Question

How many 860nm photos are contained in a 2.7uJ pulse of monochromatic radiation ?

Answer 1

Answer:$1.1683\times 10^{13}$ number of photons.

Explanation:

$\text{Energy of photon}=E= \frac{hc}{\lambda }$

$E= \frac{6.626\times 10^{-34}J s\times 3\times 10^{8}m/s}{860\times 10{-9}m}=0.02311\times 10^{-17} J$

$\text{Energy of radiation}=2.7\mu J=2.7\times 10^{-6}J(1\mu J=10^{-6}J)$  

$\text{Energy of radiation}=E\times \text{Number of photons}$

$\text{Number of photons}=\frac{\text{Energy of radiation}}{\text{Energy of photon}(E)}=\frac{2.7\times 10^{-6}J}{0.02311\times 10^{-17} J}=1.1683\times 10^{13}\text{photons}$

$1.1683\times 10^{13}$ numbers of photons will be contained by $2.7\mu J$ pulse of monochromatic radiation.