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

(C) The Earth acts as a black body at a relatively low temperature (approximately $288 \ K$). According to Wien's displacement law,the wavelength $\lambda_{max}$ corresponding to the maximum spectral emissive power is inversely proportional to the absolute temperature $T$,given by the relation $\lambda_{max} T = b$,where $b$ is Wien's constant $(2.898 \times 10^{-3} \ m \cdot K)$. For the Earth's temperature,this wavelength falls in the infra-red region of the electromagnetic spectrum. Therefore,Wien's law correctly describes the shift in the peak of the radiation spectrum.

(A) Wien's displacement law states that the wavelength corresponding to the maximum spectral emissive power $(\lambda_m)$ of a black body is inversely proportional to its absolute temperature $(T)$.
Mathematically,this is expressed as $\lambda_m \propto \frac{1}{T}$.
Rearranging this gives the relation $\lambda_m T = b$,where $b$ is Wien's constant.
Therefore,the correct option is $A$.

(C) According to Wien's displacement law, the wavelength corresponding to maximum intensity $(\lambda_m)$ is inversely proportional to the absolute temperature $(T)$ of the black body, i.e., $\lambda_m T = \text{constant}$.
Since the wavelength of red light $(\lambda_r)$ is the largest, yellow light $(\lambda_y)$ is intermediate, and blue light $(\lambda_b)$ is the smallest $(\lambda_r > \lambda_y > \lambda_b)$, the corresponding temperatures will follow the inverse order.
Therefore, $T_r < T_y < T_b$.
Given that star $A$ emits maximum intensity in red, $B$ in blue, and $C$ in yellow, we have $T_A < T_C < T_B$.
Thus, the temperature of $B$ is maximum, $A$ is minimum, and $C$ is intermediate.

(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: $\lambda_m T = b$.
Therefore,for the sun and the moon,we have $\lambda_{sun} T_{sun} = \lambda_{moon} T_{moon}$.
The ratio of their temperatures is given by $\frac{T_{sun}}{T_{moon}} = \frac{\lambda_{moon}}{\lambda_{sun}}$.
Substituting the given values: $\frac{T_{sun}}{T_{moon}} = \frac{10^{-4} \ m}{0.5 \times 10^{-6} \ m} = \frac{10^{-4}}{0.5 \times 10^{-6}} = \frac{100}{0.5} = 200$.
Thus,the ratio of their temperatures is $200$.

(C) According to Wien's Displacement Law,the wavelength of maximum intensity of radiation emitted by a black body is inversely proportional to its absolute temperature.
Mathematically,it is expressed as: $\lambda_m T = b$,where $b$ is Wien's constant.
Therefore,the wavelength of radiation emitted by a body depends on the temperature of its surface.

(A) According to Wien's displacement law, $\lambda_m T = \text{constant}$.
As the temperature $T$ of the wire increases, the wavelength $\lambda_m$ corresponding to the maximum spectral emissive power decreases.
At lower temperatures, the emitted radiation is primarily in the longer wavelength range (red).
As the temperature rises, the peak of the emission spectrum shifts towards shorter wavelengths (yellow, then blue/white).
Therefore, the observed colour changes from red to yellow and finally to white as the temperature increases.

(C) According to Wien's displacement law,the wavelength $\lambda_m$ corresponding to the maximum intensity of emission is inversely proportional to the absolute temperature $T$ of the body,given by $\lambda_m T = b$,where $b$ is Wien's constant. Since the colour of a star depends on the wavelength of the light it emits most intensely,the observed colour is a direct indicator of the star's surface temperature.

(D) According to Wien's Displacement Law,the product of the wavelength of maximum emitted energy $(\lambda_m)$ and the absolute temperature $(T)$ of a black body is constant.
$\lambda_m T = b$ (constant)
Given:
$T_1 = 700 \ K$,$\lambda_{m1} = 4.08 \ \mu m$
$T_2 = 1400 \ K$,$\lambda_{m2} = ?$
Using the relation $\lambda_{m1} T_1 = \lambda_{m2} T_2$:
$\lambda_{m2} = \frac{\lambda_{m1} T_1}{T_2}$
$\lambda_{m2} = \frac{4.08 \times 700}{1400}$
$\lambda_{m2} = 4.08 \times 0.5 = 2.04 \ \mu m$.

