A professor was sent a promotional photoelectric effect demonstration by a compa

Question

A professor was sent a promotional photoelectric effect demonstration by a company hoping to get more of the university's business. He assembles the two plates, closes the vacuum chamber around them and attaches the battery and ammeter. However, when he uses the provided source of blue visible light to illuminate the plate, no current flows. Frustrated, he breaks the device down, reassembles it, and again it doesn't work. Finally, he has an idea, and has the chemistry department identify the material of the plate. He finds that instead of Calcium, the plate has instead been made of Platinum. Why does this cause the device to fail? What intensity of blue light would be needed in order to ensure that current would flow?

Explanation / Answer

thershold energy of electron in platinum is more as campare to calcium due to this low intensity blue visible light ray could excite the free electron of platinum hence device has failed.

work function of platinum = 6.35 eV

I*hC/lambda = 6.35*1.6*10^-19

I = 6.35*1.6*10^-19*475*10^-9 / (6.62*10^-34*3*10^8)

i = 2.43

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