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Power radiated by a back body is P0 and ...

Power radiated by a back body is `P_0` and the wavelength corresponding to maximum energy is around `lamda_0` . On changing the temperature of the back body its was observed that the power radiated becomes `(256)/(81)P_0` . The shift wavelength corresponding to the the maximum energy will be

A

`+(lambda_(0))/(4)`

B

`+(lambda_(0))/(2)`

C

`-(lambda_(0))/(4)`

D

`-(lambda_(0))/(2)`

Text Solution

Verified by Experts

According to Wien's law, `lambda_(0)T_(0)=lambdaT`……….`(1)`
According to Stefan's law, `(P_(0))/(P)=((T_(0))/(T))^(4)`……`(2)`
Now, `(P)/(P_(0))=(256)/(81)=((4)/(3))^(4)`
`:. (T_(0))/(T)=(3)/(4)=(lambda)/(lambda_(0))`,
or `lambda=(3)/(4)lambda_(0)`
Wavelength shift `Deltalambda=lambda-lambda_(0)=`
`(3)/(4)lambda_(0)-lambda_(0)=-(lambda_(0))/(4)`
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