Spectral Emissive Power and Wien's Displacement Law Questions in English

(D) According to Wien's displacement law,the wavelength $\lambda_{\max}$ at which the radiant intensity is maximum is inversely proportional to the absolute temperature $T$ of the black body:
$\lambda_{\max} T = b$ (where $b$ is Wien's constant).
For object $O_1$: $\lambda_1 = 200\,nm$,so $T_1 = \frac{b}{200}$.
For object $O_2$: $\lambda_2 = 600\,nm$,so $T_2 = \frac{b}{600}$.
According to the Stefan-Boltzmann law,the power emitted per unit area (emissive power) $E$ is proportional to the fourth power of the absolute temperature:
$E \propto T^4$.
Therefore,the ratio of power emitted per unit area by $O_1$ to that of $O_2$ is:
$\frac{E_1}{E_2} = \left( \frac{T_1}{T_2} \right)^4 = \left( \frac{b/200}{b/600} \right)^4 = \left( \frac{600}{200} \right)^4 = (3)^4 = 81$.
Thus,the ratio is $81:1$.

(A) According to Wien's displacement law,$\lambda_m T = \text{constant}$,so $\lambda_A T_A = \lambda_B T_B$.
Given $\lambda_A = 500 \,nm$ and $\lambda_B = 1500 \,nm$,we have $500 T_A = 1500 T_B$,which implies $T_A / T_B = 3$.
The rate of total energy radiated by a black body is given by $P = \sigma A T^4 = \sigma (4 \pi r^2) T^4$.
Therefore,the ratio of the power radiated by $A$ to that of $B$ is $\frac{P_A}{P_B} = \left( \frac{r_A}{r_B} \right)^2 \left( \frac{T_A}{T_B} \right)^4$.
Substituting the values: $\frac{P_A}{P_B} = \left( \frac{6}{18} \right)^2 \times (3)^4 = \left( \frac{1}{3} \right)^2 \times 81 = \frac{1}{9} \times 81 = 9$.

(B) According to the Stefan-Boltzmann law,the power radiated by a blackbody is given by $P = \sigma A T^4 = \sigma (4\pi R^2) T^4$.
Given that $R_A = 400 R_B$ and $P_A = 10^4 P_B$.
Substituting these into the power ratio equation: $\frac{P_A}{P_B} = \left(\frac{R_A}{R_B}\right)^2 \left(\frac{T_A}{T_B}\right)^4$.
$10^4 = (400)^2 \left(\frac{T_A}{T_B}\right)^4$.
$10^4 = 160000 \left(\frac{T_A}{T_B}\right)^4 = 1.6 \times 10^5 \left(\frac{T_A}{T_B}\right)^4$.
$\left(\frac{T_A}{T_B}\right)^4 = \frac{10^4}{1.6 \times 10^5} = \frac{1}{16}$.
Taking the fourth root,$\frac{T_A}{T_B} = \frac{1}{2}$,which implies $T_B = 2 T_A$.
According to Wien's displacement law,$\lambda T = \text{constant}$,so $\lambda_A T_A = \lambda_B T_B$.
Therefore,$\frac{\lambda_A}{\lambda_B} = \frac{T_B}{T_A} = 2$.

(A) According to Wien's displacement law,$\lambda_{\max} T = b$ (constant),which implies $T \propto \frac{1}{\lambda}$.
According to Stefan-Boltzmann's law,the emissive power $E$ (or power radiated per unit area) is given by $E = \sigma T^4$.
Substituting $T \propto \frac{1}{\lambda}$ into the Stefan-Boltzmann law,we get $E \propto \left(\frac{1}{\lambda}\right)^4$,or $E \propto \lambda^{-4}$.
Let the initial state be $(\lambda_1, E_1)$ and the final state be $(\lambda_2, E_2)$.
Given $\lambda_1 = \lambda$ and $\lambda_2 = \frac{2\lambda}{3}$.
Then,$\frac{E_2}{E_1} = \left(\frac{\lambda_1}{\lambda_2}\right)^4 = \left(\frac{\lambda}{\frac{2\lambda}{3}}\right)^4 = \left(\frac{3}{2}\right)^4 = \frac{81}{16}$.
Therefore,$E_2 = \frac{81}{16} E$.