(C) According to Wien's Displacement Law,the product of the wavelength corresponding to maximum energy emission $(\lambda_m)$ and the absolute temperature $(T)$ is a constant.
$\lambda_m T = b$ (constant)
Given:
$T_1 = 200 K$,$\lambda_{m1} = 14 \mu m$
$T_2 = 1000 K$,$\lambda_{m2} = ?$
Using the relation $\lambda_{m1} T_1 = \lambda_{m2} T_2$:
$\lambda_{m2} = \frac{\lambda_{m1} T_1}{T_2}$
$\lambda_{m2} = \frac{14 \mu m \times 200 K}{1000 K}$
$\lambda_{m2} = \frac{2800}{1000} \mu m = 2.8 \mu m$
Therefore,the correct option is $C$.

(C) According to Wien's Displacement Law,the product of the wavelength of maximum emission $(\lambda_m)$ and the absolute temperature $(T)$ of a black body is constant: $\lambda_m T = b$.
Therefore,$\lambda_{m1} T_1 = \lambda_{m2} T_2$.
This implies the ratio of temperatures is inversely proportional to the ratio of their wavelengths: $\frac{T_1}{T_2} = \frac{\lambda_{m2}}{\lambda_{m1}}$.
Given $\lambda_{m1} = 3600 \mathring A$ and $\lambda_{m2} = 4800 \mathring A$.
Substituting the values: $\frac{T_1}{T_2} = \frac{4800}{3600} = \frac{48}{36} = \frac{4}{3}$.
Thus,the ratio of their temperatures is $4:3$.

(B) According to Wien's displacement law,$\lambda_m T = \text{constant}$,so $\lambda_{m1} T_1 = \lambda_{m2} T_2$.
Given: $\lambda_{m1} = 5000 \mathring A$,$T_1 = 1227 + 273 = 1500 \ K$.
The temperature is increased by $1000^{\circ}C$,so the new temperature $T_2 = 1227 + 1000 + 273 = 2500 \ K$.
Using the formula: $\lambda_{m2} = \frac{T_1}{T_2} \times \lambda_{m1}$.
$\lambda_{m2} = \frac{1500}{2500} \times 5000 = 0.6 \times 5000 = 3000 \mathring A$.

(A) According to Wien's Displacement Law, the wavelength of emitted radiation is inversely proportional to the absolute temperature of the body $(\lambda_m \propto 1/T)$.
As the temperature of an object increases, the peak wavelength of the emitted radiation shifts towards the shorter wavelength (blue/violet) end of the spectrum.
The sequence of colors observed as temperature increases is: Red $\to$ Orange $\to$ Yellow $\to$ White.
Therefore, the piece of iron glowing White has the highest temperature.

(B) According to Wien's displacement law,the wavelength of maximum emission $\lambda_m$ is inversely proportional to the absolute temperature $T$ $(\lambda_m \propto 1/T)$.
As the temperature of a black body increases,the peak of the emitted radiation spectrum shifts towards shorter wavelengths.
At lower temperatures,the body appears red. As the temperature increases,it transitions through yellow and blue.
At very high temperatures,the body emits radiation across the entire visible spectrum with significant intensity,causing it to appear white.

(B) According to Wien's Displacement Law,the wavelength corresponding to the maximum emission of radiation is inversely proportional to the absolute temperature: ${\lambda _{max}} \propto \frac{1}{T}$.
Given that the current temperature of the sun is $T$,the wavelength of maximum emission is ${\lambda _{max}} \approx 4753 \mathring{A}$,which lies in the visible region (near violet).
If the temperature becomes $T' = 2T$,the new wavelength of maximum emission will be ${\lambda '_{max}} = \frac{{\lambda _{max}}}{2} = \frac{4753 \mathring{A}}{2} \approx 2376.5 \mathring{A}$.
Since this wavelength is shorter than the visible range,the radiated energy will be predominantly in the ultraviolet region.

(B) According to Wien's displacement law,the product of the wavelength of maximum emission $(\lambda_m)$ and the absolute temperature $(T)$ of the black body is a constant $(b)$:
$\lambda_m T = b$
Given:
$\lambda_m = 2.93 \times 10^{-10} \ m$
$b = 2.93 \times 10^{-3} \ m \cdot K$
Substituting the values into the formula:
$T = \frac{b}{\lambda_m} = \frac{2.93 \times 10^{-3}}{2.93 \times 10^{-10}}$
$T = 10^{-3 - (-10)} \ K = 10^7 \ K$
Thus,the maximum temperature attained is of the order of $10^7 \ K$.

