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Questions
With the help of a circuit diagram describe the experiment to study the characteristics of the photoelectric effect. Hence discuss any 2 characteristics of the photoelectric effect.
With a neatly labelled circuit diagram, describe an experiment to study the characteristics of the photoelectric effect.
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Solution
- A laboratory experimental set-up for the photoelectric effect consists of an evacuated glass tube with a quartz window.
- The glass tube contains photosensitive metal plates. One is the emitter E and another plate is the collector C.
Schematic of experimental set-up for the photoelectric effect - The emitter and collector are connected to a voltage source whose voltage can be changed and to an ammeter to measure the current in the circuit.
- A potential difference of V, as measured by the voltmeter, is maintained between the emitter E and collector C. Generally, C (the anode) is at a positive potential with respect to the emitter E (the cathode). This potential difference can be varied and C can even be at a negative potential with respect to E.
- When the anode potential (V) is positive, it accelerates the electrons. This potential is called accelerating potential. When the anode potential (V) is negative, it retards the flow of electrons. This potential is known as retarding potential.
- A source S of monochromatic light of sufficiently high frequency (short wavelength ≤ 10–7 m) is used.
Two characteristics of the photoelectric effect:
- The photoelectric work function `phi_0` is constant for a given emitter. Hence if the frequency ‘ν’ of the incident radiation is decreased, the maximum kinetic energy of the emitted photoelectrons decreases, till it becomes zero for a certain frequency ν0.
Therefore, from Einstein’s equation,
0 = `"hv"_0 - phi_0`
∴ `phi_0 = "hv"_0` ........(1)
This shows that the threshold frequency is related to the work function of the metal and hence it has different values for different metals. - The photoelectric equation is,
`1/2"mv"_"max"^2 = "hv" - phi_0` ........(2)
where, hν = energy of the photon of incident radiation.
`phi_0 = "hv"_0` = photoelectric work function of the metal.
Thus, both the terms on the R.H.S of equation (2) depend on the frequency and not on the intensity of radiation. Hence the maximum kinetic energy with which photoelectrons are emitted is independent of the intensity of radiation. However, since `phi_0` and h are constants, the maximum kinetic energy of the photoelectrons is directly proportional to the frequency.
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