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

Answer

B

Explanation

Efficient power `=25/100xx200 = 50W`
Let n photons be emitted per second
Energy of n photons , `E=(nhc)/lamda`
`implies50=(nxx6.63xx10^(-34)xx3xx10^8)/(0.6xx10^(-6))`
`impliesn = 1.5 xx10^(20)`