Step 1: Understanding the Concept:
Wien's Displacement Law states that the wavelength corresponding to maximum spectral emissive power of a black body is inversely proportional to its absolute temperature.
Therefore, a hotter body will emit its maximum energy at a shorter wavelength.
Step 2: Key Formula or Approach:
Wien's Displacement Law is expressed as:
\[ \lambda_m \cdot T = \text{constant} (b) \]
From this, we can relate two bodies A and B:
\[ \lambda_A \cdot T_A = \lambda_B \cdot T_B \implies \frac{\lambda_A}{\lambda_B} = \frac{T_B}{T_A} \]
Step 3: Detailed Explanation:
We are given that the absolute temperature of body A is 3 times that of B:
\[ T_A = 3 T_B \]
Using the relation from Wien's law:
\[ \frac{\lambda_A}{\lambda_B} = \frac{T_B}{3 T_B} = \frac{1}{3} \]
This means $\lambda_A = \frac{\lambda_B}{3}$.
Since $T_A>T_B$, body A must have a shorter peak wavelength ($\lambda_A<\lambda_B$).
The difference in their wavelengths is given as $4\mu \text{m}$. We set up the equation for the difference:
\[ \lambda_B - \lambda_A = 4\mu \text{m} \]
Substitute $\lambda_A = \frac{\lambda_B}{3}$ into the difference equation:
\[ \lambda_B - \frac{\lambda_B}{3} = 4 \]
\[ \frac{3\lambda_B - \lambda_B}{3} = 4 \]
\[ \frac{2\lambda_B}{3} = 4 \]
Solve for $\lambda_B$:
\[ 2\lambda_B = 12 \]
\[ \lambda_B = 6 \mu \text{m} \]
So, the wavelength at which body B radiates maximum energy is $6\mu \text{m}$.
(For completeness, $\lambda_A = 6/3 = 2\mu \text{m}$, and the difference is indeed $6 - 2 = 4\mu \text{m}$).
Step 4: Final Answer:
The wavelength at which body $B$ radiates maximum energy is $6\mu \text{ m}$.