Why does λ_max decrease when blackbody temperature increases?

[tony873004]

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?

$$R_T = \sigma T^4 \,\,\,\, \Rightarrow \,\,\,\,T^4 = \frac{{R_T }}{\sigma }\,\,\,\, \Rightarrow \,\,\,\,T = \sqrt[4]{{\frac{{R_T }}{\sigma }}}\,\,\,$$

$$\lambda _{{\rm{max}}} = \frac{\alpha }{T}$$

$$\lambda _{{\rm{max,2}}} = \frac{\alpha }{{T_2 }} = \frac{\alpha }{{\sqrt[4]{{\frac{{R_{T,2} }}{\sigma }}}}} = \frac{\alpha }{{\sqrt[4]{{\frac{{2R_{T,1} }}{\sigma }}}}}$$

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?

 

[tony873004]

It just dawned on me. Shoudn't the frequency to up, causing the wavelength to shorten, in which case, I might have done it correctly? Can someone double check that I did this right?

 

[gabbagabbahey]

It just dawned on me. Shoudn't the frequency to up, causing the wavelength to shorten, in which case, I might have done it correctly? Can someone double check that I did this right?

Yes, higher energies mean higher frequencies and shorter wavelengths; it looks good to me