(D) The graph shows the spectral emissive power $(E_{\lambda})$ on the y-axis and wavelength $(\lambda)$ on the x-axis.
The area under the curve is given by the integral $\int_{0}^{\infty} E_{\lambda} d\lambda$.
According to the definition of spectral emissive power,this integral represents the total radiant energy emitted per unit time per unit surface area of the black body across all possible wavelengths.
This is consistent with the Stefan-Boltzmann Law,which states that the total power radiated per unit area is proportional to the fourth power of the absolute temperature $(E = \sigma T^4)$.

(B) According to Wien's displacement law,the wavelength $\lambda_m$ at which a black body radiates maximum energy is inversely proportional to its absolute temperature $T$,i.e.,$\lambda_m T = b$ (constant).
Let $\lambda_A$ and $\lambda_B$ be the wavelengths of maximum emission for bodies $A$ and $B$ respectively,and $T_A$ and $T_B$ be their temperatures.
Given: $T_A = 3T_B$.
From Wien's law: $\lambda_A T_A = \lambda_B T_B$.
Substituting $T_A$: $\lambda_A (3T_B) = \lambda_B T_B$,which implies $\lambda_B = 3\lambda_A$.
Given the difference in wavelengths: $\lambda_B - \lambda_A = 4 \mu m$.
Substituting $\lambda_B = 3\lambda_A$: $3\lambda_A - \lambda_A = 4 \mu m$,so $2\lambda_A = 4 \mu m$,which gives $\lambda_A = 2 \mu m$.
Therefore,$\lambda_B = 3 \times 2 \mu m = 6 \mu m$.

(A) According to Wien's Displacement Law,the wavelength $\lambda_m$ corresponding to maximum intensity of radiation emitted by a black body is inversely proportional to its absolute temperature $T$.
Mathematically,$\lambda_m T = \text{constant}$,or $\lambda_1 T_1 = \lambda_2 T_2$.
Given: $\lambda_1 = \lambda$,$T_1 = T$,and $T_2 = 1.5 \ T = \frac{3}{2} \ T$.
Substituting these values into the equation:
$\lambda \cdot T = \lambda_2 \cdot (1.5 \ T)$
$\lambda_2 = \frac{\lambda \cdot T}{1.5 \ T} = \frac{\lambda}{1.5} = \frac{\lambda}{3/2} = \frac{2 \lambda}{3}$.
Therefore,the new wavelength is $\frac{2 \lambda}{3}$.

(D) According to Wien's displacement law,the product of the temperature $T$ and the wavelength $\lambda_{m}$ corresponding to maximum spectral emissive power is a constant.
$\lambda_{m} T = b$ (constant)
Given:
$T_1 = 2000 \ K$
$T_2 = 3000 \ K$
Let the wavelength at $T_2$ be $\lambda'_{m}$.
Then,$\lambda_{m} T_1 = \lambda'_{m} T_2$
$\lambda'_{m} = \lambda_{m} \times \frac{T_1}{T_2}$
$\lambda'_{m} = \lambda_{m} \times \frac{2000}{3000}$
$\lambda'_{m} = \frac{2}{3} \lambda_{m}$
Therefore,the correct option is $D$.

(C) According to Wien's displacement law,the wavelength $\lambda_m$ corresponding to maximum energy emission is inversely proportional to the absolute temperature $T$ of the black body:
$\lambda_m \propto \frac{1}{T}$
Since the frequency $\nu_m$ is related to the wavelength by the relation $\nu_m = \frac{c}{\lambda_m}$,where $c$ is the speed of light,we can substitute $\lambda_m = \frac{c}{\nu_m}$ into the proportionality:
$\frac{c}{\nu_m} \propto \frac{1}{T}$
$\nu_m \propto T$
This indicates that the frequency $\nu_m$ is directly proportional to the temperature $T$. $A$ direct proportionality between two variables is represented by a straight line passing through the origin. In the given figure,the straight line $B$ represents this linear relationship.

