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

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 ?

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 = 2.892 xx 10^-3 mk ?

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

The temperature of one of the two heated black bodies is T_(1) = 2500K . Find the temperature of the other body if the wavelength corresponding to its maximum emissive capacity exceeds by Delta lambda = 0.50mu m the wavelength corresponding to the maximum emissive capacity of the first black body.

When the temperature of a black body increases, it is observed that the wavelength corresponding to maximum energy changes from 0.26 mu m . The ratio of the emissive powers of the body at the respective temperature is:

The maximum radiations from two bodies correspond to 560 nm and 420 nm respectively. The ratio of their temperature is:

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

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