Question

Monochromatic light of wavelength 667 nm is produced by a helium neon laser. The power emitted is 9mW. The number of photos arriving per sec. on the average at a target irradiated by this beam is:
A) $3\times {10}^{19}$
B) $9\times {10}^{17}$
C) $3\times {10}^{16}$
D) $3\times {10}^{15}$

Answer 1

Hint: The concept of photon originates from Planck-Einstein's relation, which is the most fundamental concept in quantum mechanics. The Planck’s relation must be applied directly to calculate the number of photons.

Complete step by step solution:
The light energy was originally considered to be in the form of straight rays. Later, several experiments showed different phenomena of light behaving as a wave such as interference, diffraction and polarisation.
However, the light wave was originally believed to be a continuous wave. It was not until Einstein’s photoelectric effect experiment that they discovered that the light wave is not continuous. Einstein’s hypothesis proved that light energy is transferred in discrete packets of energy wherein each packet has the energy equivalent to its frequency.
Energy possessed by a photon,
$E\propto \upsilon$
where $\upsilon$ = frequency of light.
By removing proportionality and introducing the Planck’s constant, we obtain an expression known as the Planck-Einstein relation given by –
$E=h\upsilon$
where h = Planck’s constant whose value is equal to $6.626\times {10}^{-34}J{s}^{-1}$
Light travels at a constant velocity of $c=3\times {10}^{8}m{s}^{-1}$ . Since we have considered the light as a wave, the velocity of the wave is equal to the product of frequency and wavelength of the wave.
$c=\upsilon \lambda$
Thus, we have –
$$E={\displaystyle \frac{hc}{\lambda }}$$
This is the energy possessed by one photon of light. When a monochromatic light of wavelength 667 nm gives rise to $n$ photons, the power emitted is given by, -
$$E=n{\displaystyle \frac{hc}{\lambda }}$$
Thus, the number of photons emitted is given by –
$n={\displaystyle \frac{E}{\left({\displaystyle \frac{hc}{\lambda }}\right)}}$
Given,
Power emitted, $E=9mW=9\times {10}^{-3}W$
Wavelength of light, $\lambda =667\times {10}^{-9}m$
Substituting the values of the above values and the constants, we obtain –
$n={\displaystyle \frac{9\times {10}^{-3}}{\left({\displaystyle \frac{6.626\times {10}^{-34}\times 3\times {10}^8}{667\times {10}^{-9}}}\right)}}$
$\Rightarrow n={\displaystyle \frac{9\times {10}^{-3}\times 667\times {10}^{-9}}{6.626\times {10}^{-34}\times 3\times {10}^8}}$
$\Rightarrow n={\displaystyle \frac{6003\times {10}^{-12}}{19.878\times {10}^{-26}}}$
$\Rightarrow n=301.99\times {10}^{26-12}$
$\Rightarrow n=301.99\times {10}^{14}$
$\Rightarrow n=3.0199\times {10}^{16}\approx 3\times {10}^{16}$

Therefore, the number of photons emitted per second is equal to $3\times {10}^{16}$.

Note: The students might get confused here, since I have considered the energy as watt while the unit of energy must be clearly joule. However, that is permitted in this problem, since they are asking the number of photons emitted per second. Since they are asking the rate of photons emitted per second, we can consider the units of power as units of energy in our solution.