Question

An argon laser emits light of wavelength 488 nm at power of 100 mW (milliwatt). Calculate the energy of the photons that are emitted. If one focus the light onto a surface using a microscope objective to a circle with diameter 5 μm (micrometer), what is the power density of the light at unit W/m2 . Calculate the number of photons emitted by the laser in ten seconds. What is the photon flux of the focused beam in 1 Å2 , roughly the area of one atom, in unit photon/s/Å2

I got 245x10^17 photons/second
Having trouble with the rest!

Answer 1

a

`E = 4.0733 *10^(-19)  \  J`

b

`n  =  2.45 *10^(16)`

c

tex]k = 2.45 *10^{19} \ photon / second[/tex]

Step-by-step explanation:

From the question we are told that

The wavelength is
`\lambda  =  488 \ nm  =  488*10^(-9) \  m`

The power is
`P  =  100 \  mW   =  100 *10^(-3) \  W`

The diameter of the circle is
`d =  5 \mu m =  5.0*10^(-6)\ m`

The time taken is
`t =  10 \  s`

Generally the energy of the photon is mathematically represented as

`E =  (hc  )/(\lambda)`

substituting
`6.626*10^(-34) \  J\cdot s` for h ,
`3.0*10^8   \ m/s` for c

So

`E =  ( 6.626 *10^(-34) *  3*10^8  )/(488 *10^(-9))`

=>
`E = 4.0733 *10^(-19)  \  J`

Generally the number of photons emitted is mathematically

`n  =  (P)/(E)`

`n  =  (10 *10^(-3))/(4.07733 *10^(-19))`

`n  =  2.45 *10^(16)`

Generally the number of photons emitted in by the laser in ten seconds is mathematically

`k = (10)/(E)`

=>
`k = (10)/(4.0733 *10^(-19))`

=>
`k =  2.45 *10^(19) \ photon /  second`