Question

An isotropic point source emits light with wavelength $500nm$. The radiation power of the source is $P=10W$. Find the number of photons passing through unit area per second at a distance of $3m$ from the source.

• A. $5.92\times {10}^{17}/m^{2}s$
• B. $2.23\times {10}^{17}/m^{2}s$
• C. $2.23\times {10}^{18}/m^{2}s$
• D. $5.92\times {10}^{18}/m^{2}s$

Answer 1

The correct option is B $2.23\times {10}^{17}/m^{2}s$

Given, $\lambda =500nm=500\times {10}^{-9}m$
As, $P=n_0\frac{hc}{\lambda }$
$n_0=\frac{P\lambda }{hc}$ .....(i)
where, $n_0$ is number of photons per second
At distance $r$ from point source, number of photons / area / time
$n^{\prime }=\frac{n_0}{4\pi r^2}=\frac{P\lambda }{hc\cdot 4\pi r^2}$ [from equation (i)]
$=\frac{10\times 500\times {10}^{-9}}{6.6\times {10}^{-34}\times 3\times {10}^8\times 4\pi {\left(3\right)}^2}$
$=2.23\times {10}^{17}/m^{2}s$