Light of frequency 1.5 times the threshold frequency is incident on photosensitive material. If the frequency is halved and intensity is doubled, the photo current becomes

When the frequency of incident light is halved of its original value i.e. \$1.5v_0\$ then it becomes less than the threshold value. In that case no photoelectric effect takes place. No photoelectrons would be emitted. The photoelectric current becomes zero.

Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity.

The above graph is showing the variation of photoelectric current with the applied voltage. They have different intensities but same frequency.Here the upper one is for high intensity and lower one is for low intensity.

Write the experimental observations of Photoelectric effect.

(1) For a given metal, there exists a certain minimum frequency above which the photoelectric effect takes place which is known as threshold frequency.(2) Maximum kinetic energy of the emitted photoelectrons increases as the frequency of the incident light is increased keeping the number of incident photons to be fixed.(3) Above threshold frequency, maximum kinetic energy of the photoelectrons depends only on the frequency of incident light but is independent of the intensity of incident light.(4) For a given metal and frequency of incident light, the rate at which the photoelectrons ejected is directly proportional to the intensity of incident light. An increase in the magnitude of the incident light increases the magnitude of the photoelectric current.(5) The time lag between the incidence of photons and radiation of photoelectric effect is very small, nearly \$10^{-9}\$ second.

The threshold frequency for certain metal is \$3.3\times { 10 }^{ 14 } Hz\$. If light of \$8.2\times { 10 }^{ 14 }Hz\$ frequency is incident on the metal, then the cut-off voltage of the photoelectric current will be :

In photoelectric effect, energy is conserved.In photoelectric effect, Einstein's equation is given byPhoton energy \$=\$ kinetic energy of electron \$+\$ work functioni.e. \$hv=e{ V }_{ s }+h{ v }_{ 0 }\$where, \${ V }_{ s }\$ is the stopping potential and \${ v }_{ 0 }\$ is the threshold frequency.\$\therefore { V }_{ s }=\dfrac { h }{ e } \left( v-{ v }_{ 0 } \right)\$ .....(i) Given, \${ v }_{ 0 }=3.3\times { 10 }^{ 14 }Hz\$, \$v=8.2\times { 10 }^{ 14 }Hz\$ \$h=6.6\times { 10 }^{ -34 }J-s\$, \$e=1.6\times { 10 }^{ -19 }C\$Substituting these values in equation (i), we get\${ V }_{ s }=\dfrac { 6.6\times { 10 }^{ -34 }\left( 8.2\times { 10 }^{ 14 }-3.3\times { 10 }^{ 14 } \right) }{ 1.6\times { 10 }^{ -19 } } \$\${ V }_{ s }=\dfrac { 6.6\times 4.9 }{ 1.6 } \times { 10 }^{ -1 }=2V\$\$\Rightarrow { V }_{ s }=2V\$

Write four laws of photoelectric effect.

Photo electric effect: When some photosensitive substances are exposed to ultraviolet light, electrons are ejected from their surface which are called photoelectrons. This process is called photoelectric emission.Laws of photoelectric emission:(i) There is a definite cut off value of frequency below which electrons cannot be ejected by any substance.(ii) Number of emitted electrons are directly proportional to the intensity of light incident.(iii) Kinetic energy of emitted electrons depends on the frequency of incident light on substance.(iv) There is no time logging between the incident of light and emission of electrons.

(a) Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.(b) Write the basis features of photon picture of electron magnetic radiation on which Einstein's photoelectric equation is based.

