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

Answer» Power of the bulb = 100 watt `= 100 Js^(-1)`
Energy of one photon
`= 4.969 xx 10^(-19)J`
Number of photons emitted
`=(100 Js^(-1))/(4.969 xx 10^(-19)J) = 2.012 xx 10^(20) s^(-1)`


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