Question

What is the peak wavelength of the radiation emitted by a blackbody at 35° C?

Answer 1

Data: t = 35°C, b = 2.898 × 10⁻³ m.K

The absolute temperature of the blackbody,

T = t + 273

= 35 + 273

= 308 K

By Wien’s displacement law, λ_m T = b

∴ The peak wavelength,

log 2.898    =  0.4621
log 308     = − 2.4886
                      ${\displaystyle \underline{\overline{3}.9735}}$
antilog  ${\displaystyle \overline{3}.9735}$ = 9.408 × 10⁻³

${\displaystyle {\lambda }_m=\frac{b}{T}=\frac{2.898\times {10}^{-3}}{308}}$

= 9.408 × 10⁻⁶ m

= 9.408 μm