Question

A 200W sodium street lamp emits yellow light of wavelength $0.6\mu m$. Assuming it to be 25% efficient converting electrical energy to light, the number of photons of yellow light it emits per second is :

• A. $62\times {10}^{20}$
• B. $3\times {10}^{19}$
• C. $1.5\times {10}^{20}$
• D. $6\times {10}^{18}$

Answer 1

The correct option is C $1.5\times {10}^{20}$

The number of photons per sec $,n=$ amount of power converted to light / energy of photon

Power converted to light $P_l=200\left(25/100\right)=50W$

Energy of photon, $E_p=\frac{hc}{\lambda }=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{0.6\times {10}^{-6}}=33.1\times {10}^{-20}J$

Thus, $n=\frac{50}{33.1\times {10}^{20}}=1.5\times {10}^{20}$