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