A 100 watt buble emits monochromatic light of wavelength 400 nm. Then the number of photons emitted per seccond by the buble is nearly -

Options

A. `2.012 xx 10^(20)`
B. `3.475xx10^(20)`
C. `7.860xx10^(20)`
D. `5.786xx10^(20)`

Answer

A

Explanation

Number of photons emitted per second
`= ("Energy emitted per second, i.e., power of the bulb")/("Enegry of one photon")`
Power of the bulb `= 100 W = 100 Js^(-1)`
Enegry of one photon,
`E = hv`
`=((6.626 xx10^(-34)Js)(3xx10^(8)ms^(-1)))/((400xx10^(-9)m))`
`= 4.969 xx 10^(-19)J`
Thus, the number of photons emitted per second
`= (100J s^(-1))/(4.969xx10^(-19)m) = 2.92xx10^(20) Js^(-1)`