(B) According to Wien's displacement law,$\lambda_{\text{max}} T = \text{constant}$.
Therefore,$\frac{T_1}{T_2} = \frac{\lambda_{\text{max}2}}{\lambda_{\text{max}1}} = \frac{\lambda_0 / 4}{\lambda_0} = \frac{1}{4}$,which implies $T_2 = 4T_1$.
According to the Stefan-Boltzmann law,the power radiated by a black body is $P = \sigma A T^4$.
Therefore,the ratio of power radiated is $\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^4$.
Substituting the values,$\frac{P_2}{P_1} = (4)^4 = 256$.
Thus,the new power radiated is $P_2 = 256 P$.

(B) According to Wien's displacement law,the wavelength $\lambda_m$ at which a body radiates maximum energy is inversely proportional to its absolute temperature $T$,given by $\lambda_m T = b$ (where $b$ is Wien's constant).
For bodies $P$ and $Q$,we have $\lambda_P T_P = \lambda_Q T_Q$.
Given that $T_P = 4 T_Q$,we substitute this into the equation:
$\lambda_P (4 T_Q) = \lambda_Q T_Q \implies \lambda_Q = 4 \lambda_P$ (Equation $1$).
We are given the difference in wavelengths as $\lambda_Q - \lambda_P = 3 \mu m$ (Equation $2$).
Substituting Equation $1$ into Equation $2$:
$4 \lambda_P - \lambda_P = 3 \mu m$
$3 \lambda_P = 3 \mu m$
$\lambda_P = 1 \mu m$.
Now,finding $\lambda_Q$ using Equation $1$:
$\lambda_Q = 4 \times 1 \mu m = 4 \mu m$.

(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 $x: P_x \propto 1 \times (6)^4 = 1296$.
For $y: P_y \propto 4 \times (4)^4 = 4 \times 256 = 1024$.
For $z: P_z \propto 9 \times (3)^4 = 9 \times 81 = 729$.
Thus,$P_x > P_y > P_z$,which implies $E_x > E_y > E_z$.

(B) According to Wien's Displacement Law,$\lambda_{\max} T = b$,where $b$ is Wien's constant. Therefore,$T = \frac{b}{\lambda_{\max}}$.
According to the Stefan-Boltzmann Law,the total emissive power $E$ of a black body is proportional to the fourth power of its absolute temperature: $E = \sigma T^4$.
Substituting the expression for $T$,we get $E = \sigma \left( \frac{b}{\lambda_{\max}} \right)^4$.
Let the initial wavelength be $\lambda_1 = \lambda$ and the final wavelength be $\lambda_2 = \frac{2 \lambda}{3}$.
Since $E \propto \frac{1}{\lambda_{\max}^4}$,we can write the ratio of the emissive powers as:
$\frac{E'}{E} = \left( \frac{\lambda_1}{\lambda_2} \right)^4$.
Substituting the values: $\frac{E'}{E} = \left( \frac{\lambda}{\frac{2 \lambda}{3}} \right)^4 = \left( \frac{3}{2} \right)^4$.
Calculating the value: $\frac{E'}{E} = \frac{81}{16}$.
Therefore,the new emissive power is $E' = \frac{81}{16} E$.

(D) According to Wien's displacement law,$\lambda_m T = b$,where $b$ is Wien's constant.
If $v_m$ is the frequency corresponding to the wavelength $\lambda_m$,then $\lambda_m = \frac{c}{v_m}$.
Substituting this into the displacement law,we get $\left(\frac{c}{v_m}\right) T = b$.
Rearranging the terms,we find $v_m = \left(\frac{c}{b}\right) T$.
Since $c$ (speed of light) and $b$ (Wien's constant) are constants,$v_m \propto T$.
This represents a linear relationship passing through the origin,which corresponds to the straight line labeled $B$ in the graph.
Therefore,option $D$ is the correct answer.

(D) According to Wien's displacement law,the product of the wavelength at which the radiation energy density is maximum $(\lambda_m)$ and the absolute temperature $(T)$ is a constant.
$\lambda_m T = \text{constant}$
Given that at temperature $T$,the maximum is at $\lambda_0$,we have:
$\lambda_0 T = \lambda' T'$
Here,$T' = 2T$. Substituting this into the equation:
$\lambda_0 T = \lambda' (2T)$
$\lambda' = \frac{\lambda_0 T}{2T} = \frac{\lambda_0}{2}$
Therefore,at temperature $2T$,the maximum energy density occurs at $\frac{\lambda_0}{2}$.

