To find the total average energy density of an electromagnetic wave, we must consider both the electric field and the magnetic field components. For a plane electromagnetic wave, the energy density can be given by:
\(u = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \mu_0 H^2\)
However, in a vacuum, the energy density contributed by the electric field is equal to that contributed by the magnetic field. Thus, the total average energy density can be defined using only the electric field:
\(u = \epsilon_0 E^2\)
Given:
Substituting the values, the average energy density is calculated as follows:
\(u = \frac{1}{2} \times 8.85 \times 10^{-12} \times (50)^2\)
We'll proceed with the calculations:
\(u = \frac{1}{2} \times 8.85 \times 10^{-12} \times 2500\)
\(u = \frac{1}{2} \times 2.2125 \times 10^{-8}\)
Finally:
\(u = 1.106 \times 10^{-8} \, \text{Jm}^{-3}\)
This matches the given correct answer. Therefore, the total average energy density is:
Hence, the correct option is:
\(< 1.106 \times 10^{-8} \, \text{Jm}^{-3} \)
The average energy density of the electric field is given by:
\[ U_E = \frac{1}{2} \epsilon_0 E^2 \]
Substituting the given values:
\[ U_E = \frac{1}{2} \times 8.85 \times 10^{-12} \times (50)^2 \]
Calculating:
\[ U_E = 1.106 \times 10^{-8} \, \text{Jm}^{-3} \]
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)