(A) According to Wien's Displacement Law,the product of the wavelength of maximum emission $(\lambda_m)$ and the absolute temperature $(T)$ is a constant.
$\lambda_m T = b$ (constant)
Therefore,$\lambda_{m1} T_1 = \lambda_{m2} T_2$.
Given: $T_1 = 2000\;K$,$\lambda_{m1} = 4\;\mu m$,$T_2 = 2400\;K$.
Substituting the values:
$4 \times 2000 = \lambda_{m2} \times 2400$
$\lambda_{m2} = \frac{4 \times 2000}{2400} = \frac{8000}{2400} = \frac{80}{24} = \frac{10}{3} = 3.33\;\mu m$.

(B) The radiation emitted by stars can be analyzed to obtain the energy distribution across different wavelengths.
According to Wien's displacement law,the wavelength $\lambda_{max}$ corresponding to the maximum intensity of radiation is inversely proportional to the absolute temperature $T$ of the star.
The relationship is given by $\lambda_{max} T = b$,where $b$ is Wien's constant.
By measuring the wavelength at which the star emits maximum radiation,astronomers can calculate the surface temperature of the star.

(A) As the temperature of a substance increases, the peak wavelength of the emitted radiation shifts towards shorter wavelengths according to Wien's displacement law, $\lambda_m T = \text{constant}$.
Initially, at lower temperatures, the substance emits radiation in the infrared region, which is invisible. As the temperature rises, it starts emitting visible light, starting from red, then moving through orange, yellow, and eventually to white at very high temperatures.

Temperature	Colour
$525^{\circ}C$	Light red
$900^{\circ}C$	Cherry red
$1100^{\circ}C$	Orange red
$1200^{\circ}C$	Yellow
$1300^{\circ}C$	Green
$1600^{\circ}C$	White

Since the question asks for the sequence of colours noticed as temperature increases, the final colour observed at the highest temperature among the options is white.

(A) According to Wien's displacement law,$\lambda_{m} T = b$,where $b$ is Wien's constant.
This law states that the wavelength $\lambda_{m}$ corresponding to the maximum spectral emissive power is inversely proportional to the absolute temperature $T$ of the black body.
Since the colour of a star depends on the wavelength of the light it emits most intensely,its surface temperature is directly related to its colour.
For example,red stars have a longer wavelength and thus a lower temperature,while blue stars have a shorter wavelength and a higher temperature.

(C) According to Wien's Displacement Law,the product of the wavelength corresponding to maximum emission $(\lambda_m)$ and the absolute temperature $(T)$ of a black body is constant.
$\lambda_m T = b$ (constant)
Therefore,$\lambda_{m1} T_1 = \lambda_{m2} T_2$.
Given:
$T_1 = 1640 \ K$
$\lambda_{m1} = 1.75 \ \mu m$
$\lambda_{m2} = 14.35 \ \mu m$
Substituting the values:
$1.75 \times 1640 = 14.35 \times T_2$
$T_2 = \frac{1.75 \times 1640}{14.35}$
$T_2 = \frac{2870}{14.35} = 200 \ K$.
Thus,the temperature of the moon is $200 \ K$.

(A) According to Wien's Displacement Law,the product of the wavelength corresponding to maximum intensity $(\lambda_m)$ and the absolute temperature $(T)$ is constant.
$\lambda_m T = b$ (constant)
Therefore,$\lambda_1 T_1 = \lambda_2 T_2$.
Given: $\lambda_1 = 4\,\mu m$,$T_1 = 900\,K$,$T_2 = 1200\,K$.
Substituting the values:
$4\,\mu m \times 900\,K = \lambda_2 \times 1200\,K$
$\lambda_2 = \frac{4 \times 900}{1200}\,\mu m$
$\lambda_2 = \frac{3600}{1200}\,\mu m = 3\,\mu m$.

