Step 1: Use Compton shift formula.
\[
\Delta \lambda = \lambda_C (1 - \cos\theta)
\]
The longest wavelength occurs at \(\theta = 180^\circ\):
\[
\Delta \lambda_{\max} = 2\lambda_C = 2(0.024) = 0.048\, \text{Å}
\]
\[
\lambda'_{\max} = \lambda + \Delta\lambda = 0.500 + 0.048 = 0.548\, \text{Å}
\]
Step 2: Conclusion.
Hence, the longest possible scattered wavelength = 0.548 Å.

At a particular temperature T, Planck's energy density of black body radiation in terms of frequency is \(\rho_T(\nu) = 8 \times 10^{-18} \text{ J/m}^3 \text{ Hz}^{-1}\) at \(\nu = 3 \times 10^{14}\) Hz. Then Planck's energy density \(\rho_T(\lambda)\) at the corresponding wavelength (\(\lambda\)) has the value \rule{1cm}{0.15mm} \(\times 10^2 \text{ J/m}^4\). (in integer)
[Speed of light \(c = 3 \times 10^8\) m/s]
(Note: The unit for \(\rho_T(\nu)\) in the original problem was given as J/m³, which is dimensionally incorrect for a spectral density. The correct unit J/(m³·Hz) or J·s/m³ is used here for the solution.)