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The maximum energy in thermal radiation from a source occurs at the wavelength 4000Å. The effective temperature of the sourceA. 7000 KB. 80000 KC. `10^(4)K`D. `10^(6)K` |
Answer» Correct Answer - A `lamda_(m)=(b)/(T) " "therefore T=(b)/(lamda_(m))=(2.93xx10^(-3))/(4000xx10^(-10))=7325K`. |
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