-450
From the exponential term \( e^{i(\omega t + ax - 600y)} \), the wave vector \( \mathbf{k} \) is: \[ \mathbf{k} = a\hat{x} - 600\hat{y}. \] The direction of propagation of the electromagnetic wave is given by the wave vector \( \mathbf{k} \), and it must be perpendicular to the direction of the electric field \( \mathbf{E} \) because \( \mathbf{E} \cdot \mathbf{k} = 0 \) (transverse wave condition).
The electric field is: \[ \mathbf{E} = 4\hat{x} + 3\hat{y}. \] The dot product \( \mathbf{E} \cdot \mathbf{k} = 0 \) gives: \[ (4)(a) + (3)(-600) = 0. \] Simplify: \[ 4a - 1800 = 0. \] Solve for \( a \): \[ a = \frac{1800}{4} = 450. \]
The value of \( a \) is: 450.
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/mΒ³, which is dimensionally incorrect for a spectral density. The correct unit J/(mΒ³Β·Hz) or JΒ·s/mΒ³ is used here for the solution.)