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 step by step, we will break it down into two parts as per the question.

### Given Data:
- Emissivity of body A, \( \epsilon_A = 0.01 \)
- Emissivity of body B, \( \epsilon_B = 0.81 \)
- Temperature of body A, \( T_A = 5802 \, K \)
- Wavelength difference, \( \Delta \lambda = \lambda_B - \lambda_A = 1 \, \mu m \)

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