Mean Current of Photomultiplier at anode

  • Thread starter sunrah
  • Start date
  • #1
199
22

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:

Answers and Replies

  • #2
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.
 

Related Threads on Mean Current of Photomultiplier at anode

Replies
1
Views
3K
Replies
3
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
1
Views
16K
  • Last Post
Replies
9
Views
5K
Replies
3
Views
1K
Replies
12
Views
4K
  • Last Post
Replies
8
Views
4K
  • Last Post
Replies
4
Views
12K
Top