(a)Wave nature of radiation cannot explain the photo electric effects because of :(i) The immediate ejection of photo electrons (ii) The presence of threshold frequency for a metal surface.(iii) The fact that kinetic energy of the emitted electrons is independent of the intensity of light and depends upon its frequency. Thus, the photoelectric effect cannot be explained on the basis of wave nature of light (b) Photon picture of electron magnetic radiation on which Einstein's photoelectric equation is based on particle nature of light . Its basic features are:(i) In interaction with matter radiation behaves as if it is made up of particles called photons.(ii) Each photon has energy \$(E=hv)\$, momentum \$p=\dfrac{h\nu}{c}\$(p=and speed c, the speed of light(iii) All photon of light of a particular frequency v, or wavelength λλ , have the same energy \$E=\dfrac{hc}{\lambda}\$and momentum \$p=\dfrac{h\nu }{c}\$(p=hvc). (iv)By increasing the intensity of light of given wavelength there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy Thus, photon energy is independent of intensity of radiation (v) Photons are electrically neutral and are not deflected by electric and magnetic fields.(vi) In a photon-particle collision (such as photon-electron collision ),the total energy and total momentum are conserved. However number of Photon may not be observed.

What is photoelectric effect? Explain the effect of increase (i) frequency (ii) intensity of incident radiation on photoelectric current with suitable graphs.

Photoelectric effect is a phenomenon which is based on the idea that electromagnetic radiation is made of a series of particles called photons.When a photon hits an electron on a metal surface,the electron can be emitted.The emitted electrons are called photoelectrons.(i)On increasing the frequency the saturation currents remains the same as seen in the first figure.(ii)On increasing the Intensity the saturation current also increases.

In photoelectric effect, the slope of the straight line graph between stopping potential and frequency of the incident light gives the ratio of Planck's constant to

\$hv=hv_{0}+K.E\$at stopping potential K.E\$=\$eV\$\Rightarrow hv=hv_{0}+eV\$\$V=\dfrac{h}{e}(v-v_{0})\$The slope of graph b/w stopping potential and frequency is \$\dfrac{h}{e}\$, where e is charge of electron.

In photoelectric effect, the photo current.

When intensity of incident photon increases number of electrons emitted from the surface increases due to which current increases. Frequency of incident photon only limits the maximum kinetic energy of the photoelectron emitted and so the current is independent of frequency of photon.

Observations on Photo Electric Effect | Definition, Examples, Diagrams

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Hertz experiment in production of electromagnetic waves

To test Maxwell's hypothesis, Hertz used an oscillator made of polished brass knobs, each connected to an induction coil and separated by a tiny gap over which sparks could leap. To confirm this, Hertz made a simple receiver of looped wire,its ends separated by a tiny gap. The receiver was placed several yards from the oscillator. Hertz reasoned that, if Maxwell's predictions were correct, electromagnetic waves would be transmitted during each series of sparks,that would induce a current in the loop that would send sparks across the gap. This occurred when Hertz turned on the oscillator, producing the first transmission and reception of electromagnetic waves. Hertz also noted that electrical conductors reflect the waves and that they can be focused by concave reflectors. He found that nonconductors allow most of the waves to pass through.

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Hertz's observation

The phenomenon of photoelectric emission was discovered in 1887 by Heinrich Hertz, during electromagnetic wave experiments.He observed that when light falls on a metal surface, some electrons near the surface absorb enough energy from the incident radiation to overcome the attraction of the positive ions in the material of the surface.

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Lenard's experiment

Wilhelm Hallwachs and Philipp Lenard investigated the phenomenon of photoelectric emission in detail during 1886-1902.
Their observations can be summarised as below:

When light is incident on a metal plate, it may cause the electrons to come out of the metal.
Hallwachs and Lenard also observed that when the light fell on the emitter plate, no electrons were emitted at all when the frequency of the incident light was smaller than a certain minimum value. This frequency is called threshold frequency.
He also observed that different metals have different threshold frequencies.

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Experimental study of photoelectric effect

Let us study an experimental setup as shown in the figure. Here a photosensitive C plate is irradiated with light of particular frequency. The plate A is maintained at some potential difference with plate C. Let us study the effects of different factors such as: 1) intensity of light 2) frequency of light 3) potential difference between the plates

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Effect of intensity of light on photocurrent

let us say that the plate A is given a positive potential with respect to the plate C. When the light is incident on the plate C, electrons will be attracted to plate A. More the intensity of the light, more will be the electrons ejected and more will be the photocurrent.

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Effect of potential on photoelectric current

More the potential difference between the plate C and A is increased, the phoelectric current will also increase.

