Question

The spectral emissive power Eλ for a body at temperature T1 is plotted against the wavelength and area under the curve is found to be A. At a different temperature T2 the area is found to be 9A. Then λ1/λ2 =

• A. 3
• B. 1/3
• C. 1/√3
• D. √3

Answer 1

Answer: (D)

√3

Answer 2

The area under the spectral emissive power curve $E_\lambda$ versus wavelength $\lambda$ represents the total emissive power $E$ of the body. According to Stefan-Boltzmann's law, the total emissive power of a black body is proportional to the fourth power of its absolute temperature: $E = \sigma T^4$.

Given that the area at temperature $T_1$ is $A$ and at temperature $T_2$ is $9A$, we can write: $E_1 = A = \sigma T_1^4$ $E_2 = 9A = \sigma T_2^4$

Dividing the second equation by the first: $\frac{\sigma T_2^4}{\sigma T_1^4} = \frac{9A}{A}$ $\left(\frac{T_2}{T_1}\right)^4 = 9$ Taking the fourth root of both sides: $\frac{T_2}{T_1} = 9^{1/4} = (3^2)^{1/4} = 3^{1/2} = \sqrt{3}$

Wien's displacement law states that the wavelength at which the spectral emissive power is maximum ($\lambda_{max}$) is inversely proportional to the absolute temperature: $\lambda_{max} T = b$, where $b$ is Wien's displacement constant.

Assuming $\lambda_1$ and $\lambda_2$ refer to the peak wavelengths at temperatures $T_1$ and $T_2$ respectively: $\lambda_1 T_1 = b$ $\lambda_2 T_2 = b$

Equating these two expressions: $\lambda_1 T_1 = \lambda_2 T_2$

Rearranging to find the ratio $\lambda_1/\lambda_2$: $\frac{\lambda_1}{\lambda_2} = \frac{T_2}{T_1}$

Substituting the ratio of temperatures we found: $\frac{\lambda_1}{\lambda_2} = \sqrt{3}$