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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 beA. `(lambda_(0))/(4)`B. `(lambda_(0))/(2)`C. `(lambda_(0))/(4)`D. `(lambda_(0))/(2)` |
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Answer» Correct Answer - C `lambda_(m)T =` const and `P = sigmaAT^(4), Deltalambda = lambda - lambda_(0)` |
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