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

Power radiated by a black body is `P_0` and the wavelength corresponding to maximum energy is around `lamda_0`, On changing the temperature of the black body, it was observed that the power radiated becames `(256)/(81)P_0`. The shift in wavelength corresponding to 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

The correct Answer is:
C
`lambda_(m)T =` const and `P = sigmaAT^(4), Deltalambda = lambda - lambda_(0)`
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