Question 756: Thermodynamics & KTG

Question

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

Accepted Answer

The black discs $\mathrm{x}$, $\mathrm{y}$, and $\mathrm{z}$ have radii of $1 \, 	ext{m}$, $2 \, 	ext{m}$, and $3 \, 	ext{m}$, respectively. The wavelengths corresponding to maximum intensity are $200 \, 	ext{nm}$, $300 \, 	ext{nm}$, and $400 \, 	ext{nm}$, respectively. Here, we explore the relationship between the emissive power $E_x$, $E_y$, and $E_z$ for these discs.
Emissive Power
The formula for emissive power is given by:
$  E = \sigma A T^4 \quad \Rightarrow \quad E \propto A T^4  $
Where:
$A$ is the area of the disc.
$T$ is the absolute temperature.
$\sigma$ is the Stefan-Boltzmann constant.
Area Calculation
The area $A$ of a disc is given by $A = \pi R^2$. Therefore:
$ A \propto R^2 $
Given:
$R_1 = 1 \, 	ext{m}$
$R_2 = 2 \, 	ext{m}$
$R_3 = 3 \, 	ext{m}$
The areas are proportional as follows:
$  A_x : A_y : A_z \quad 	ext{proportional to} \quad 1 : 4 : 9 \quad 	ext{... (i)}  $
Temperature Analysis Using Wien's Displacement Law
By Wien's displacement law, we have:
$  \lambda_{\max} \, T = 	ext{constant} \quad \Rightarrow \quad T \propto \frac{1}{\lambda}  $
Given maximum wavelengths:
$\lambda_{\max 1} = 200 \, 	ext{nm}$
$\lambda_{\max 2} = 300 \, 	ext{nm}$
$\lambda_{\max 3} = 400 \, 	ext{nm}$
The wavelengths are proportional as follows:
$  \lambda_x : \lambda_y : \lambda_z \quad 	ext{proportional to} \quad 2 : 3 : 4  $
Thus the temperatures are:
$  \frac{1}{T_x} : \frac{1}{T_y} : \frac{1}{T_z} \quad 	ext{proportional to} \quad 2 : 3 : 4  $
Therefore, the temperatures are proportional as:
$  T_x : T_y : T_z \quad 	ext{proportional to} \quad \frac{1}{2} : \frac{1}{3} : \frac{1}{4}  $
Simplified, this gives:
$  T_x : T_y : T_z \quad 	ext{proportional to} \quad 6 : 4 : 3 \quad 	ext{... (ii)}  $
Comparison of Emissive Powers
We compare the product $A T^4$ for the discs:
For disc $\mathrm{x}$: $A_x T_x^4 = 1 	imes (6)^4 = 1296$
For disc $\mathrm{y}$: $A_y T_y^4 = 4 	imes (4)^4 = 1024$
For disc $\mathrm{z}$: $A_z T_z^4 = 9 	imes (3)^4 = 729$
Thus, the relationship between the emissive powers is:
$  E_x > E_y > E_z  $

Answer

$\mathrm{E}_{\mathrm{x}}>\mathrm{E}_{\mathrm{y}}>\mathrm{E}_{\mathrm{z}}$

Answer

$\mathrm{E}_{\mathrm{x}}<\mathrm{E}_{\mathrm{y}}<\mathrm{E}_{\mathrm{z}}$

Answer

$\mathrm{E}_{\mathrm{x}}=\mathrm{E}_{\mathrm{y}}=\mathrm{E}_{\mathrm{z}}$

Answer

$\mathrm{E}_{\mathrm{y}}>\mathrm{E}_{\mathrm{x}}<\mathrm{E}_{\mathrm{z}}$

Thermodynamics & KTG

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

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