Question

A photo multiplier has 12 plates including photo cathode and anode. Assume each diode doubles the electrons. If a 10 $W$ source of 300 $nm$ is placed 2 $m$ away then find the photo current shown by the photo multiplier tube. The dark current is 160 $\mu A.$Assume efficiency 10% and flux reaching the photo cathode is ${10}^{-4}$ of the original.

Answer 1

$I_p\propto \frac{1}{r^2}E\left(eV\right)=\frac{1240}{300}=4.13$ $eV$
$\therefore$ number of photons per second
$N=\frac{P}{hf}=\frac{2.5\times {10}^{-4}}{4.13\times 1.6\times {10}^{-19}}=3.8\times {10}^{14}$
Number of photo electrons emitted per second $Ne=3.8\times {10}^{14}\times \frac{10}{100}=3.8\times \frac{{10}^{13}}{s^{-l}}$
Current shown by photo multiplier $I_p=3.8\times {10}^{13}\times 1.6\times {10}^{-19}\times 2^{10}=6.1mA$
Exact Current $=$ 6.1 $mA$ -160 $\mu A=$ 5.94 $mA$.