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A 25-watt bulb emits monochromatic yellow light of the wavelength of 0.57μm. Calculate the rate of emission of quanta per second.

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#### Solution

Power of bulb, P = 25 Watt = 25 Js^{–1}

Energy of one photon, E = hν = `"hc"/lambda`

Substituting the values in the given expression of E:

`"E" = ((6.626 xx 10^(-34))(3xx10^8))/(0.57xx10^(-6)) = 34..87xx10^(-20) "J"`

E = 34.87 × 10^{–20} J

Rate of emission of quanta per second

`= 25/(34.87 xx 10^(-20)) = 7.169 xx 10^(19) "s"^(-1)`

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