(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Mean Current of Photomultiplier at anode

**Physics Forums | Science Articles, Homework Help, Discussion**