# Mean Current of Photomultiplier at anode

## 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 108 and a 50% quantum efficiency for the photoelectric effect.

## Homework Equations

$E_{kin} = h\nu - W$

## The Attempt at a Solution

No. of incident photons = No. of electrons emitted ??
$N_{\gamma} = N_{e}$
$\frac{P * t * λ}{h * c} = \frac{66.2pW * 1s * 600nm}{h*c} = 1.66E29$ (in one second)

average current produced at first dynode = I0

$I_{0} = \frac{dq}{dt} = \frac{N_{e} * e}{1s} = 3.201E-11A$

Then this is amplified by a factor 108 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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