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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. - Physics

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प्रश्न

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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उत्तर

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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अध्याय 11: Dual Nature Of Radiation And Matter - Exercises [पृष्ठ ७२]

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एनसीईआरटी एक्झांप्लर Physics [English] Class 12
अध्याय 11 Dual Nature Of Radiation And Matter
Exercises | Q 11.22 | पृष्ठ ७२

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