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
                      `underline(overline(3).9735)`
antilog  `overline(3).9735` = 9.408 × 10⁻³

`λ_m = b /T = (2.898 xx 10^-3)/308`

= 9.408 × 10⁻⁶ m

= 9.408 μm