# how many photons are emitted per second by a 5mw laser source operating at 632.8nm?

Energy, E of the photon is given by -

E = $\frac{hc}{\lambda}$

So, E = $\frac{6.63x{10}^{-34}Jsx3x{10}^{8}m/s}{632.8x{10}^{-9}m}=3.14x{10}^{-19}J$

Now, the energy emitted per second by the laser source = 5 mW or 5 x 10

^{-3}J

Thus, the total number of photons emitted per second = $\frac{5x{10}^{-3}J}{3.14x{10}^{-19}J}=1.6x{10}^{16}$

Regards

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