Question

A laser used for many applications of hard surface dental work emits 2780-nm wavelength pulses of variable energy (0-300 mJ) about 20 times per second.part a. Determine the number of photons in one 80-mJ pulse.part b. Determine the average power of photons in one 80-mJ pulse during 1 s.

Answer 1

Answer:

a

    $n = 1.119 *10^{18} \ photons$

b

  $P = 1.6 \ W$

Explanation:

From the question we are told that

    The wavelength is  $\lambda = 2780 nm = 2780 *10^{-9} \ m$

     The  energy  is  $E = 80 mJ = 80 *10^{-3} \ J$

This energy is mathematically represented as

     $E = \frac{n * h * c }{\lambda }$

Where  c is the speed of light with a value  $c = 3.0 *10^{8} \ m/s$

             h is the Planck's  constant with the value  $h = 6.626 *10^{-34} \ J \cdot s$

             n is the number of pulses

So

      $n = \frac{E * \lambda }{h * c }$

substituting values

       $n = \frac{80 *10^{-3} * 2780 *10^{-9}}{6.626 *10^{-34} * 3.0 *10^{8} }$

       $n = 1.119 *10^{18} \ photons$

Given that the pulses where emitted 20 times in one second then the period of the pulse is

       $T = \frac{1}{20}$

      $T = 0.05 \ s$

Hence the average power of photons in one 80-mJ pulse during 1 s is mathematically represented as

       $P = \frac{E}{T}$

substituting values

       $P = \frac{ 80 *10^{-3}}{0.05}$

        $P = 1.6 \ W$