My textbook goes on explaining variations of photo current/stopping potential and frequency/intensity of the light used.
My doubt is what effect does the amplitude of light wave have on photo current and stopping potential?
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$\begingroup$ Amplitude is equivalent to intensity, changing amplitude will bring the same results as changing intensity. $\endgroup$– TofiCommented Feb 19, 2017 at 10:22
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1 Answer
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The intensity of a wave is directly proportional to the square of the amplitude of the wave. Mathematically speaking, $I \propto a^2$. So, increasing the amplitude increases the intensity of the wave. And the increase in intensity of the incident radiation increases the number of photo-electrons, but it does not affect the stopping potential of the concerned metal surface.
answered Feb 19, 2017 at 11:13
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$\begingroup$ But my teacher told us that intensity is related to number of photons per second. Is that also correct? $\endgroup$ Commented Feb 19, 2017 at 11:40
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1$\begingroup$ Yes, it can be taken as an analogy at max. That is, you are relating the wave property of light (intensity) with its particle property (no. of photons). This is possible since light has dual nature. But in general, one should take this thing strictly or rigorously. $\endgroup$ Commented Feb 19, 2017 at 11:44
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$\begingroup$ Yes amplitude of light is the number of photons. Light is made up of individual photons. $\endgroup$ Commented Feb 19, 2017 at 18:06