Question

Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons. The second number tells you why our eye can never count photons, even in barely detectable light.

(a) The number of photons emitted per second by a medium wave transmitter of 10 kW power, emitting radio waves of wave length 500 m.

(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive $\left({10}^{10}W{m}^2\right)$. Take the area of the pupil to be about $0.4c{m}^2$, and the average frequency of white light to be about $6\times {10}^{14}Hz$.

Answer 1

(a). $P={10}^{4}W$

$E_1=\frac{hc}{\lambda }$

$\Rightarrow \frac{6.6\times {10}^{-34}\times 3\times {10}^8}{500}$

$=3.96\times {10}^{-28}$

$P=nE$

$\Rightarrow n=\frac{{10}^4}{3.96\times {10}^{-28}}=3\times {10}^{31}photons/s$

We see that the energy of a radio photon is exceedingly small, and the number of photons emitted per second in a radio beam is enormously large. There is, therefore, the negligible error involved in ignoring the existence of a minimum quantum of energy (photon) and treating the total energy of a radio wave is continuous.

(b). $I={10}^{-10}W{m}^{-2}$

$A=0.4c{m}^2$

$E=h\nu$

$\Rightarrow 6.6\times {10}^{-34}\times 6\times {10}^{14}=3.96\times {10}^{-19}$

$I=nE$

$n=\frac{{10}^{10}}{3.96\times {10}^{-19}}=2.52\times {10}^8{m}^2/s$

$n_A=n\times A$

$\Rightarrow 2.52\times {10}^8\times 0.4\times {10}^{-4}={10}^4{s}^{-1}$

Though this number is not as large as in (a) above, it is large enough for us never to sense or count individual photons by our eye.