Question

Light of 200 microwatt is incident on a photodetector. If the light is red, 600nm, how many photons are striking the surface per second?

Answer 1

`n=6.0332* 10^(14)per second` photon per second

Step-by-step explanation:

We have given power of the incident light
`P=200* 10^(-6)=2* 10^(-4)W`

Time = 1 sec

We know that energy E = power × time =
`2* 10^(-4)* 1=2* 10^(-4)j`

Wavelength of light
`\lambda =600nm=600* 10^(-9)m`

We know that energy pf each photon is given by
`E=(hc)/(\lambda )`

Let there are n numbers of photon
`(nhc)/(\lambda )=2* 10^(-4)`

`(n* 6.6* 10^(-34)* 3* 10^8)/(600* 10^(-9) )=2* 10^(-4)`

`n=6.0332* 10^(14)per second`