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

Answer

a

Explanation

Here, `P=200W, lambda=0.6 mum=0.6xx10^(-6)`,
Energy converted to light `=25%`
Energy of one photon of yellow light `=(hc)/lambda`
`=((6.6xx10^(-34))xx(3xx10^(8)))/(0.6xx10^(-6))=33xx10^(-20)J`
Energy radiated per second as yellow light
`=200xx25/100=50"watt"`
Number of photons of yellow light emitted per second `=50/(33xx10^(-20))=1.7xx10^(20)`
The closest value is `1.5xx10^(20)`