Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are same. The two bodies emit total radiant power at the same rate. The wavelength `lambda_B` corresponding to maximum spectral radiancy from B is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A by 1.0 `mum`. If the temperature of A is 5802 K, calculate (a) the temperature of B, (b) wavelength `lambda_B`.

Explanation

To solve the problem, we need to follow a systematic approach using the information provided about the thermal emissivities, temperatures, and the relationship between the wavelengths of the two bodies.

### Step-by-Step Solution:

**Step 1: Set up the equation for power emitted.**

Given that the two bodies emit total radiant power at the same rate, we can express this using the Stefan-Boltzmann law:
\[ E_A \cdot A \cdot T_A^4 = E_B \cdot A \cdot T_B^4 \]
Since the areas \(A\) are the same, we can cancel them out:
\[ E_A \cdot T_A^4 = E_B \cdot T_B^4 \]

**Step 2: Substitute the emissivity values.**

From the problem, we know:
- \(E_A = 0.01\)
- \(E_B = 0.81\)
- \(T_A = 5802 \, K\)

Substituting these values into the equation gives:
\[ 0.01 \cdot (5802)^4 = 0.81 \cdot T_B^4 \]

**Step 3: Rearrange to find \(T_B\).**

Rearranging the equation to solve for \(T_B\):
\[ T_B^4 = \frac{0.01 \cdot (5802)^4}{0.81} \]

**Step 4: Calculate \(T_B\).**

Calculating \(5802^4\):
\[ 5802^4 = 11316421296016 \]
Now substituting this value back into the equation:
\[ T_B^4 = \frac{0.01 \cdot 11316421296016}{0.81} = \frac{113164212960.16}{0.81} \approx 139,000,000,000 \]
Taking the fourth root to find \(T_B\):
\[ T_B = \left(139000000000\right)^{1/4} \approx 1934 \, K \]

**Step 5: Calculate the wavelength \(\lambda_B\) using Wien's Displacement Law.**

Wien's Displacement Law states:
\[ \lambda_{\text{max}} = \frac{b}{T} \]
where \(b\) is Wien's displacement constant, approximately \(0.3 \, m \cdot K\).

Substituting \(T_B\) into the equation:
\[ \lambda_B = \frac{0.3 \, m \cdot K}{1934 \, K} \approx 0.000155 \, m = 1.55 \, \mu m \]

**Step 6: Adjust for the shift in wavelength.**

The problem states that the wavelength \(\lambda_B\) is shifted from \(\lambda_A\) by \(1.0 \, \mu m\):
\[ \lambda_A = \lambda_B + 1.0 \, \mu m = 1.55 \, \mu m + 1.0 \, \mu m = 2.55 \, \mu m \]

### Final Answers:
- (a) The temperature of body B \(T_B\) is approximately \(1934 \, K\).
- (b) The wavelength \(\lambda_B\) is approximately \(1.55 \, \mu m\).