Wein Displacement Law

Wien's Displacement Law is a foundational principle in the study of thermal radiation physics. It asserts that the wavelength at which a blackbody emits radiation most intensely is inversely proportional to its absolute temperature. This law is  essential for comprehending concepts like thermal radiation and blackbody emission spectra, offering insights into the energy distribution of objects based on their temperatures.

It pertains specifically to blackbodies, theoretical objects that absorb all incident radiation and emit radiation at all wavelengths. This Law plays a pivotal role in deducing the temperatures of stars by analyzing the properties of their emitted radiation.

1.0Statement Of Wien's Displacement Law

The wavelength (m) corresponds to the maximum emission of radiation decreasing with increasing temperature. This is known as Wien's Displacement Law.

${\lambda }_m\propto \frac{1}{T}$

${\lambda }_m\cdot \mid T=Weinconstant(b)$

• ${\lambda }_{max}$ is the wavelength of maximum emission.
• T represents the Absolute temperature of the blackbody in Kelvin
•  Where b is Wein’s Constant = 2.89 $\times 10^{-3}\mathrm{mK}$

2.0Spectral Emissive Power(E)

It is the emissive power about a particular wavelength. The amount of heat radiation emitted by unit area of the body in one second in unit spectral region at a given wavelength.

• For GB(General Body or Gray): $E_{\lambda }=\frac{{\left(Q_{\lambda \mathring{A}}\right)}_{GB}}{At}$
• For IBB(Ideal Black Body): $E_{\lambda }=\frac{{\left(Q_{\lambda A}\right)}_{IBB}}{At}$
• SI Unit- W/m² - Å
• Spectral emissive power, also known as spectral radiance, quantifies the radiant flux emitted per unit area of the emitting surface, per unit solid angle and per unit wavelength. It quantifies how much radiant energy is emitted from a surface at a specific wavelength and direction.
• Spectral emissive power is essential in understanding the detailed characteristics of thermal radiation emitted by objects.

3.0Blackbody Radiations

Electromagnetic radiation emitted by a theoretical idealized object known as a blackbody characterizes its unique nature. A blackbody absorbs all incident radiation without reflecting  at any wavelength and emits radiation across all wavelengths with optimal efficiency. While real objects do not perfectly match these conditions, blackbodies are vital theoretical constructs. They emit radiation over a continuous spectrum of wavelengths spanning from infrared through visible light and ultraviolet, extending further into microwave and radio frequencies at lower temperatures, creating a seamless and uninterrupted emission spectrum.

4.0Characteristics of an Ideal Black Body

• The nature of emitted radiations from the surface of an ideal black body depends only on its temperature.
• The radiations emitted from the surface of an ideal black body are called either full or white radiations.
• At any temperature the spectral energy distribution curve for the surface of an ideal black body is always continuous and according to this concept if the spectrum of a heat source obtained is continuous then it must be an ideal black body.
• For an ideal black body, $a=a_{\lambda }=1;r=t=0;e_r=1$
• At low temperature, the surface of the ideal black body is a perfect absorber and at a high temperature it proves to be a perfect emitter.

Examples of Ideal Black Body

There are two experimentally ideal black body

(a).Wein’s Ideal Black Body

(b).Ferry’s Ideal Black Body

Note: An ideal black body need not be of black color(eg. Sun)

5.0Wein’s Displacement Formula

Wien's Displacement Law states that as the temperature of a blackbody increases, the wavelength at which it emits radiation most intensely decreases.

• ${\lambda }_m\propto \frac{1}{T}\Rightarrow {\lambda }_m=\frac{b}{T}$
• For numerical $\Rightarrow {\lambda }_m\mathrm{T}=\mathrm{b}$ (Wein Constant =2.89 $\times 10^{-3}\mathrm{mK}$)
• ${\lambda }_{m1}{T}_1={\lambda }_{m2}{T}_2$
• Relation between frequency and temperature $v_m=\frac{c}{b}\mathrm{T}[\therefore \mathrm{c}=\lambda \times \mathrm{v}]$

6.0Variation of Spectral Radiation Density with wavelength

• At any temperature the spectral energy distribution curve for the surface of an ideal black body is always continuous.
• Area = ${\int }_0^{\infty }{E}_{\lambda }d\lambda =\mathrm{E}=\sigma T^4$

$\mathrm{E}\propto T^4$       $\left(\therefore \text{ Area }\propto T^4\right)$

Hence, $\frac{A_1}{A_2}={\left[\frac{T_1}{T_2}\right]}^4$

• If the temperature of a black body is increased the wavelength corresponding to maximum emission of radiation is shifted from longer wavelength to shorter wavelength, due to this color of the body changes.

