Answer
(a) $15.88\times 10^{10}Hz$
(b) $1.9mm$
Work Step by Step
(a) We can find the peak frequency as
$f_{peak}=(5.88\times 10^{10}s^{-1}K^{-1})(2.7K)$
$f_{peak}=15.88\times 10^{10}Hz$
(b) The required wavelength can be determined as
$\lambda=\frac{v}{f}$
We plug in the known values to obtain:
$\lambda=\frac{3\times 10^8}{15.88\times 10^{10}}$
$\lambda=1.9\times 10^{-3}m=1.9mm$
