Question

A monochromatic light of frequency $v=\frac{1}{6.63}\times 10^{16}Hz$ is produced by a laser. The power emitted is $P=10^{-2}W.$ The average number of photos per second emitted by the source is

• A. $\frac{1}{(6.63{)}^2}\times 10^{16}$
• B. $(6.63{)}^2\times 10^{20}$
• C. $(6.63{)}^2\times 10^{16}$
• D. $10^{20}$
• E. $10^{16}$

Answer 1

Answer: (E)

Given, frequency of EM wave, $v=\frac{1}{6.63}\times 10^{16}Hz$
and power emitted $P=10^{-2}W$.
Energy of a photon $E=hv$
$E=6.63\times 10^{-34}\times \frac{10^{16}}{6.63}=10^{-18}J$
So, number of photons per second $=\frac{\text{ Power }}{E}$
$=\frac{10^{-2}}{10^{-18}}=10^{16}$