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

Answer

C

Explanation

Efficient power `P=(N)/(t)xx(hc)/(lamda)=200xx0.25`
On Solving `(N)/(t)=50xx(lamda)/(hc)=1.5xx10^(20)`