Question

How many photons are emitted per second by a 5 mW laser operating at 620 Angstrom

Answer 1

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Wavelength = 620 nm
Power = 5 mili watt

Energy of one photon = hc/lemda

where , h is plank's constant
c is velocity of light and lemda is wavelength .
after putting value of h and c

Energy of one photon = 12375/lambda( 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, number of photon per second = 1.56 × 10^16

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