The frequency and intensity of a light source are doubled. Consider the following statements.
(A) The saturation photocurrent remains almost the same.
(B) The maximum kinetic energy of the photoelectrons is doubled.
(a) A and B are true.
(b) A is true but B is false.
(c) A is false but B is true.
(d) A and B are false.
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I don't know this but I will try to know that of the answer
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Therefore, as the frequency of light increases, so will the stopping potential. This will push the present down. The combined result of these 2 is that the current remains identical . Therefore A is right.
When the luminous source frequency doubles, electron kinetic energy increases. Yet, it's getting more than double. Thus, B is wrong.
Explanation:
Option (b) is correct.
- Saturated current is directly correlated with the light intensity. When the light intensity increases, a significant number of photons come down on the metal surface. A large proportion of electrons therefore interfere with the photons.
- As a consequence, the number of electrons released increases, and hence the current increases as well.
- At same time, the frequency of the source of light also increases. Therefore, as the frequency of light increases, so will the stopping potential. This will push the present down. The combined result of these 2 is that the current remains identical . Therefore A is right.
- From the Photoelectric Equation of the Einstein.
=hv−φ
here
= electron kinetic energy
v = light frequency
φ = Metal work function
- This is apparent from the equation above. When the luminous source frequency doubles, electron kinetic energy increases. Yet, it's getting more than double. Thus, B is wrong.
To know more about frequency and intensity of a light, visit:
How is the photoelectric current affected on increasing the frequency and intensity of incident radiation and why?
https://brainly.in/question/7767015
The intensity of light depends on -
1) amplitude
2) frequency
3) wavelength
4)none
https://brainly.in/question/7129668
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