Question

A ruby laser produces radiation of wavelengths, 662.6 nm in pulse duration of \[{10^{ - 6}}\,{\text{s}}\]. If the laser produces 0.39 J of energy per pulse, how many photons are produced in each pulse?
A. \[1.3 \times {10^9}\]
B. \[1.3 \times {10^{18}}\]
C. \[1.3 \times {10^{27}}\]
D. \[3.9 \times {10^{18}}\]

Answer 1

Hint: Recall the energy of the photon of wavelength \[\lambda \]. In each pulse, the number of photons will be produced and the energy of one pulse is given in the question. Express the energy of the n photons and then determine the number of photons n.

Formula used:
Energy, \[E = \dfrac{{hc}}{\lambda }\]
Here, h is the Planck’s constant, c is the speed of light and \[\lambda \] is the wavelength of the photon.

Complete Step by Step Answer:
We have, the energy of a photon is expressed as,
\[E = \dfrac{{hc}}{\lambda }\]
Here, h is the Planck’s constant, c is the speed of light and \[\lambda \] is the wavelength of the photon.
In each pulse, the number of photons will be produced and the energy of one pulse is given. Let n be the number of photons produced per pulse. Then the energy of the pulse is,
\[E = n\dfrac{{hc}}{\lambda }\]
Rearranging the above equation for n, we get,
\[n = \dfrac{{E\lambda }}{{hc}}\]
Substituting \[E = 0.39\,{\text{J}}\], \[\lambda = 662.6\,{\text{nm}}\], \[h = 6.626 \times {10^{ - 34}}\,{\text{Js}}\] and \[c = 3 \times {10^8}\,{\text{m/s}}\] in the above equation, we get,
\[n = \dfrac{{\left( {0.39} \right)\left( {662.6 \times {{10}^{ - 9}}} \right)}}{{\left( {6.626 \times {{10}^{ - 34}}} \right)\left( {3 \times {{10}^8}} \right)}}\]
\[ \Rightarrow n = \dfrac{{2.584 \times {{10}^{ - 7}}}}{{1.9878 \times {{10}^{ - 25}}}}\]
\[ \therefore n = 1.3 \times {10^{18}}\]

Thus, in each pulse, there will be \[1.3 \times {10^{18}}\] photons.

Note:The formula for energy of the photon is \[E = h\nu \], where, \[\nu \] is the frequency of the photon. The frequency of the photon is given as, \[\nu = \dfrac{c}{\lambda }\]. Students must remember the values of speed of light and Planck’s constant to solve this question. The energy of the photon should be in joules if the wavelength is in nm or meter.