The black discs x, y and z have radii 1 m, 2 | vedclass

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(A) The emissive power $E$ of a black body is given by the Stefan-Boltzmann law: $E = \sigma T^4$. However,the total power radiated by a surface is $P

Vedclass · Accepted Answer

$E_{x} > E_{y} > E_{z}$

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(A) The emissive power $E$ of a black body is given by the Stefan-Boltzmann law: $E = \sigma T^4$. However,the total power radiated by a surface is $P = A \sigma T^4$. Assuming the question asks for the total power radiated (often referred to as emissive power in this context): $P \propto A T^4$. Since $A = \pi R^2$,we have $A \propto R^2$. Given radii $R_x = 1 \ m, R_y = 2 \ m, R_z = 3 \ m$,the areas are in ratio $A_x : A_y : A_z = 1^2 : 2^2 : 3^2 = 1 : 4 : 9$. By Wien's displacement law,$\lambda_{max} T = b$ (constant),so $T \propto \frac{1}{\lambda_{max}}$. Given $\lambda_{max,x} = 200 \ nm, \lambda_{max,y} = 300 \ nm, \lambda_{max,z} = 400 \ nm$,the temperatures are in ratio $T_x : T_y : T_z = \frac{1}{200} : \frac{1}{300} : \frac{1}{400} = \frac{1}{2} : \frac{1}{3} : \frac{1}{4} = 6 : 4 : 3$. Now,calculating $P \propto A T^4$: For $x: P_x \propto 1 	imes (6)^4 = 1296$. For $y: P_y \propto 4 	imes (4)^4 = 4 	imes 256 = 1024$. For $z: P_z \propto 9 	imes (3)^4 = 9 	imes 81 = 729$. Thus,$P_x > P_y > P_z$,which implies $E_x > E_y > E_z$.

The black discs $x, y$ and $z$ have radii $1 \ m, 2 \ m$ and $3 \ m$ respectively. The wavelengths corresponding to maximum intensity are $200 \ nm, 300 \ nm$ and $400 \ nm$ respectively. The relation between emissive power $E_{x}, E_{y}$ and $E_{z}$ is

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