Mean Current of Photomultiplier at anode

[sunrah]

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?

 

[ananthsaran7]

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 calcualte that, i.e no. of photoelectrons=no of incident photons *Quantum efficiency . It is usually a function of the incident wavelength.