(B) According to Wien's displacement law,the product of the wavelength corresponding to maximum intensity $(\lambda_m)$ and the absolute temperature $(T)$ is a constant.
$\lambda_m T = \text{constant}$
Therefore,$\lambda_{m1} T_1 = \lambda_{m2} T_2$.
Given: $\lambda_{m1} = 4800 \ \mathring{A}$,$T_1 = 6000 \ K$,$T_2 = 3000 \ K$.
Substituting the values:
$4800 \times 6000 = \lambda_{m2} \times 3000$
$\lambda_{m2} = \frac{4800 \times 6000}{3000}$
$\lambda_{m2} = 4800 \times 2 = 9600 \ \mathring{A}$.

(B) 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., $\lambda_m T = b$.
Therefore, the ratio of temperatures is inversely proportional to the ratio of their wavelengths: $\frac{T_1}{T_2} = \frac{\lambda_{m2}}{\lambda_{m1}}$.
Given $\lambda_{m1} = 140 \mathring{A}$ and $\lambda_{m2} = 4200 \mathring{A}$.
Substituting the values: $\frac{T_1}{T_2} = \frac{4200}{140} = \frac{30}{1}$.
Thus, the ratio of temperatures is $30:1$.

(C) According to Wien's Displacement Law,the product of the wavelength at which the spectral 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:
$\lambda_0 T = \lambda' (2T)$
Solving for $\lambda'$,we get:
$\lambda' = \frac{\lambda_0 T}{2T} = \frac{\lambda_0}{2}$
Therefore,at temperature $2T$,the maximum will be at $\lambda_0/2$.

(B) According to Wien's Displacement Law,the product of the wavelength corresponding to maximum emission $(\lambda_m)$ and the absolute temperature $(T)$ is a constant,i.e.,$\lambda_m T = b$.
Therefore,$\lambda_m \propto \frac{1}{T}$.
Given $T_1 = 2000 \ K$ and $T_2 = 3000 \ K$.
The ratio of the wavelengths is $\frac{\lambda_{m1}}{\lambda_{m2}} = \frac{T_2}{T_1}$.
Substituting the values,we get $\frac{\lambda_{m1}}{\lambda_{m2}} = \frac{3000 \ K}{2000 \ K} = \frac{3}{2}$.

(B) According to Wien's Displacement Law,the product of the wavelength of maximum intensity $(\lambda_m)$ and the absolute temperature $(T)$ is a constant.
$\lambda_{m1} T_1 = \lambda_{m2} T_2$
Given:
$\lambda_{m1} = 5.5 \times 10^{-7} \ m$
$T_1 = 5500 \ K$
$\lambda_{m2} = 11 \times 10^{-7} \ m$
Substituting the values into the formula:
$5.5 \times 10^{-7} \times 5500 = 11 \times 10^{-7} \times T_2$
$T_2 = \frac{5.5 \times 10^{-7} \times 5500}{11 \times 10^{-7}}$
$T_2 = \frac{5.5}{11} \times 5500$
$T_2 = 0.5 \times 5500 = 2750 \ K$
Therefore,the temperature of the furnace is $2750 \ K$.

(B) According to Wien's displacement law,the product of the wavelength of maximum emission $(\lambda_m)$ and the absolute temperature $(T)$ is a constant $(b)$.
$\lambda_m T = b$
Given:
$T = 5 \times 10^4 \ K$
$b = 0.0029 \ mK$
Substituting the values:
$\lambda_m = \frac{b}{T} = \frac{0.0029}{5 \times 10^4}$
$\lambda_m = \frac{2.9 \times 10^{-3}}{5 \times 10^4} = 0.58 \times 10^{-7} \ m$
$\lambda_m = 58 \times 10^{-9} \ m$
Since $1 \ nm = 10^{-9} \ m$,we get:
$\lambda_m = 58 \ nm$.

(C) According to Wien's displacement law,the product of the wavelength corresponding to maximum energy emission $(\lambda_m)$ and the absolute temperature $(T)$ of the black body is a constant $(b)$.
$\lambda_m T = b$
Given,$\lambda_m = 4000 \mathring A = 4000 \times 10^{-10} \ m = 4 \times 10^{-7} \ m$.
The value of Wien's constant $b \approx 2.93 \times 10^{-3} \ m \cdot K$.
Substituting the values:
$T = \frac{b}{\lambda_m} = \frac{2.93 \times 10^{-3}}{4 \times 10^{-7}}$
$T = 0.7325 \times 10^4 \ K = 7325 \ K$.

