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Use the formula lambda_(m)T=0.29 cm K 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 :We know , every body at given temperature `T_(1)` emits radiations of all wavelengths in CERTAIN range .For a black body, the wavelength corresponding to maximum intensity of radiation at a given temperature . `lambda_(m)T=0.29 CMK ` or `T=(0.29)/(lambda_(m))` For `lambda_(m)=10^(-6)m =10^(-4) cm , T=(0.29)/(10^(-4))=2900 K` Temperature for other wavelengths can be SIMILARLY found. These NUMBERS tell us the temperature ranges required for OBTAINED radiations in different parts of e.m spectrum . Thus to obtain visible radiation, say, `lambda_(m)=5xx10^(-5)` cm, the source should have temperature `T=(0.29)/(5xx10^(-5))=6000 K` It is to be noted that , a body at lower temperature willalso proudce this wavelength but not with maximum intensity. |
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