Question

how many photons are emitted per second by a 5 mw laser operating at 620 nm?

Answer 1

The photon energy is equal to speed times the Plnck constant divided by the wavelength of the photons.
$e= \frac{hc}{l}$  (l is the wavelength, lambda) (Mark as 1)

The energy of all the photons emitted at one second is 5mW = 0.005J/s
Also, the energy of the photons is equal to amount of emitted photons times the energy of one photon
$E=ne$
$E=n \frac{hc}{l}$
$0.005=n \frac{6.626*10^{-34}*3*10^8}{620*10^{-9}}$
$n=1.55*10^{16}$ photons

Answer 2

given ,

wavelength = 620 nm
power = 5 mili watt

we know,
Energy of one photon = hc/lemda

where , h is plank' constant
c is velocity of light and lemda is wavelength .
after putting value of h and c
Energy of one photon = 12375/lemda( in A° )

Energy of one photon = 12375/( 6200 A°)
= 1.9959 ev
Energy of one photon = 1.9959 × 1.6× 10^-19 joule = 3.19 × 10^-19 Joule

now Let no of photon = n in one second

so, no of photon in one second = Power × one second/energy of one photon

= 5 × 10^-3/3.19 × 10^-19

= 1.56 × 10^(-3 +19 )

= 1.56 × 10^16

hence, no of photon per second = 1.56 × 10^16