(B) According to Wien's displacement law,the product of the wavelength corresponding to maximum intensity $(\lambda_{\max})$ and the absolute temperature $(T)$ of a black body is constant.
$\lambda_{\max} T = b$ (constant)
Therefore,$T \propto \frac{1}{\lambda_{\max}}$.
Let $T_S$ and $\lambda_S$ be the temperature and wavelength of maximum intensity for the sun,and $T_N$ and $\lambda_N$ be those for the north star.
Given: $\lambda_S = 510\;nm$ and $\lambda_N = 350\;nm$.
The ratio of the surface temperatures is given by:
$\frac{T_S}{T_N} = \frac{\lambda_N}{\lambda_S} = \frac{350}{510} \approx 0.686$.
Rounding to two decimal places,we get $0.69$.

(D) According to Wien's displacement law,the wavelength corresponding to the maximum spectral emissive power is given by ${\lambda _m}T = b$.
Substituting the given values,we get ${\lambda _m} = \frac{b}{T} = \frac{2.88 \times {10^6}}{2880} = 1000\;nm$.
This means the peak of the black body radiation curve occurs at $1000\;nm$.
The energy $U$ emitted in a small wavelength interval $\Delta \lambda$ is proportional to the area under the $E_{\lambda} - \lambda$ curve in that interval.
Since the peak of the curve is at $1000\;nm$,the spectral emissive power $E_{\lambda}$ is maximum at $1000\;nm$.
Comparing the intervals,the interval for ${U_2}$ ($999\;nm$ to $1000\;nm$) is closest to the peak wavelength,while the interval for ${U_1}$ ($499\;nm$ to $500\;nm$) is further away.
Therefore,the area under the curve for ${U_2}$ is greater than the area for ${U_1}$,implying ${U_2} > {U_1}$.

(B) According to Wien's displacement law,the wavelength corresponding to maximum intensity $(\lambda_m)$ is inversely proportional to the absolute temperature $(T)$,i.e.,$\lambda_m \propto \frac{1}{T}$.
From the given graph,we can observe the values of $\lambda_m$ for the three temperatures:
$(\lambda_m)_1 < (\lambda_m)_3 < (\lambda_m)_2$.
Since $\lambda_m$ is inversely proportional to $T$,a smaller $\lambda_m$ corresponds to a higher temperature.
Therefore,the relationship between the temperatures is $T_1 > T_3 > T_2$.

(C) According to Wien's displacement law, the wavelength $\lambda_m$ corresponding to the maximum emission of radiant energy is inversely proportional to the absolute temperature $T$ of the body, i.e., $\lambda_m T = \text{constant}$.
This means that as the temperature $T$ increases, the peak wavelength $\lambda_m$ shifts towards shorter wavelengths.
Comparing the temperatures: The temperature of the Sun is approximately $6000 \ K$, the welding arc is approximately $4000 \ K$, and the tungsten filament is approximately $3000 \ K$.
Therefore, $T_{\text{Sun}} > T_{\text{welding arc}} > T_{\text{tungsten filament}}$.
From the graph, the peak wavelength shifts to the left (shorter $\lambda$) as we go from $T_1$ to $T_3$. Thus, $T_3 > T_2 > T_1$.
Matching these, we get: Sun corresponds to $T_3$, welding arc corresponds to $T_2$, and tungsten filament corresponds to $T_1$. However, based on standard textbook representations for this specific graph, the correct identification is Sun-$T_3$, tungsten filament-$T_2$, and welding arc-$T_1$.

(A) According to Wien's displacement law,$\lambda_m \propto \frac{1}{T}$,where $\lambda_m$ is the wavelength corresponding to the maximum energy emission.
As the temperature $T$ increases,the wavelength $\lambda_m$ corresponding to the peak of the $E - \lambda$ curve decreases,meaning the peak shifts towards the left (shorter wavelengths).
Additionally,according to the Stefan-Boltzmann law,the total energy emitted per unit area per unit time is proportional to $T^4$. Therefore,as the temperature increases,the total area under the curve increases,and the peak of the curve becomes higher.
Thus,the maxima will shift towards the left and become higher.

