Mean Current of Photomultiplier at anode

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SUMMARY

The mean current at the anode of a photomultiplier tube with a cascade of 8 dynodes, when exposed to a weak light source of 600 nm wavelength and 66.2 pW mean power, is calculated to be 3.201 mA. This calculation assumes an amplification factor of 10^8 and a quantum efficiency of 50%. The initial number of incident photons is determined using the formula N_{γ} = (P * t * λ) / (h * c), resulting in approximately 1.66E29 photons per second. The error identified in the discussion is the assumption that the number of incident photons equals the number of emitted electrons; instead, the quantum efficiency must be applied to determine the actual number of photoelectrons emitted.

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  • Familiarity with the equation E_{kin} = hν - W
  • Basic principles of photon and electron interactions
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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 108 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 = I0

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

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?
 
Last edited:
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The mistake is right in the first line i.e the number of incident photons =no of electrons emitted. You should use the quantum efficiency to calculate that, i.e no. of photoelectrons=no of incident photons *Quantum efficiency . It is usually a function of the incident wavelength.
 

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