Effect of Intensity of the Incident Radiation
We have already discussed the photoelectric effect in detail. When the ultra violet rays shine upon a zinc plate it causes the plate to emit electrons. These emitted electrons are termed as photoelectrons since the photons are responsible for their emission. The number of electrons ejected from the material in photoelectric effect depends on various factors.
The electrode C i.e. collecting electrode is made positive with respect to D. Keeping the frequency of light and the potentials fixed, the intensity (amount of energy falling per unit area per second) of incident light is varied and the photoelectric current (i) is measured in ammeter. The photoelectric current is directly proportional to the intensity of light. The current gives an account of number of photoelectrons ejected per second.
Effect of p.d. between C & D
Keeping the intensity and frequency of light constant, the positive potential of C is increased gradually. The photoelectric current increases with increase in voltage (accelerating voltage) till, for a certain positive potential of plate C, the current becomes maximum beyond which it does not increase for any increase in the accelerating voltage. This maximum value of the current is called as saturation current.
Make the potential of C as zero and make it increasingly negative. The photoelectric current decrease as the potential is made increasingly negative (retarding potential), till for a sharply defined negative potential Vc of C, the current becomes zero. The retarding potential for which the photoelectric current becomes zero is called as cut-off or stopping potential (Vc).
eVC = 1/2 mv2max
When light of same frequency is used at higher intensity, the value of saturation current is found to be greater, but the stopping potential remains the same. Hence the stopping potential is independent of intensity of incident radiations of same frequency.
We further discuss some of the factors which have an impact on the photoelectric current.
- In case the electroscope is negatively charged, it discharges slowly and the gold leaves fall down.
- If the intensity or the amplitude of the ultra violet radiations is increased, it augments the emission of photoelectrons per second and hence the electrode discharges even more quickly. The number of photons released or emitted per second is proportional to the intensity of the radiation. This hold true in case of any metal.
- In case the electroscope is positively charged, no loss of charge takes place. Since the zinc plate has a positive charge so the frère electrons present on it require much more energy to leave it.
- Another point to be noted is that can visible light be used to cause photoelectric effect in zinc? Since eth visible light is below the threshold frequency of zinc, so it would not be able to cause any emission of electrons. Moreover, it is also not possible to increase the intensity of light to cause photo electricity. Availability of more light but of same frequency is not useful as it does not increase the energy of the light photons.
- In order to escape the potential of the metal, it is necessary for an electron to do a certain amount of work. Hence, a certain amount of energy must be transferred to the electron by the incident radiation.
- Different metals have different work functions. It is due to this reason that in metals with lower work function even a low frequency radiation fulfills the need by providing adequate energy to the electrons to escape. Similar is the case with metals having higher work function.
askIITians provides all inclusive study material which explains all the topics of IIT JEE Physics in detail. All the areas and topics like the effect of intensity on saturation current or the double intensity current incident radiation have been explained quite well. The concepts are well supported with various solved examples and figures.