Question

A $100W$ sodium lamp is radiating light of wavelength $5890\stackrel{o}{A}$, uniformly in all directions. At what distance from the lamp, the average flux is 1 photon $(c{m}^2-s{)}^{-1}$?

• A. 48600 km
• B. 48860 km
• C. 50000 km
• D. 58900 km

Answer 1

The correct option is B 48860 km

Power of the beam $P=100W$

Wavelength $\lambda =5890A^o=5890\times {10}^{-10}m$

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

Number of photons absorbed per second $N=\frac{P}{E}=\frac{P\lambda }{hc}$

$\therefore$ $N=\frac{100\times 5890\times {10}^{-10}}{6.62\times {10}^{-34}\times 3\times {10}^8}=3\times {10}^{20}$ photons per second

Photon flux at a distance $r$ is ${\varphi }_{flux}=\frac{N}{4\pi r^2}$

where ${\varphi }_{flux}=1photon/c{m}^2/s={10}^{4}photon/m^2/s$

$\therefore$ ${10}^4=\frac{3\times {10}^{20}}{4\pi r^2}$

We get $r^2=0.239\times {10}^{16}$

$⟹r=0.48860\times {10}^{8}m=48860km$