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hichkok12 [17]
4 years ago
10

A laser used for many applications of hard surface dental work emits 2780-nm wavelength pulses of variable energy (0-300 mJ) abo

ut 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.
Physics
1 answer:
Bess [88]4 years ago
6 0

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

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