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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. |
Answer» For the first black body `(lambda_(min))_(1) = (b)/(T_(1))` Then `(lambda_(min))_(2) = (b)/(T_(1)) + Delta lambda = (b)/(T_(2))` Hence `T_(2) = (b)/((b)/(T_(1))+Delta lambda) = (bT_(1))/(b+T_(1)Delta lambda) = 1.747kK` |
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