(C) According to Wien's displacement law,the wavelength corresponding to maximum spectral intensity $\lambda_m$ is inversely proportional to the absolute temperature $T$:
$\lambda_m \propto \frac{1}{T}$
Since the frequency $f$ is related to wavelength $\lambda$ by $f = \frac{c}{\lambda}$,where $c$ is the speed of light,we have $\lambda \propto \frac{1}{f}$.
Substituting this into Wien's law:
$\frac{1}{f} \propto \frac{1}{T} \Rightarrow f \propto T$
Therefore,if the temperature $T$ is doubled,the frequency $f$ at which the spectral intensity becomes maximum will also be doubled.

(A) According to Stefan-Boltzmann Law,the total emissive power $E$ of a black body is given by $E = \sigma T^4$,where $\sigma$ is the Stefan-Boltzmann constant and $T$ is the absolute temperature.
According to Wien's Displacement Law,the wavelength $\lambda_m$ corresponding to maximum intensity is inversely proportional to the absolute temperature $T$,i.e.,$T = \frac{b}{\lambda_m}$,where $b$ is Wien's constant.
Substituting $T$ in the expression for $E$,we get $E \propto (\frac{1}{\lambda_m})^4 = \frac{1}{\lambda_m^4}$.
Note that the emissive power $E$ of a black body depends only on its temperature,not on its surface area or radius.
Given $\lambda_x = 200 \ nm$,$\lambda_y = 300 \ nm$,and $\lambda_z = 400 \ nm$.
Since $E \propto \frac{1}{\lambda_m^4}$,we have:
$E_x \propto \frac{1}{200^4}$
$E_y \propto \frac{1}{300^4}$
$E_z \propto \frac{1}{400^4}$
Comparing the values,since $200 < 300 < 400$,it follows that $\frac{1}{200^4} > \frac{1}{300^4} > \frac{1}{400^4}$.
Therefore,$E_x > E_y > E_z$,which means $E_x$ is the maximum.

(C) According to Wien's displacement law,the wavelength $\lambda_m$ corresponding to maximum energy emission is inversely proportional to the absolute temperature $T$:
$\lambda_m T = b$ (where $b$ is Wien's constant).
Therefore,$\lambda_{mA} T_A = \lambda_{mB} T_B$.
Given that $T_A = 3 T_B$,we substitute this into the equation:
$\lambda_{mA} (3 T_B) = \lambda_{mB} T_B \implies \lambda_{mB} = 3 \lambda_{mA}$.
We are also given the difference in wavelengths:
$\lambda_{mB} - \lambda_{mA} = 4 \mu m$.
Substituting $\lambda_{mB} = 3 \lambda_{mA}$ into the difference equation:
$3 \lambda_{mA} - \lambda_{mA} = 4 \mu m \implies 2 \lambda_{mA} = 4 \mu m \implies \lambda_{mA} = 2 \mu m$.
Now,calculate $\lambda_{mB}$:
$\lambda_{mB} = 3 \lambda_{mA} = 3(2 \mu m) = 6 \mu m$.

(D) According to Wien's displacement law,the product of the wavelength of maximum emission $(\lambda_{m})$ and the absolute temperature $(T)$ is a constant,i.e.,$\lambda_{m} T = b$.
This implies that $T \propto \frac{1}{\lambda_{m}}$.
The wavelength of blue light is shorter than the wavelength of yellow light $(\lambda_{\text{blue}} < \lambda_{\text{yellow}})$.
Since the star '$Q$' emits blue light and star '$P$' emits yellow light,we have $\lambda_{Q} < \lambda_{P}$.
Therefore,$T_{Q} > T_{P}$,which can also be written as $T_{P} < T_{Q}$.

(B) According to Wien's displacement law,the product of the wavelength of maximum emission and the absolute temperature is a constant:
$\lambda_m T = \text{constant}$
Therefore,$\lambda_m \propto \frac{1}{T}$.
Given $T_1 = 2000 \ K$ and $T_2 = 3000 \ K$,and the initial wavelength is $\lambda_m$:
$\frac{\lambda_2}{\lambda_m} = \frac{T_1}{T_2}$
$\frac{\lambda_2}{\lambda_m} = \frac{2000}{3000} = \frac{2}{3}$
$\lambda_2 = \frac{2}{3} \lambda_m$
Thus,the corresponding wavelength at $3000 \ K$ is $\frac{2}{3} \lambda_m$.

