Question

$A 160$ W light source is radiating light of wavelength $6200\,\mathring{A}$ uniformly in all directions. The photon flux at a distance of $1.8\, m$ is of the order of (Planck’s con stant $= 6.63\times 10^{-34} J-s$)

• A. $10^{2}m^{-2} s^{-1}$
• B. $10^{12}m^{-2} s^{-1}$
• C. $10^{19}m^{-2} s^{-1}$
• D. $10^{25}m^{-2} s^{-1}$

Answer 1

Answer: (C)

Photon flux $=$ Number of photons per unit area per unit time
$=\frac{I}{\left(\frac{hc}{\lambda}\right)}=\frac{I\lambda}{hc}=\frac{P\lambda}{4\pi r^{2}hc}$
$=\frac{160\times6200\times10^{-10}}{4\times\pi\times\left(1.8\right)^{2}\times6.63\times10^{-34}\times3\times10^{8}}$
$=1.22\times10^{19} \,m^{-2}\,s^{-1} $