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

a

`n = 1.119 *10^(18) \ photons`

b

`P = 1.6 \ W`

Step-by-step 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 = (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 = (E * \lambda )/(h * c )`

substituting values

`n = (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 = (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 = (E)/(T)`

substituting values

`P = ( 80 *10^(-3))/(0.05)`

`P = 1.6 \ W`