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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 block 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

Verified by Experts

In the bulb filament given, the energy radiated per sec per `m^(2)` of its surface area is given as
`E=(P)/(A)=(100)/(0.25 xx 10^(-4))=4 xx 10^(6) J//s m^(2)`
If T is the temperature of the filament then according to stefen's law, we have
`E= sigmaT^(4)`
or `4xx 10^(5) = 5.67 xx 10^(-8) xT^(4)`
or `T^(4) =(4 xx 10^(6))/(5.67 xx 10^(-8)) = 7.055 xx 10^(13)`
or `T=[7.055 xx 10^(13)]^(14)` = 2898.14K
If the filament radiates the maximum energy at a wavelength `lambda_(m)`, from Wein's displacement law, we have
`lambda_(m) T=b`
or `lambda_(m) =(b)/(T)` =`(2.89 xx 10^(-3))/(2898.14) = 9971.9 dotA`
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Knowledge Check

  • Assuming a filament in a 100W light bulb acts like a perfect blackbody, what is the temperature of the hottest portion of the filament if it has a surface area of 6.3 xx 10^(-5) m^(2) ? The Stefan-Boltzmann constant is 5.67 xx 10^(-8) W //(m^(2).K^(2)) .

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    B
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    5800K

  • Calculate the temperature at which a perfect black body radiates at the rate of 1 W cm^(-2) , value of Stefan's constant, sigma = 5.67 xx 10^(-8) W m^(-2)K^(-4)

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    576 K
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  • Calculate the temperature at which a perfect black body radiates at the rate of 1 W cm^(-2) , value of Stefan's constant, sigma = 5.67 xx 10^(-5) W m^(-2) K^(-8)

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