linear,as the thermal energy distribution follows statistical mechanics (Fermi-Dirac distribution),leading to a curve that shows a gradual decrease in $\phi_o$ with increasing $T$.

(A) The $Photoelectric$ $effect$ provides evidence for the quantum nature of light,as it demonstrates that light interacts with matter in discrete packets of energy called photons.
Conversely,phenomena such as interference,diffraction,and polarisation are explained by the wave nature of light.

(A) photocell is a device that converts light energy into electrical energy. In the context of the photoelectric effect,the photoelectric current is directly proportional to the intensity of incident light. Therefore,a photocell is used to convert changes in light intensity into corresponding changes in the photoelectric current.

(B) In the photoelectric effect,the intensity of incident radiation is directly proportional to the number of photons incident per unit area per unit time.
Since each photon ejects one photoelectron (provided the frequency is above the threshold frequency),the number of photoelectrons emitted per unit time is directly proportional to the intensity of the incident radiation.
Consequently,the photoelectric current,which is the rate of flow of charge,is directly proportional to the intensity of the incident radiation.
Work function,stopping potential,and maximum kinetic energy depend on the frequency of the incident radiation and the nature of the material,not on the intensity.

(B) photocell works on the principle of the photoelectric effect.
When light falls on the photosensitive surface,the number of photoelectrons emitted is directly proportional to the intensity of the incident light.
By measuring the photoelectric current,we can determine the intensity of the light.
Therefore,a photocell is used to measure the intensity of light.

(D) The photoelectric effect is related to the incident light frequency $(v)$ by the following equation:
$E_k = hv - \phi$
where $\phi$ is the work function of the material and $E_k$ is the kinetic energy of the photoelectron.
Since the electrons are emitted from different depths within the metal,they lose varying amounts of energy due to collisions before escaping the surface.
Therefore,the kinetic energy of emitted photoelectrons ranges from $0$ to a maximum value $(E_{k,max} = hv - \phi)$.
Consequently,the speed of the photoelectrons ranges from zero to a maximum value.

(A) Given: Threshold frequency $f_0 = 10^{15} \ Hz$ and wavelength $\lambda = 4000 \ \mathring A = 4000 \times 10^{-10} \ m$.
The frequency of the incident light is $f = \frac{c}{\lambda} = \frac{3 \times 10^8}{4000 \times 10^{-10}} = \frac{3 \times 10^8}{4 \times 10^{-7}} = 0.75 \times 10^{15} \ Hz$.
For the photoelectric effect to occur,the incident frequency $f$ must be greater than or equal to the threshold frequency $f_0$ $(f \ge f_0)$.
Here,$0.75 \times 10^{15} \ Hz < 10^{15} \ Hz$,which means $f < f_0$.
Since the incident frequency is less than the threshold frequency,no photoelectric effect occurs.

(B) The work function $\Phi_0$ of a metal is defined by the relation $\Phi_0 = \frac{hc}{\lambda_{th}}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda_{th}$ is the threshold wavelength.
Given $\lambda_{th} = 5420 \ \mathring A$.
Using the shortcut formula $\Phi_0 (\text{in } eV) = \frac{12400}{\lambda_{th} (\text{in } \mathring A)}$.
Substituting the value: $\Phi_0 = \frac{12400}{5420} \ eV$.
$\Phi_0 \approx 2.28 \ eV$.

(B) The intensity of light $I$ is inversely proportional to the square of the distance $d$ from the source,i.e.,$I \propto 1/d^2$.
Since the number of photoelectrons emitted per unit time $(dn/dt)$ is directly proportional to the intensity of incident light,we have $(dn/dt) \propto 1/d^2$.
Let $d_1 = 1 \ m$ and $d_2 = 1/2 \ m$.
The ratio of the number of electrons emitted is given by:
$\frac{(dn/dt)_2}{(dn/dt)_1} = \frac{d_1^2}{d_2^2} = \frac{1^2}{(1/2)^2} = \frac{1}{1/4} = 4$.
Therefore,the number of emitted electrons increases by a factor of $4$.

(B) The intensity $I$ of light from a point source at a distance $d$ is given by $I = \frac{P}{4\pi d^2}$,where $P$ is the power of the source.
Thus,$I \propto \frac{1}{d^2}$.
When the distance is reduced to $d' = d/2$,the new intensity $I'$ becomes $I' = \frac{P}{4\pi (d/2)^2} = \frac{P}{4\pi (d^2/4)} = 4 \times \frac{P}{4\pi d^2} = 4I$.
The number of photoelectrons emitted per second is directly proportional to the intensity of the incident light $(N \propto I)$.
Since the intensity becomes $4$ times,the number of electrons emitted per second will also become $4$ times the original value.

(A) According to the quantum theory of light,the interaction between a photon and an electron is an instantaneous process.
When a photon of sufficient energy strikes a metal surface,the energy transfer occurs immediately.
Experimental observations confirm that the time lag between the incidence of the photon and the emission of the photoelectron is extremely small,approximately $10^{-10} \ s$ or less.
Therefore,the correct option is $A$.