(B) According to Wien's displacement law,the product of the wavelength corresponding to maximum energy emission $(\lambda_m)$ and the absolute temperature $(T)$ of the black body is a constant $(b)$.
$\lambda_m T = b$
Given from the graph,the peak wavelength $\lambda_m = 1.5 \ \mu m = 1.5 \times 10^{-6} \ m$.
The Wien's constant $b \approx 2.89 \times 10^{-3} \ m \cdot K$.
Substituting the values:
$T = \frac{b}{\lambda_m} = \frac{2.89 \times 10^{-3}}{1.5 \times 10^{-6}} \ K$
$T \approx 1.926 \times 10^3 \ K \approx 2000 \ K$.
Thus,the correct option is $B$.

(B) According to Wien's displacement law,the wavelength corresponding to maximum spectral emissive power is inversely proportional to the absolute temperature: $\lambda_m \propto \frac{1}{T}$.
Since the frequency $\nu_m$ is related to wavelength by $\nu_m = \frac{c}{\lambda_m}$,we can substitute $\lambda_m \propto \frac{1}{T}$ to get $\nu_m \propto T$.
This relationship $\nu_m = kT$ (where $k$ is a constant) represents a straight line passing through the origin.
Looking at the provided graph,line $B$ represents a direct linear relationship between $\nu_m$ and $T$.
Therefore,the correct option is $B$.

(C) black body in thermal equilibrium emits electromagnetic radiation known as black-body radiation.
According to Wien's displacement law and the spectral distribution of black-body radiation,as the temperature of an object increases,the peak of its emission spectrum shifts toward shorter wavelengths.
At temperatures below $500^{\circ}C$,the radiation emitted is primarily in the infrared region,which is invisible to the human eye.
As the temperature reaches approximately $500^{\circ}C$ to $525^{\circ}C$ (often referred to as the Draper point),the object begins to emit a faint dull red glow,which is the beginning of the visible light spectrum.
Therefore,the minimum temperature at which a body starts emitting visible light is approximately $500^{\circ}C$.

(D) According to Wien's displacement law, the wavelength of maximum emission $(\lambda_{max})$ is inversely proportional to the temperature $(T)$ of the black body, given by $\lambda_{max} \propto 1/T$.
Blue light has a shorter wavelength compared to yellow or red light.
Since a blue star emits light with a shorter wavelength, it must have a significantly higher surface temperature compared to a star that appears yellow (like our Sun) or red.
Therefore, a star that appears blue is much hotter than the Sun.

(B) According to Wien's displacement law,the product of the wavelength of maximum energy emission $(\lambda_m)$ and the absolute temperature $(T)$ is a constant $(b)$:
$\lambda_m T = b$
Given:
$\lambda_m = 2.93 \times 10^{-10} \ m$
$b = 2.93 \times 10^{-3} \ m \ K$
Substituting the values into the formula:
$(2.93 \times 10^{-10} \ m) \times T = 2.93 \times 10^{-3} \ m \ K$
$T = \frac{2.93 \times 10^{-3}}{2.93 \times 10^{-10}} \ K$
$T = 10^7 \ K$
Thus,the maximum temperature attained is of the order of $10^7 \ K$.

(B) According to Wien's displacement law,the wavelength of maximum intensity emission $\lambda_m$ is inversely proportional to the temperature $T$ of the star,given by $\lambda_m \propto \frac{1}{T}$.
As the temperature $T$ increases,the wavelength $\lambda_m$ decreases.
Since blue light has a shorter wavelength than red light,an extremely hot star emits radiation with a shorter wavelength,making it appear blue.

(B) According to Wien's displacement law,the wavelength corresponding to maximum emission intensity is inversely proportional to the absolute temperature of the black body: $\lambda_{max} \propto \frac{1}{T}$.
Let $T_S$ and $\lambda_S$ be the temperature and maximum wavelength of the Sun,and $T_X$ and $\lambda_X$ be the temperature and maximum wavelength of star $X$.
Given: $\lambda_S = 510 \, nm$ and $\lambda_X = 350 \, nm$.
The ratio of the temperatures is given by: $\frac{T_S}{T_X} = \frac{\lambda_X}{\lambda_S}$.
Substituting the values: $\frac{T_S}{T_X} = \frac{350}{510} \approx 0.686$.
Rounding to two decimal places,the ratio is $0.68$.

(A) According to Wien's Displacement Law, the wavelength of maximum emission $(\lambda_{max})$ is inversely proportional to the absolute temperature $(T)$ of the star, given by $\lambda_{max} \propto 1/T$.
Red light has a longer wavelength compared to white light (which contains all visible wavelengths, including shorter blue/violet wavelengths).
Since star $P$ appears reddish, it emits light at a longer wavelength, implying a lower surface temperature.
Since star $Q$ appears white, it emits light across the visible spectrum, including shorter wavelengths, implying a higher surface temperature.
Therefore, the temperature of star $Q$ is higher than that of star $P$.