$T_3>T_2>T_1:{\lambda }_{m3}<{\lambda }_{m2}<{\lambda }_{m1}$

7.0Key Points Of Wien Displacement Law

• Spectral energy distribution curves are continuous. At any temperature in all possible wavelengths radiation between (0 to ∞) are emitted but quantity of radiation is different for different wavelengths.
• As the wavelength increases, the amount of radiation emitted first increases, becomes maximum and then decreases.

8.0Sample Questions in Wien's Displacement Law

Q-1. Temperatures of two stars are in ratio 3:2. If the wavelength of maximum radiation from the first body is 4000 Å, what is the corresponding wavelength of the second body?

Solution:  

${\lambda }_{\max }\propto \frac{1}{T}\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=\frac{T_1}{T_2}$

$|\frac{{\lambda }_2}{4000}=\frac{3}{2}$

${\lambda }_2=6000\mathring{A}$

Q-2. Two spherical ideal black bodies of radii r1and r2 are having surface temperature $T_1$ and $T_2$ respectively, if both radiate the same power. Then calculate the ratio of $T_1$$T_2$.

Solution:

$P_1=P_2$

$\mid A_1\sigma T_1^4=A_2\sigma T_2^4$

$\pi r_1^2\sigma T_1^4=\pi r_2^2\sigma T_2^4$

$r_1^2{T}_1^4=r_2^2{T}_2^4$

$\frac{T_1}{T_2}=\sqrt{\frac{r_2}{r_1}}$

Q-3. Why does an optical pyrometer calibrated for ideal blackbody radiation provide an inaccurate temperature reading for a red-hot iron piece in open air, but give an accurate reading when the same piece is placed inside a furnace?

Solution:

Let T denote the temperature of the heated iron in the furnace. Heat radiated per second per unit area

$\Rightarrow \mathrm{E}=\sigma T^4$

When the body is placed in the open at temperature $T_0$, then $heat/radiated/second/area,\Rightarrow E^{\prime }=\sigma$

$\Rightarrow \left(T^4-T_0^4\right)$

Since ( E' < E ), it is evident that the optical pyrometer provides a temperature reading that is too low when used in an open environment.

Q-4.Two bodies A and B have thermal emissivity of 0.01 and 0.81 respectively. The outer surface areas of two bodies are equal, and both bodies emit radiant power at the same rate. The wavelength ${\lambda }_b$ corresponding to the maximum spectral radiance of body B is shifted by 1 meter from the wavelength corresponding to the maximum spectral radiance of body A. Given that the temperature of body A is 5802 K, calculate:

1. The temperature of B
2. Wavelength b

Solution:

1. As the bodies A and B having same radiant power

$\therefore P_A=P_B\Rightarrow e_A\sigma A_A{T}_A^4=e_B\sigma A_B{T}_B^4$

$\Rightarrow (0.01)\sigma A{T}_A^4=0.81\sigma A{T}_B^4$

$\Rightarrow T_B={\left(\frac{0.01}{0.81}\right)}^{1/4}{T}_A$

$\Rightarrow T_B=\frac{T_A}{3}=\frac{5802}{3}=1934\mathrm{K}$

2. According to Wien's displacement Law

${\lambda }_A{T}_A={\lambda }_B{T}_B\Rightarrow {\lambda }_B=\left(\frac{5802}{1934}\right){\lambda }_A$

$\Rightarrow {\lambda }_B=3{\lambda }_A$

As ${\lambda }_B-{\lambda }_A=1\mathrm{m}$

$\Rightarrow {\lambda }_B-\frac{{\lambda }_B}{3}=1\mathrm{m}$

$\Rightarrow \frac{2{\lambda }_B}{3}=1\mathrm{m}$

$\Rightarrow {\lambda }_B=1.5\mathrm{m}$

Q-5. The figure depicts the spectral energy distribution of radiation emitted by a blackbody at a certain temperature. Determine the potential temperature of the blackbody is $\left(b=3\times 10^{-3}\mathrm{mK}\right)$.

Solution:

Using Wien's Law,                                                                         

$\Rightarrow \left(1.5\times 10^{-6}\right)\mathrm{T}=3\times 10^{-3}$        

$\Rightarrow \left(1.5\times 10^{-6}\right)\mathrm{T}=3\times 10^{-3}$

$\Rightarrow T=\frac{3\times 10^{-3}}{1.5\times 10^{-6}}=2\times 10^3=2000\mathrm{K}$