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Use the formula lambda_(m)T=0.29cmK to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you? |
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Answer» Solution :According to Wien.s displacement law, for a black body maintained at a temperature T K the WAVELENGTH `lambda_(m)` corresponding to maximum INTENSITY of radiation is given by the relation `lambda_(m)T=0.29cmK=0.29xx10^(-2)mK=2.9xx10^(-3)mK` `rArr""T=(2.9xx10^(-3))/(lambda_(m)" (in m)")K` If we consider `lambda_(m)=1mum=10^(-6)m," then T"=(2.9xx10^(3))/(10^(-6))=2900K` Similarly, temperatures for other VALUES of wavelength may also be found. These figures tells us the temperature range required for obtaining radiations of different regions of electromagnetic spectrum. As an EXAMPLE, if we example, if we want to obtain visible LIGHT of `6000Å=6xx10^(-7)m`, then `T=(2.9xx10^(3))/(6xx10^(-7))=4833K` |
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