(A) The emissive power $E$ is defined as the heat energy $Q$ radiated per unit area $A$ per unit time $t$, given by the formula $E = \frac{Q}{A \cdot t}$.
To find the total heat radiated $Q$, we rearrange the formula: $Q = E \cdot A \cdot t$.
Given values are $E = 0.7 \,kcal \,s^{-1} \,m^{-2}$, $A = 0.04 \,m^2$, and $t = 20 \,s$.
Substituting these values: $Q = 0.7 \times 0.04 \times 20$.
$Q = 0.7 \times 0.8 = 0.56 \,kcal$.
Therefore, the amount of heat radiated is $0.56 \,kcal$.

(A) Given: Surface temperature $T = 6000 \,K$,Wien's constant $b = 2.897 \times 10^{-3} \,m-K$.
According to Wien's displacement law,$\lambda_{max} T = b$.
Therefore,$\lambda_{max} = \frac{b}{T} = \frac{2.897 \times 10^{-3}}{6000} \,m$.
$\lambda_{max} = 4.828 \times 10^{-7} \,m$.
Since $1 \,Å = 10^{-10} \,m$,we have $4.828 \times 10^{-7} \,m = 4828 \times 10^{-10} \,m = 4828 \,Å$.

(B) According to Wien's displacement law,$\lambda_{max} T = b$ (constant).
Let $\lambda_A$ and $\lambda_B$ be the wavelengths of maximum energy for bodies $A$ and $B$,and $T_A$ and $T_B$ be their absolute temperatures.
Given: $T_A = 3 T_B$ and $\lambda_B - \lambda_A = 4 \mu m$ (since $T_A > T_B$,$\lambda_A < \lambda_B$).
From Wien's law: $\lambda_A T_A = \lambda_B T_B$.
Substituting $T_A = 3 T_B$: $\lambda_A (3 T_B) = \lambda_B T_B$,which gives $\lambda_B = 3 \lambda_A$.
Substitute this into the difference equation: $3 \lambda_A - \lambda_A = 4 \mu m$.
$2 \lambda_A = 4 \mu m \implies \lambda_A = 2 \mu m$.
Therefore,$\lambda_B = 3 \times 2 \mu m = 6 \mu m$.

(D) According to Wien's displacement law,the product of the wavelength of maximum intensity $\lambda_{m}$ and the absolute temperature $T$ is a constant.
$\lambda_{m} T = \text{constant}$
Given:
$T_{1} = 2200 \ K$
$T_{2} = 3300 \ K$
Let the new wavelength be $\lambda_{m}^{\prime}$.
Using the relation $\lambda_{m} T_{1} = \lambda_{m}^{\prime} T_{2}$:
$\lambda_{m}^{\prime} = \lambda_{m} \times \frac{T_{1}}{T_{2}}$
$\lambda_{m}^{\prime} = \lambda_{m} \times \frac{2200}{3300}$
$\lambda_{m}^{\prime} = \frac{2}{3} \lambda_{m}$

(D) According to Wien's displacement law,$\lambda_{m} T = b$ (constant),so $T \propto \frac{1}{\lambda_{m}}$.
According to Stefan-Boltzmann law,the total emissive power $E$ of a black body is given by $E = \sigma T^{4}$.
Substituting $T \propto \frac{1}{\lambda_{m}}$,we get $E \propto \left(\frac{1}{\lambda_{m}}\right)^{4} = \frac{1}{\lambda_{m}^{4}}$.
Note that the emissive power $E$ is independent of the surface area $A$ of the black body.
Therefore,the ratio of emissive powers is $Ex : Ey : Ez = \frac{1}{\lambda_{x}^{4}} : \frac{1}{\lambda_{y}^{4}} : \frac{1}{\lambda_{z}^{4}}$.
Substituting the given values: $Ex : Ey : Ez = \frac{1}{(200)^{4}} : \frac{1}{(300)^{4}} : \frac{1}{(400)^{4}}$.
Since $200 < 300 < 400$,it follows that $\frac{1}{200^{4}} > \frac{1}{300^{4}} > \frac{1}{400^{4}}$.
Thus,$Ex > Ey > Ez$.