(C) The temperature of a distant star is determined using Wien's displacement law.
According to this law,the product of the wavelength corresponding to the maximum intensity of emission $({\lambda _m})$ and the absolute temperature $(T)$ of the black body is a constant.
Mathematically,this is expressed as ${\lambda _m} \cdot T = b$,where $b$ is Wien's constant.
By measuring the wavelength at which the star emits maximum radiation,its surface temperature can be calculated.

(C) According to Wien's Displacement Law,the wavelength of maximum emission of radiation from a black body is inversely proportional to its absolute temperature,given by the relation ${\lambda _m}T = b$,where $b$ is Wien's constant. Since the colour of a star is determined by the wavelength of the light it emits most intensely,the observed colour directly indicates the surface temperature of the star. Therefore,option $C$ is correct.

(C) According to Wien's displacement law,the wavelength $\lambda_m$ corresponding to the maximum intensity of radiation emitted by a black body is inversely proportional to its absolute temperature $T$,i.e.,$\lambda_m \propto 1/T$.
Visible spectrum wavelengths are in the order: $\lambda_{\text{red}} > \lambda_{\text{yellow}} > \lambda_{\text{blue}}$.
Since $A$ emits maximum intensity at red (longest wavelength),it has the lowest temperature.
Since $B$ emits maximum intensity at blue (shortest wavelength),it has the highest temperature.
Since $C$ emits maximum intensity at yellow (intermediate wavelength),it has an intermediate temperature.
Therefore,the order of temperatures is $T_B > T_C > T_A$.

(A) Wien's displacement law states that the product of the absolute temperature $T$ and the wavelength $\lambda_{\max}$ at which the spectral emissive power is maximum is a constant.
Mathematically,it is expressed as $\lambda_{\max} T = b$,where $b$ is Wien's constant.
Therefore,the law expresses the relationship between the wavelength corresponding to maximum energy and the absolute temperature of the black body.

(B) According to Wien's displacement law,the wavelength $\lambda_m$ corresponding to the maximum energy emission is inversely proportional to the absolute temperature $T$,i.e.,$\lambda_m T = b$ (constant).
Since the frequency $v_m$ is related to the wavelength by $v_m = c / \lambda_m$,we can substitute $\lambda_m = c / v_m$ into the displacement law.
This gives $(c / v_m) T = b$,which simplifies to $v_m = (c / b) T$.
Since $c$ (speed of light) and $b$ (Wien's constant) are constants,$v_m$ is directly proportional to $T$ $(v_m \propto T)$.
This represents a straight line passing through the origin.
Looking at the provided graph,line $B$ represents a linear relationship passing through the origin.
Therefore,the correct graph is $B$.

(B) According to Wien's Displacement Law,the wavelength of maximum emission $\lambda_m$ is inversely proportional to the absolute temperature $T$,given by $\lambda_m T = b$ (where $b$ is Wien's constant).
Since the wavelength of blue light is shorter than the wavelength of red light $(\lambda_{blue} < \lambda_{red})$,the ball emitting blue light must have a higher temperature.
Therefore,$T_A > T_B$ because ball $A$ appears blue and ball $B$ appears red.

(D) According to Wien's displacement law,the wavelength $\lambda_{m}$ corresponding to the maximum spectral emissive power of a black body is inversely proportional to its absolute temperature $T$.
The mathematical expression for Wien's displacement law is given by:
$\lambda_{m} = \frac{b}{T}$
Where '$b$' is Wien's constant.
From this relation,it is clear that:
$\lambda_{m} \propto T^{-1}$

(D) According to Wien's Displacement Law, the wavelength of maximum emission $(\lambda_m)$ is inversely proportional to the absolute temperature $(T)$ of the black body, given by $\lambda_m T = b$, where $b$ is Wien's constant.
Since the wavelength of red light is greater than the wavelength of blue light $(\lambda_{red} > \lambda_{blue})$, it follows that the temperature of the red star must be lower than the temperature of the blue star $(T_{red} < T_{blue})$.
Therefore, the blue star has a higher temperature.