A modern 200 W sodium street lamp emits yellow light of wavelength 0.6 `mum`. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is

Options

A. `1.5 xx10^(20)`
B. `6 xx10^(18)`
C. `62 xx10^(20)`
D. `3 xx10^(19)`

Answer

A

Explanation

Power (P) `=200 xx25/100= 50 ` Watt
Number of photon `=P/E = (Plamda)/(hc)`
`=(50 xx0.6xx10^(-6))/(6.6 xx10^(-34) xx3 xx10^8)=1.5 xx10^(20)`