Please show all work! Thanks 3. A photomultiplier tube has incident light at AJ4

Question

Please show all work! Thanks

3. A photomultiplier tube has incident light at AJ450 nm o n its cathode at an intensity of 1.2 pw/cm The cathode has an effective area of 0.2 cm The effective area accounts for the angle o incidence of the light. The f photomultiplier tube has 10 identical dynodes for the purpose of secondany emission gain. The anode current, s measured to b 40HA. Determine: (10 points) a. The number of photons per second that are incident on the cathode b. The secondary emission ratio of the dynodes. Assume a 100% generation rate of electrons for each photon incident on the cathode Photomultiplier Tube Cathode Anode nodes. Electron Paths Incident Photon

Explanation / Answer

a)

The total incident power is

P =intensity*surface = 1.2*10^-12*0.2 =0.24*10^-12 W =0.24 pW

The energy of one incident photon is

E=h*F = h*c/lambda = 6.62*10^-34*3*10^8/450*10^-9 =4.413*10^-19 J

The number of photons incident on cathode (in 1 second) is

N0 = P/E = 0.24*10^-12/4.413*10^-19 =543806 photons/second =5.44*10^5 (1/s)

b)

The number of electrons that arrive at the anode (in one second) is

N1 = I/e = 40*10^-6/1.6*10^-19 =2.5*10^14 (electrons/secomd)

There are 10 dinodes. The rate of generation (electrons generated/electrons incident) of one dinode is R

R^10 = N1/N0

R = log(10) (N1/N0) =log (2.5*10^14/5.44*10^5) =8.66

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