Physical Electronics design question http://i.imgur.com/BVBFxbe.jpg Design a pho

Question

Physical Electronics design question

http://i.imgur.com/BVBFxbe.jpg

Design a photomultiplier tube (PMT) to have an incident light at lambda = 450 nm on its cathode at an intensity of 1.2 pW/cm^2. The cathode needs to have an effective area of 0.2 cm^2. The effective area accounts for the angle of incidence of the light. The photomultiplier tube needs to have 10 identical dynodes for the purpose of secondary emission gain. The anode current, I_A,needs to be 40 mu A. Then, determine: The number of photons per second that are incident on the cathode The secondary emission ratio of the dynodes. Assume a reasonable percentage of generation rate of electrons for each photon incident on the cathode. The designed photomultiplier tube should look like the following.

Explanation / Answer

a. power incident on the cathode(P)= Intensity*Area= 0.24 pW

Energy of a photon(E)= hc/l ; l-wavelength

So,Number of photons incident per secons= P/E =5.433*105

b. Number of electrons ejected from cathod per second(N0)= 5.433*105

Take r- secondary emission ratio. So,

number of electrons emitted from the last dynode(N)= N0*r10

Anode current= charge per second= e*N=40*10-6 ; e-elctronic charge

So, e*N0*r10=40*10-6

r=7.364

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