(C) Given,maximum wavelength,$\lambda_m = 289.8 \ nm = 289.8 \times 10^{-9} \ m = 2.898 \times 10^{-7} \ m$.
Stefan's constant,$\sigma = 5.67 \times 10^{-8} \ W \ m^{-2} \ K^{-4}$.
Wien's constant,$b = 2898 \ \mu m \ K = 2898 \times 10^{-6} \ m \ K$.
According to Wien's displacement law,$\lambda_m = \frac{b}{T}$,so $T = \frac{b}{\lambda_m}$.
Substituting the values,$T = \frac{2898 \times 10^{-6}}{289.8 \times 10^{-9}} = \frac{2898 \times 10^{-6}}{2.898 \times 10^{-7}} = 10^4 \ K$.
According to Stefan-Boltzmann law,the intensity of radiation $I$ is given by $I = \sigma T^4$ (assuming the star acts as a black body,$e=1$).
Substituting the values,$I = (5.67 \times 10^{-8}) \times (10^4)^4 = 5.67 \times 10^{-8} \times 10^{16} = 5.67 \times 10^8 \ W \ m^{-2}$.

(D) According to Wien's displacement law,the product of the wavelength $\lambda_{m}$ corresponding to the maximum intensity of radiation and the absolute temperature $T$ of the body is a constant.
Mathematically,$\lambda_{m} T = b$.
Here,$\lambda_{m}$ is measured in meters $(m)$ and $T$ is measured in Kelvin $(K)$.
Therefore,the unit of Wien's constant $b$ is the product of the units of wavelength and temperature,which is $m K$.

(C) All bodies at a temperature above absolute zero emit electromagnetic radiation. The human body,being at a temperature of approximately $37^{\circ}C$ $(310 \ K)$,emits thermal radiation.
According to Wien's displacement law,the peak wavelength of this radiation corresponds to the infrared region of the electromagnetic spectrum.
The wavelength range for human body radiation is typically in the infrared region,specifically around $10 \ \mu m$.
Therefore,the radiation emitted by the human body is in the infrared region.

(C) According to Wien's displacement law,the product of the wavelength of maximum emission and the absolute temperature is constant: $\lambda T = b$.
Therefore,$\lambda_A T_A = \lambda_B T_B$.
This implies the ratio of temperatures is inversely proportional to the ratio of wavelengths: $\frac{T_A}{T_B} = \frac{\lambda_B}{\lambda_A}$.
Given $\lambda_A = 360 \ nm$ and $\lambda_B = 480 \ nm$.
Substituting the values: $\frac{T_A}{T_B} = \frac{480}{360} = \frac{4}{3}$.
Thus,the ratio of the surface temperatures of $A$ and $B$ is $4: 3$.

(C) According to Wien's displacement law, the wavelength corresponding to maximum intensity $(\lambda_{m})$ is inversely proportional to the absolute temperature $(T)$, i.e., $\lambda_{m} T = \text{constant}$.
Since the frequency $\nu_{m}$ is related to wavelength by $\nu_{m} = c / \lambda_{m}$, we can substitute $\lambda_{m} = c / \nu_{m}$ into the displacement law.
This gives $(c / \nu_{m}) T = \text{constant}$, which simplifies to $\nu_{m} / T = \text{constant}'$ or $\nu_{m} \propto T$.
Therefore, the graph of $\nu_{m}$ versus $T$ for a perfectly black body is a straight line passing through the origin, which corresponds to line $C$ in the given figure.

(B) The surface temperature of the stars is determined using Wien's displacement law.
According to this law,$\lambda_{m} T = b$,where $\lambda_{m}$ is the wavelength corresponding to the maximum spectral emissive power,$T$ is the absolute temperature of the black body,and $b$ is Wien's constant.
The value of Wien's constant is approximately $2.898 \times 10^{-3} \ m \cdot K$.
By measuring the wavelength $\lambda_{m}$ at which the star emits maximum radiation,the surface temperature $T$ can be calculated as $T = b / \lambda_{m}$.

