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

• A

  $5.92 \times 10^{17} / \mathrm{m}^2 \mathrm{~s}$
• Correct answer
  B

  $2.23 \times 10^{17} / \mathrm{m}^2 \mathrm{~s}$
• C

  $2.23 \times 10^{18} / \mathrm{m}^2 \mathrm{~s}$
• D

  $5.92 \times 10^{18} / \mathrm{m}^2 \mathrm{~s}$

Solution

Given, $\lambda=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}$, As, $\begin{aligned} P & =n_0 \frac{h c}{\lambda} \\ n_0 & =\frac{P \lambda}{h c}...(i)\end{aligned}$ where, $n_0$ is number of photons per second At distance $r$ from point source, number of photons/area/time $n^{\prime}=\frac{n_0}{4 \pi r^2}=\frac{P \lambda}{h c \cdot 4 \pi r^2} \quad[$ from Eq. (i)] $\begin{aligned} & =\frac{10 \times 500 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8 \times 4 \pi(3)^2} \\ & =2.23 \times 10^{17} / \mathrm{m}^2 . \mathrm{s}\end{aligned}$

Hence, the correct answer is option: 'B'.