Question

A 100 W point source emits monochromatic light of wavelength $6000\stackrel{o}{A}$. Calculate the photon flux (in SI unit) at a distance of 5 m from the source. Given $h=6.6\times {10}^{34}$ J s and $C=3\times {10}^{8}m{s}^{-1}$

• A. ${10}^{15}$
• B. ${10}^{18}$
• C. ${10}^{20}$
• D. ${10}^{22}$

Answer 1

Answer: (B)

${10}^{18}$

The photon flux is defined as the number of photons per second per unit area.

Thus, photon flux $\varphi =$ number of photons per second / area $=\frac{n}{A}$

Energy of photon $=\frac{hc}{\lambda }=\frac{(6.62\times {10}^{-34})(3\times {10}^8)}{6000\times {10}^{-10}}=3.31\times {10}^{-19}J$

Given, the energy per second is $100J$

Thus number of photons per sec, $n=\frac{100}{3.31\times {10}^{-19}}=3.02\times {10}^{20}$

Area $A=4\pi \times 5^2=314m^2$

Thus, $\varphi =\frac{3.02\times {10}^{20}}{314}=9.6\times {10}^{17}\sim {10}^{18}$