Question

How many photons are emitted per second by a 5 megawatt laser operating at 620 nm?

Answer 1

HERE IS YOUR ANSWER BUDDY!

Given in the question,

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

CONGO! :)