A 200 W sodium street lamp emits yellow light of wavelength `0.6 mu m`. Assuming it to be 25% efficient in coverting 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

Effective power `=(25)/(100)xx200W`
`=50W`
Now, `50=nhv=(nhc)/(lamda)`
`n=(50lamda)/(hc)`
`=(50xx0.6xx10^(-6))/(6.6xx10^(-34)xx3xx10^(8))=1.5xx10^(20)`