At some stage, all the photoelectrons ejected by the metal will be accumulated by the collector plate and a saturation in current occurs. This current is called saturation current. After this stage the photocurrent does not increase no matter how much intensity of light is used.
If we increase the intensity of the light the saturation current also increases.Let initial intensity be $I_{1}$ and now it is increased to $I_{2}$ .Hence the saturation current of light with intensity $I_{2}$ is more than that of $I_{1}$
We now apply a negative (retarding) potential to the plate A with respect to the plate C and make it increasingly negative gradually. When the polarity is reversed, the electrons are repelled and only the most energetic electrons are able to reach the collector A. The photocurrent is found to decrease rapidly until it drops to zero at a certain sharply defined, critical value of the negative potential $V_{0}$ on the plate A. For a particular frequency of incident radiation, the minimum negative (retarding) potential $V_{0}$ given to the plate A for which the photocurrent stops or becomes zero is called the cut-off or stopping potential.

Observations on Photo Electric Effect

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Hertz experiment in production of electromagnetic waves

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Hertz's observation

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Lenard's experiment

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Experimental study of photoelectric effect

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Effect of intensity of light on photocurrent

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Effect of potential on photoelectric current

REVISE WITH CONCEPTS

Experimental Study of Photoelectric Effect

Light of frequency 1.5 times the threshold frequency is incident on photosensitive material. If the frequency is halved and intensity is doubled, the photo current becomes

Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity.

Write the experimental observations of Photoelectric effect.

The threshold frequency for certain metal is $3.3 \times 10^{14} H z$ . If light of $8.2 \times 10^{14} H z$ frequency is incident on the metal, then the cut-off voltage of the photoelectric current will be :

Write four laws of photoelectric effect.

Explain how does (i) photoelectric current and (ii) kinetic energy of the photoelectrons emitted in a photocell vary if the frequency of incident radiation is doubled, but keeping the intensity same? Show the graphical variation in the above two cases.

(a) Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.
(b) Write the basis features of photon picture of electron magnetic radiation on which Einstein's photoelectric equation is based.

What is photoelectric effect? Explain the effect of increase (i) frequency (ii) intensity of incident radiation on photoelectric current with suitable graphs.

In photoelectric effect, the slope of the straight line graph between stopping potential and frequency of the incident light gives the ratio of Planck's constant to

In photoelectric effect, the photo current.

definition

Hertz experiment in production of electromagnetic waves

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Hertz's observation

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Lenard's experiment

definition

Experimental study of photoelectric effect

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Effect of intensity of light on photocurrent

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Effect of potential on photoelectric current

REVISE WITH CONCEPTS

Experimental Study of Photoelectric Effect

Light of frequency 1.5 times the threshold frequency is incident on photosensitive material. If the frequency is halved and intensity is doubled, the photo current becomes

Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity.

Write the experimental observations of Photoelectric effect.

The threshold frequency for certain metal is 3.3×1014Hz. If light of 8.2×1014Hz frequency is incident on the metal, then the cut-off voltage of the photoelectric current will be :

Write four laws of photoelectric effect.

Explain how does (i) photoelectric current and (ii) kinetic energy of the photoelectrons emitted in a photocell vary if the frequency of incident radiation is doubled, but keeping the intensity same? Show the graphical variation in the above two cases.

(a) Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons. (b) Write the basis features of photon picture of electron magnetic radiation on which Einstein's photoelectric equation is based.

What is photoelectric effect? Explain the effect of increase (i) frequency (ii) intensity of incident radiation on photoelectric current with suitable graphs.

In photoelectric effect, the slope of the straight line graph between stopping potential and frequency of the incident light gives the ratio of Planck's constant to

In photoelectric effect, the photo current.

The threshold frequency for certain metal is $3.3 \times 10^{14} H z$ . If light of $8.2 \times 10^{14} H z$ frequency is incident on the metal, then the cut-off voltage of the photoelectric current will be :

(a) Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.
(b) Write the basis features of photon picture of electron magnetic radiation on which Einstein's photoelectric equation is based.