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1800-102-2727Christian Huygen proposed propagation of light in the form of waves. But it was unable to describe the motion of light in vacuum which was later explained by wave theory proposed by Maxwell. Wave theory further was unable to describe the photoelectric effect discovered by Hertz. Max Plank described the propagation of light in the form of packets of energy called quanta. Einstein further used this theory to propose the photoelectric equation which we will discuss here.
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In 1905, Einstein (1879-1955) proposed a completely different picture of electromagnetic waves. He picturises the EM waves as energy discrete units called photons and photoelectric emission does not take place by continuous absorption of energy from radiation. These discrete units are also called quanta of energy of radiation. Each quantum of energy is h, where is the frequency of light and h is Planck's constant.
In photoelectric effect, an electron absorbs a photon (hv) of radiation. If this absorbed energy exceeds the threshold (work function of the metal), the most loosely bound electron will emerge with maximum kinetic energy, and the most tightly bound electron will emerge with almost zero kinetic energy.
Einstein's photoelectric equation,
The observations of a photoelectric effect can be explained easily using the above equation,
where vo is threshold frequency. No photoelectric emission is possible below , even if the incident radiation of high intensity and long duration falls on the surface.
The maximum wavelength of the incident radiation above which photoelectric emission is not possible. If the photoelectric effect has to happen then the energy of the photon must be greater than the work function of metal.
Maximum K.E of the electrons emitted can be calculated as,
The minimum frequency of the incident radiation below which photoelectric emission is not possible. If the photoelectric effect has to happen then the energy of the photon must be greater than the work function of metal.
Maximum KE of the emitted electrons will be,
Q1. In a photoelectric experiment, the collector plate is at 2.0 𝑉 with respect to the emitter plate made of copper (𝜙=4.5 𝑒𝑉). The emitter is illuminated by a source of monochromatic light of wavelength 200 𝑛𝑚. What will be the minimum and maximum K.E of the photoelectrons that are reaching the collector.
Answer: The maximum K.E. with which the photoelectrons can leave the emitter plate is
Therefore, the K.E. of photoelectrons leaving the emitter plate lies between
0<K<1.7 eV
The direction of the electric field is from A to C, this means that the force acting on the electrons is in the opposite direction to the direction of the electric field i.e. towards A due to which the electrons get accelerated in the direction of force. The electron is going to gain kinetic energy and lose potential energy.
The change in potential energy when electron goes from C to A, U=-2eV-0
The increase in K.E is equal to this decrease in P.E., K=2eV
As, the K.E. of photoelectrons leaving the emitter plate lies between 0<K<1.7 eV
Therefore, the K.E. of photoelectrons reaching the collector plate lies between
2 eV <Kcollector<3.7 eV
KEmin=2eV KEmax=3.7 eV
Q2. The threshold frequency of a metal for photoelectric effect to take place is 𝑓o. The maximum velocity of electrons emitted is 𝑣1, if the light incident on the metal plate is having frequency 2𝑓o. When the frequency of the incident radiation is increased to 5𝑓o, the maximum velocity of electrons emitted is 𝑣2. Find the ratio of 𝑣1 to 𝑣2.
Answer: The maximum K.E. of emitted electrons is
Q3. The photoelectric threshold wavelength of silver is 3250×10−10 𝑚. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536×10−10 𝑚 is
[Given ℎ=4.14×10−15 𝑒𝑉 𝑠 & 𝑐=3×108 𝑚 /s]
A. 0.6×105 𝑚/s
B. 61×103 𝑚/s
C. 0.3×106 𝑚/s
D. 6×105 𝑚/s
Answer: We know that the maximum K.E. of a photoelectron is given by
Q4. The graph is showing variation between stopping potential (𝑉o)with frequency (𝜈) for two photosensitive metals 𝑃 and 𝑄. Which metal will be having a smaller threshold wavelength and why?
Answer:
The x-intercept on Vo vs 𝜈 graph gives the threshold frequency of the metal.
Hence metal Q has smaller threshold wavelength.
Question 1. On what factor does the work function of a metal depend upon?
Answer: Work function depends on the nature of the metal surface.
Question 2. If a light of frequency less than threshold frequency is incident on a metal surface, what will be the kinetic energy of the electrons emitted?
Answer: Zero. Because no electron will get emitted for frequency less than threshold frequency.
Question 3. How does the kinetic energy of the electrons emitted during a photoelectric effect vary with the frequency of light?
Answer: Kinetic energy varies linearly with the frequency of the incident light.
Question 4. What is the dimension of work function?
Answer: It is same as that of energy i.e. [M1L2T-2].