(C) According to Wien's displacement law,the product of the wavelength of maximum emission $(\lambda_m)$ and the absolute temperature $(T)$ is constant:
$\lambda_{m1} T_1 = \lambda_{m2} T_2$
Given:
$\lambda_{m1} = 500 \,nm$,$T_1 = 6000 \,K$
$\lambda_{m2} = 400 \,nm$,$T_2 = ?$
Substituting the values into the equation:
$500 \,nm \times 6000 \,K = 400 \,nm \times T_2$
$T_2 = \frac{500 \times 6000}{400}$
$T_2 = 5 \times 1500$
$T_2 = 7500 \,K$
Therefore,the temperature of the star is $7500 \,K$.

(D) According to Wien's displacement law,the product of the wavelength of maximum intensity $(\lambda_m)$ and the absolute temperature $(T)$ of a black body is constant: $\lambda_m T = b$ (where $b$ is Wien's constant).
Therefore,$T \propto \frac{1}{\lambda_m}$.
Given: $\lambda_A = 0.5 \mu m = 0.5 \times 10^{-6} \ m$ and $\lambda_B = 0.1 \ mm = 0.1 \times 10^{-3} \ m = 10^{-4} \ m$.
The ratio of temperatures is $\frac{T_A}{T_B} = \frac{\lambda_B}{\lambda_A}$.
Substituting the values: $\frac{T_A}{T_B} = \frac{10^{-4}}{0.5 \times 10^{-6}} = \frac{10^{-4}}{5 \times 10^{-7}} = \frac{1000}{5} = 200$.
Thus,the ratio of the temperatures of the bodies $A$ and $B$ is $200$.

(D) According to Wien's displacement law, the wavelength corresponding to the maximum intensity of emission $(\lambda_m)$ is inversely proportional to the absolute temperature $(T)$ of the body, given by $\lambda_m T = b$, where $b$ is Wien's constant.
As the temperature of the iron rod increases, the wavelength of the emitted radiation decreases.
Initially, at lower temperatures, the emitted radiation is in the red part of the visible spectrum.
As the temperature increases further, the peak wavelength shifts towards shorter wavelengths (yellow, then blue, and finally white), which is a combination of all visible wavelengths.
Thus, the change in colour is explained by Wien's displacement law.

(A) According to Wien's displacement law, the wavelength $(\lambda_m)$ corresponding to the maximum energy emitted by a black body is inversely proportional to its absolute temperature $(T)$.
Mathematically, $\lambda_m \propto \frac{1}{T}$.
This can be written as $\lambda_m = \frac{b}{T}$, where $b$ is Wien's constant.
Therefore, $\lambda_m \cdot T = b$, which is a constant.

(A) Given:
Temperature of the body, $T_1 = 3000 \,K$
Wavelength of maximum energy emission for the body, $\lambda_1 = 9660 Å$
Wavelength of maximum energy emission for the sun, $\lambda_2 = 4950 Å$
According to Wien's displacement law, the product of the wavelength of maximum emission and the absolute temperature is constant:
$\lambda_1 T_1 = \lambda_2 T_2$
Rearranging to find the temperature of the sun $(T_2)$:
$T_2 = \frac{\lambda_1 T_1}{\lambda_2}$
Substituting the given values:
$T_2 = \frac{9660 Å \times 3000 \,K}{4950 Å}$
$T_2 = \frac{28980000}{4950} \,K$
$T_2 \approx 5854.54 \,K$
Rounding to the nearest whole number, we get $T_2 \approx 5855 \,K$.

(D) According to Wien's displacement law,the product of the wavelength of maximum emission $(\lambda_m)$ and the absolute temperature $(T)$ is a constant,i.e.,$\lambda_m T = b$ (constant).
Therefore,$\lambda_{m_1} T_1 = \lambda_{m_2} T_2$.
Given:
$\lambda_{m_1} = 4.08 \ \mu m$
$T_1 = 700 \ K$
$T_2 = 1400 \ K$
Substituting the values:
$4.08 \times 700 = \lambda_{m_2} \times 1400$
$\lambda_{m_2} = \frac{4.08 \times 700}{1400}$
$\lambda_{m_2} = \frac{4.08}{2} = 2.04 \ \mu m$.
Thus,the new wavelength is $2.04 \ \mu m$.

