Question

Assume that a 100 W sodium lamp radiates its energy uniformly in all directions in the form of photons with an associated wavelength of 589nm. At what distance from the lamp will the average flux of photons be \[1\text{ photons}/c{{m}^{2}}s\].

Answer 1

Hint: We have to find the distance at which the average flux of photon is \[1\text{ photons}/c{{m}^{2}}s\] for that we have to calculate the number of photons emitted per second from the sodium lamp. The number of photons per second is given by dividing the power of the bulb by the energy of photons. So first we will calculate the energy of photons then the number of photons emitted per second and finally the distance.

Formula used:
\[\begin{align}
  & E=h\nu =\dfrac{hc}{\lambda } \\
 & N=\dfrac{P}{E} \\
 & n=\dfrac{N}{4\pi {{r}^{2}}} \\
\end{align}\]

Complete answer:
Let us first calculate the energy of photons. The energy of photons is given as the product of Planck’s constant h and the frequency associated with photons and frequency is given as the speed of light divided with wavelength, therefore the equation can be given as
\[E=h\nu =\dfrac{hc}{\lambda }\]
Now, the speed of light c and Planck’s constant h are the universal constant and their values are
\[c=3\times {{10}^{8}}m/s\text{ and }h=6.62\times {{10}^{-34}}{{m}^{2}}kg/s\]respectively.
The given wavelength for the sodium lamp is \[\lambda =589nm=589\times {{10}^{-9}}m\]
Substituting the values in the above equation we will get the energy of photons emitted from sodium lamps.

\[\begin{align}
  & E=\dfrac{\left( 6.62\times {{10}^{-34}} \right)\left( 3\times {{10}^{8}} \right)}{589\times {{10}^{-9}}m} \\
 & E=\dfrac{19.86}{589}\times {{10}^{-17}} \\
 & E=0.0337\times {{10}^{-17}} \\
 & E=3.37\times {{10}^{-19}}J \\
\end{align}\]

As the lamp is radiating energy of \[100J/s\], the number of photons emitted per second can be
\[N=\dfrac{P}{E}\]
Where N is the number of photons and P is the power of the source from which photons are emitted and E is the energy of photons which we calculated above. Given \[P=100W\] and the energy we calculated above, the number of photons can be given as
\[N\simeq 3\times {{10}^{20}}photons/s\]
Now the average flux is given as the number of photons emitted per second from the surface area of the source, here the sodium lamp. Let's say the lamp is spherical in shape, then its surface area is given as \[4\pi {{r}^{2}}\]. Hence the average flux can be given as
\[n=\dfrac{N}{4\pi {{r}^{2}}}\]
According to the question the average is \[n=1\text{ photons}/c{{m}^{2}}s=1\text{ }\times {{10}^{4}}\text{photons}/{{m}^{2}}s\]. The above equation for average flux can be rewritten for the distance as
\[r=\sqrt{\dfrac{N}{4\pi n}}\]
Now substituting the values, we get
\[\begin{align}
  & r=\sqrt{\dfrac{3\times {{10}^{20}}}{4(3.14)(1\text{ }\times {{10}^{4}})}} \\
 & r=\sqrt{0.238\text{ }\times {{10}^{16}}} \\
 & r=0.487\text{ }\times {{10}^{8}} \\
 & r=48.7\times {{10}^{6}}m \\
\end{align}\]
Hence at the distance \[48.7\times {{10}^{6}}m\] the lamp will have average flux of photons \[1\text{ photons}/c{{m}^{2}}s\]

Note:
Einstein postulated that the energy travels in the form of packets of photons which is given as \[h\nu \] which is the energy of photons. It can vary with the light source as the frequency will be different or each source.
Here I have changed the unit of the average flux of the photon to avoid error although it is not necessary because the other term does not have length dimension.