Answer
$E = (4.28\times 10^{-10}~MeV/K)~T$
Work Step by Step
We can find an expression for the energy of a photon:
$E = \frac{hc}{\lambda_{max}}$
$E = \frac{hc}{(2898~\mu m~K)/T}$
$E = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(2898~\mu m~K)}~T$
$E = (6.859\times 10^{-23}~J/K)~T$
$E = (6.859\times 10^{-23}~J/K)(\frac{1~eV}{1.6\times 10^{-19}~J})~T$
$E = (4.28\times 10^{-4}~eV/K)~T$
$E = (4.28\times 10^{-10}~MeV/K)~T$
