Photomultiplier. Acoustics.

Photomultiplier

Light incident on a photomultiplier cathode has an intensity of $1.2 pW/cm^2$ and an wavelength of $\lambda=450 nm$. The cathode area for light is $0.2 cm^2$ and the photomultiplier has 10 dynodes. If the current on the anode is $40 \mu A$ please find

1. how many photons fall each second on the cathode.

2. the value of the secondary emission of dynodes.

photomultiplier

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

$N1 = I/e = 40*10^{-6}/1.6*10^{-19} =2.5*10^{14} (electrons/second)$

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

$R^{10} = N1/N0$

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

Acoustic Intensity

You are given a chamber having walls that absorbs the sound. This chamber has an opening of 2.0m x 1.0 m. Outside this chamber there is a loudspeaker situated at a distance of 84 m, towards the room. The sound that enters the room through the opening has an intensity of 56 dB. The acoustic wave is spherically symmetric uniform, and there is no reflection from the ground. What is the acoustic power of the loudspeaker, given the  hearing threshold of $1*10^{-12} W/m^2$.

Acoustics room

The decibel rating of a sound intensity is by DEFINITION

$dB =10*log (I/I0)$

where the log is base 10 and $I0 =10^{-12}W/m^2$ is the hearing threshold.

$56 =10*log(I/I0)$

$I = I0*10^{5.6} =10^{-12}*10^{5.6} =10^{-6.4} =3.981*10^{-7} W/m^2$

The intensity is by definition

$Intensity = Power/Area$

Since the loudspeaker is 84 m away from the window and is radiating uniformly, the wave front of the sound will be a sphere having radius 84m.

$S =4*\pi*R^2 =88668.3 m^2$

Hence the acoustic power of the speaker is

$P = I*S =3.981*10^{-7}*88668.3 =0.0353 W$

(verify: The acoustic power of a music concert is about 0.1 W.

Selected Sound Sources (Wiki))

A 56 dB sound intensity is only a bit higher than a normal conversation. Imagine you have a music concert 100 meters away. You will hear it as a normal conversation