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. `3 xx 10^(19)`
B. `1.5 xx 10^(20)`
C. `6 xx 10^(18)`
D. `62 xx 10^(20)`

Answer

B

Explanation

Energy of a photon ` = (hc )/(lambda)`
The number of photons emitted per second ` = (N)/(t)`
Hence power emitted `p = (N)/(t) ((hc)/(lambda))`
It is given that sodium lamp has `25%` efficient in converting electricial energy to light energy emitted
`P = 200 xx (25)/(100) = 50 W = 50 J//sec`
No. of photons `(N)/(t) = (P lambda)/(lambda)`
` = (50 xx 0.6 xx 10^(-6))/(6.6 xx 10^(-34) xx 3 xx 10^(8)) = 1.5 xx 10^(20)`