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If the filament of a 100 W bulb has an a...

If the filament of a 100 W bulb has an area `0.25cm^2` and behaves as a perfect black body. Find the wavelength corresponding to the maximum in its energy distribution. Given that Stefan's constant is `sigma=5.67xx10^(-8) J//m^(2)s K^(4)`.

Text Solution

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In the bulb filament given, the energy radiated per second `m^2` of its surface area is given as
`E=(P)/(A)=(100)/(0.25xx10^-4)=4xx10^6 J//s-m^(2)`
If `T` is the temperature of the filament then according to Stefan's law we have
`E=sigmaT^4`
or `4xx10^6=5.67xx10^-8xxT^4`
or `T^4=(4xx10^6)/(5.67xx10^-8)=7.055xx10^(13)`
or `T[7.055xx10^(13)]^(1//4)=2898.14K`
If the filament radiates the maximum energy at a wavelength `lamda_m`, from Wien's displacement law, we have
`lamda_mT=b`
or `lamda_m=(b)/(T)`
`=(2.89xx10^-3)/(2898.14)=9971.9Å`
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