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The spectral emissive power `E_(lambda)` for a body at temperature `T_(1)` is poltted againist the wavelenght and area under the curve is found to be `A.At` a different temperature `T_(2)` the area is found to be A then `lambda_(1)//lambda_(2)=` .A. 3B. `1//3`C. `1//sqrt3`D. `sqrt3` |
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Answer» Correct Answer - D Area of graph `=E` `E alpha T^(4)` and `lambda alpha (1)/(T) rArr E alpha (1)/(lambda^(4))` Area `alpha (1)/(lambda^(4))` . |
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