Power emitted per unit area of a black body is `R " watt/m"^(2)`. At what wavelength will the power radiated by black body be maximum is stefan's constant is `sigma` and Wien's constant is b then.

Options

A. `lambda = ((bsigma)/(R ))^((1)/(4))`
B. `lambda =((b)/(R ))^((1)/(4))`
C. `((bR)/(sigma ))^((1)/(4))=lambda`
D. `lambda=b ((sigma)/(R ))^((1)/(4))`

Explanation

To solve the problem of finding the wavelength at which the power radiated by a black body is maximum, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to find the wavelength (λ_max) at which the power emitted per unit area of a black body is maximum, given the power emitted (R), Stefan's constant (σ), and Wien's constant (b).

2. **Use Stefan-Boltzmann Law**: According to the Stefan-Boltzmann law, the power radiated per unit area (P) of a black body is given by:
\[ P = \sigma T^4 \]
where T is the absolute temperature of the black body.

3. **Relate Temperature to Power**: Rearranging the equation gives us the temperature in terms of power:
\[ T = \left(\frac{R}{\sigma}\right)^{1/4} \]

4. **Apply Wien's Displacement Law**: Wien's displacement law states that the product of the temperature (T) and the wavelength at which the emission is maximum (λ_max) is a constant:
\[ T \cdot \lambda_{\text{max}} = b \]
where b is Wien's constant.

5. **Substitute for Temperature**: Substitute the expression for temperature (T) from step 3 into the equation from step 4:
\[ \left(\frac{R}{\sigma}\right)^{1/4} \cdot \lambda_{\text{max}} = b \]

6. **Solve for λ_max**: Rearranging the equation to solve for λ_max gives:
\[ \lambda_{\text{max}} = \frac{b \cdot \sigma^{1/4}}{R^{1/4}} \]

### Final Expression:
Thus, the wavelength at which the power radiated by the black body is maximum is given by:
\[ \lambda_{\text{max}} = b \cdot \frac{\sigma^{1/4}}{R^{1/4}} \]

### Conclusion:
After analyzing the options provided in the question, we can conclude that option D is the correct answer.