A 2.4 mW laser ( =525 nm) shines on a cesium photocathode( ? = 1.95 eV). Assume

Question

A 2.4 mW laser ( =525 nm) shines on a cesium photocathode(? = 1.95 eV). Assume an efficiency of 10-5 forproducing photoelectrons (that is, 1 photoelectron is produced forevery 105 incident photons) and determine thephotoelectric current.

I tried to solve this by acquiring the number of photons per second(i got 6.315E15) and then simply dividing that by 10^5 since(6.315E15)/(1x10^5) should be the number of photoelectrons producedper second, and then simply converting that to amperes by dividingthose electrons by the value for one amp one amp (6.242 ×1018) but its not taking my answer :(   howcan I use the power factor to solve this?

Explanation / Answer

Power P = 2.4 mW Energy per unit time E = 2.4 mJ                                   = 2.4 * 10 ^ -3 J wavelength = 525 nm = 525 * 10 ^ -9 m we know E = nhc / from this No.of photos n = E / hc                                      = [ 2.4*10^-3J*525*10^-9 m] / [6.625*10^-34Js*3*10^ 8 m / s]                                       = 63.39 * 10 ^ 14 So, No.of photo electrons = 10 ^ -5 * 63.39 * 10 ^14                                         = 63.39* 10 ^ 9 Total charge of these electrons Q = 63.69* 10 ^ 9 * 1.6 * 10 ^-19 C                                                   = 101.433 * 10 ^ -10 C So, photo electric current i = Q /t    where t = time = 1 s Therefore i = 101.433 * 10 ^ -10 A

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