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## Homework Statement

A weak light source (wavelength 600 nm, mean power 66.2 pW) falls on a multiplier tube with a cascade of 8 dynodes. How large is the mean current at the anode with an amplification of 10

^{8}and a 50% quantum efficiency for the photoelectric effect.

## Homework Equations

[itex]E_{kin} = h\nu - W[/itex]

## The Attempt at a Solution

No. of incident photons = No. of electrons emitted ??

[itex]N_{\gamma} = N_{e}[/itex]

[itex] \frac{P * t * λ}{h * c} = \frac{66.2pW * 1s * 600nm}{h*c} = 1.66E29[/itex] (in one second)

average current produced at first dynode = I

_{0}

[itex]I_{0} = \frac{dq}{dt} = \frac{N_{e} * e}{1s} = 3.201E-11A[/itex]

Then this is amplified by a factor 10

^{8}to give final average current at anode

I = I_{0} * 1E8 = 3.201E-3A

well, that was my idea but I don't see how the 50% efficiency fits into this. spot my mistake anyone?

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