A photomultiplier tube is to be used to detect light pulses each of which consists of a small but fixed number of photons. The average photoelectric efficiency is 10%. That is photon has 10% probability of causing the emission of a detectable photoelectron. Assume the photomultiplier gain is 106 and that the output current as a function of time can be approximated as shown in figure.
Imax
when averaged over many pulses is 80 μ A. Then which of the following are true -
Text Solution
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(a, b ,c)
The total quantity of charge carried by one pulse of current is
Q = 
dt,
which is the area of the triangle in Fig. Thus
Q = 
× 20 × 10 –9 × 80 × 10 –6 = 8 × 10 –13 C
and the number of electrons carried by one pulse is
n = 
= 
= 5 × 10 6 .
Then the number of photoelectrons emitted per light pulse is
n' = n/10 6 = 5 ,
and hence the number of photons in one light pulse is
N = n'/0.1 = 50.
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