The figure below shows a cubic unit cell with lattice constant \(a\). The shaded crystallographic plane intersects the x-axis at 0.5a. The Miller indices of the shaded plane are: 
(\(\={2}10\))
Step 1: Identify intercepts.
Given that the shaded plane cuts the x-axis at \(0.5a\) and is parallel to the y-axis, it does not intersect the y-axis (intercept is ∞). It also cuts the z-axis at \(a\).
Step 2: Express intercepts in terms of \(a\).
Intercepts = (0.5a, ∞, a).
Step 3: Take reciprocals of the fractional intercepts.
\[ \text{Reciprocals:} \ (2, 0, 1) \] Since the plane cuts the x-axis on the negative side (at -0.5a), we represent it as \(\={2}10\).
Step 4: Conclusion.
The Miller indices of the shaded plane are \(\={2}10\).

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.)