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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 $10^{6}$ and that the out current as a function of time (in natiosecond) can appruximated as shown in figure.
$I_{\max}$ when averaged over many pulses is $80 μ A$ . II which of the following statements is/are true.

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The charge carried by one pulse is $8 \times 10^{- 13} C$

Number of photo electrons emitted per light p. is 5

Number of photons in one light pulse is 50

Number of electrons carried by one pulse $5 \times 10^{5}$

answer is A, B, C.

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Detailed Solution

The total quantity of charge carried by one pulse of infrent is

$Q = \int I d t$

which is the area of the triangle in figure. Thus

$Q = \frac{1}{2} (20 \times 10^{- 9}) (80 \times 10^{- 6}) = 8 \times 10^{- 13} C$

and the number of electrons carried by one pulse is

$n = \frac{Q}{e}$

$\Rightarrow n = \frac{8 \times 10^{- 13}}{1.6 \times 10^{- 19}} = 5 \times 10^{6}$

Then the number of photoelectrons emitted per light pulse is

$n^{'} = \frac{n}{10^{6}} = 5$

and hence the number of photons in one light pulse is

$N = \frac{n^{'}}{0.1} = 50$

Hence, (A), (B) and (C) are correct.

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