(A) Given: $\lambda_m = 289.8 \, nm = 289.8 \times 10^{-9} \, m$, $b = 2898 \, \mu m K = 2898 \times 10^{-6} \, m K$, $\sigma = 5.67 \times 10^{-8} \, W m^{-2} K^{-4}$.
Using Wien's displacement law: $\lambda_m T = b$.
$T = \frac{b}{\lambda_m} = \frac{2898 \times 10^{-6}}{289.8 \times 10^{-9}} = \frac{2898 \times 10^{-6}}{289.8 \times 10^{-9}} = 10 \times 10^3 = 10^4 \, K$.
Radiation intensity (emissive power) $E = \sigma T^4$.
$E = (5.67 \times 10^{-8}) \times (10^4)^4$.
$E = 5.67 \times 10^{-8} \times 10^{16} = 5.67 \times 10^8 \, W/m^2$.

(A) According to Wien's displacement law,$\lambda_m \cdot T = b$,where $\lambda_m$ is the wavelength corresponding to the maximum radiation of energy from a black body,$b$ is Wien's constant,and $T$ is the absolute temperature.
From this relation,we can see that $\lambda_m = \frac{b}{T}$.
Therefore,$\lambda_m \propto \frac{1}{T}$,which can be written as $\lambda_m \propto T^{-1}$.

(C) According to Wien's displacement law,the relationship between the wavelength of maximum emission $\lambda_m$ and the absolute temperature $T$ of a black body is given by $\lambda_m T = b$,where $b$ is Wien's constant.
Given:
$\lambda_m = 6 \ mm = 6 \times 10^{-3} \ m$
$b = 3 \times 10^{-3} \ mK$
Substituting these values into the formula:
$(6 \times 10^{-3} \ m) \times T = 3 \times 10^{-3} \ mK$
$T = \frac{3 \times 10^{-3}}{6 \times 10^{-3}} \ K$
$T = \frac{1}{2} \ K = 0.5 \ K$
Therefore,the temperature of the black body is $0.5 \ K$.

(A) According to Wien's displacement law, the relationship between the wavelength of maximum emission and the absolute temperature of a black body is given by:
$\lambda T = b$
Where:
$\lambda$ is the wavelength of radiation = $1 \,mm = 1 \times 10^{-3} \,m$
$b$ is Wien's constant = $3 \times 10^{-3} \,mK$
$T$ is the temperature of the black body.
Substituting the values into the formula:
$T = \frac{b}{\lambda} = \frac{3 \times 10^{-3} \,mK}{1 \times 10^{-3} \,m} = 3 \,K$
Therefore, the temperature of the black body is $3 \,K$.

(D) According to Stefan-Boltzmann law,the power radiated is $P = e A \sigma T^4$. Since the bodies have equal surface area and radiate at the same rate $(P_A = P_B)$,we have $e_A T_A^4 = e_B T_B^4$.
Given $e_A = 0.01$,$e_B = 0.81$,and $T_A = 5802 \ K$.
Substituting the values: $0.01 \times (5802)^4 = 0.81 \times T_B^4$.
Taking the fourth root on both sides: $T_B = T_A \times (0.01 / 0.81)^{1/4} = 5802 \times (1/81)^{1/4} = 5802 / 3 = 1934 \ K$.
From Wien's displacement law,$\lambda_A T_A = \lambda_B T_B = b$ (where $b \approx 2898 \ \mu m \cdot K$).
Thus,$\lambda_A = b / T_A = 2898 / 5802 \approx 0.5 \ \mu m$.
Given $\lambda_B - \lambda_A = 1 \ \mu m$,so $\lambda_B = 1 + 0.5 = 1.5 \ \mu m = \frac{3}{2} \ \mu m$.

(B) According to Wien's displacement law, the product of the wavelength corresponding to maximum intensity $(\lambda_m)$ and the absolute temperature $(T)$ of the black body is a constant $(b).$
Formula: $\lambda_m T = b$
Given values:
$\lambda_m = 480 \, nm = 480 \times 10^{-9} \, m$
$b = 2.88 \times 10^{-3} \, mK$
Calculation:
$T = \frac{b}{\lambda_m}$
$T = \frac{2.88 \times 10^{-3}}{480 \times 10^{-9}}$
$T = \frac{2.88}{480} \times 10^6$
$T = 0.006 \times 10^6 = 6000 \, K$
Therefore, the surface temperature of the sun is $6000 \, K$.