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 $E=h v$
$E=6.63 \times 10^{-34} \times \frac{10^{16}}{6.63}=10^{-18} J$
So, number of photons per second $=\frac{\text { Power }}{E}$
$=\frac{10^{-2}}{10^{-18}}=10^{16}$