During an experiment on the photoelectric effect, light of wavelength 400 nm illuminates a cesium plate at a rate of 5.0 W . The potential of the collector plate is kept sufficiently positive relative to the emitter to

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Quantrex Academy · Accepted Answer

Correct answer: C. 1.6 μ A

The energy of a single incident photon is given by:

E = \dfrac{hc}{\lambda}

Substituting the given values:

E = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} \approx 4.97 \times 10^{-19} \text{ J}

The number of photons incident on the plate per second is:

N = \dfrac{P}{E} = \dfrac{5.0}{4.97 \times 10^{-19}} \approx 1.0 \times 10^{19} \text{ s}^{-1}

Since one out of every

10^6

photons ejects a photoelectron, the number of photoelectrons emitted per second is:

n = N \times 10^{-6} = 1.0 \times 10^{19} \times 10^{-6} = 1.0 \times 10^{13} \text{ s}^{-1}

The saturation photocurrent is the total charge of these electrons per unit time:

I = ne = 1.0 \times 10^{13} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-6} \text{ A} = 1.6 \text{ \mu A}

During an experiment on the photoelectric effect, light of...

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