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At a given temperature, [tex]\lambda_{max}[/tex] for a blackbody cavity = 6500 angstroms. What will [tex]\lambda_{max}[/tex] be if the temperature of the cavity walls is increased so that the rate of emission of spectral radiation is doubled?
[tex] R_T = \sigma T^4 \,\,\,\, \Rightarrow \,\,\,\,T^4 = \frac{{R_T }}{\sigma }\,\,\,\, \Rightarrow \,\,\,\,T = \sqrt[4]{{\frac{{R_T }}{\sigma }}}\,\,\,[/tex]
[tex]\lambda _{{\rm{max}}} = \frac{\alpha }{T}[/tex]
[tex]\lambda _{{\rm{max,2}}} = \frac{\alpha }{{T_2 }} = \frac{\alpha }{{\sqrt[4]{{\frac{{R_{T,2} }}{\sigma }}}}} = \frac{\alpha }{{\sqrt[4]{{\frac{{2R_{T,1} }}{\sigma }}}}}[/tex]
Just looking at the formula, it seems the answer should be [tex]\frac{1}{{\sqrt[4]{2}}} = 0.84\,\lambda _{{\rm{max,1}}}[/tex]
But shouldn't the max wavelength go up if the temperature is going up?
[tex] R_T = \sigma T^4 \,\,\,\, \Rightarrow \,\,\,\,T^4 = \frac{{R_T }}{\sigma }\,\,\,\, \Rightarrow \,\,\,\,T = \sqrt[4]{{\frac{{R_T }}{\sigma }}}\,\,\,[/tex]
[tex]\lambda _{{\rm{max}}} = \frac{\alpha }{T}[/tex]
[tex]\lambda _{{\rm{max,2}}} = \frac{\alpha }{{T_2 }} = \frac{\alpha }{{\sqrt[4]{{\frac{{R_{T,2} }}{\sigma }}}}} = \frac{\alpha }{{\sqrt[4]{{\frac{{2R_{T,1} }}{\sigma }}}}}[/tex]
Just looking at the formula, it seems the answer should be [tex]\frac{1}{{\sqrt[4]{2}}} = 0.84\,\lambda _{{\rm{max,1}}}[/tex]
But shouldn't the max wavelength go up if the temperature is going up?
