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Question
Two monochromatic beams A and B of equal intensity I, hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then what inference can you make about their frequencies?
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Solution
Effect of intensity: If the intensity of light is increased (while its frequency is kept the same) the current levels off at a higher value, showing that more electrons are being emitted per unit of time. But the stopping potential V0 doesn't change, i.e. Intensity `∝` no . of incident photon no. of emitted photoelectron per time photocurrent.
Effect of frequency: If the frequency of incident light increases, (keeping intensity constant) stopping potential increases but there is no change in photoelectric current.
Let us assume nA is the number of photons falling per second of beam A and nB is the number of photons falling per second of beam B.
And it is given that the number of photons hitting the screen by beam A is twice that by beam B.nA = 2nB
The energy of the falling photon of beam A = hvA
The energy of a falling photon of beam B = hvB
Now, according to the question, the intensity of A is equal to the intensity of B.
Therefore, I = nAvA = nBvB
⇒ `v_A/v_B = n_B/n_A = n_B/(2n_B) = 1/2`
⇒ vB = 2vA
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