The wavelength corresponding to maximum spectral radiancy of a black body A is lambda_(A) =5000A .Consider another black body B whose surface area is twice that of A and total radiant energy by B is 16 times that emitted by A .The wavelength corresponding to maximum spectrum radiancy for B will be

A power radiated by a black body is P_0 . and the wavelength corresponding to the maximum energy is around lambda . On changing the temperature of the black body, It was observed that the power radiated is increased to 256/81P_0 . The shift in the wavelength corresponding to the maximum energy will be

If the temperature of a black body is increased then the wavelength corresponding to the maximum emission will

The power radiated by a blackbody is P at a certain temperature. If power radiated by this body becomes 1/16 th the times. then wavelength corresponding to maximum intensity becomes

When the temperature of black body rises, then the wavelength corresponding to the maximum intensity (lamda_(m))

The absolute temperatures of two black bodies are 2000 K and 3000 K respectively. The ratio of wavelengths corresponding to maximum emission of radiation by them will be

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

The surface temperature of a black body is 1200 K. What is the wavelength corresponding to maximum intensity of emission of radiation if Wien's constant b = 2.892xx10^(-3)mK ?