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Wein’s Displacement Law

Wein’s Displacement Law :- In 1893, the German physicist Wilhelm Wien discovered an important relationship between the temperature of a black body and the wavelength at which it emits radiation most strongly. This relationship is called Wien’s Displacement Law.

Statement of Wien’s Displacement Law

The wavelength corresponding to maximum emission of a black body is inversely proportional to its absolute temperature.

Mathematically,

$\displaystyle {{\lambda }_{m}}\propto \frac{1}{T}$

$\displaystyle {{\lambda }_{m}}=\frac{b}{T}$

$\displaystyle {{\lambda }_{m}}T=b(constant)$

where :

= wavelength at which maximum emission occurs
$T$ = absolute temperature of the black body (in Kelvin)
$b$ = Wien’s constant =
Dimensions of b = [M⁰LT⁰ θ]

Wein’s Displacement Law in terms of frequency and temperature

$\displaystyle c={{\nu }_{m}}{{\lambda }_{m}}={{\nu }_{m}}\times \frac{b}{T}$

$\displaystyle \Rightarrow {{\nu }_{m}}=\frac{c}{b}T$

Explanation

Wien’s Displacement Law establishes a fundamental connection between the color (wavelength) of the radiation and the temperature of the radiating body. It explains why heated objects change color as their temperature rises—from red hot to white hot.
As the temperature of a black body increases, the peak of the emission curve shifts toward shorter wavelengths.
This means that hotter bodies emit radiation of shorter wavelengths (towards the blue/violet region), while cooler bodies emit radiation of longer wavelengths (towards the red/infrared region).

Examples

At room temperature (), the peak wavelength lies in the infrared region, which is why we cannot see it with naked eyes but can feel it as heat.
At the temperature of the Sun’s surface (), the peak wavelength lies in the visible region (around yellow-green light), which is why sunlight appears bright to our eyes.

Example 1.

Solar radiation is found to have an intensity maximum near the wavelength range of 470 nm. Assuming the Sun’s surface behaves as a perfect absorber (a = 1), calculate the temperature of the solar surface.

Solution :

As a = 1, Sun can be assumed as a black body. From Wein’s Displacement Law,

$\displaystyle {{\lambda }_{m}}=\frac{b}{T}$

$\displaystyle \Rightarrow T=\frac{b}{{{\lambda }_{m}}}=\frac{2.89\times {{10}^{-3}}}{470\times {{10}^{-9}}}\approx 6000K$

Example 2.

The temperature of a furnace is 2000°C. The maximum intensity in its spectrum is obtained at 4000 Å. If the maximum intensity is at 2000 Å, then calculate the temperature of the furnace in °C.

Solution :

$\displaystyle {{\lambda }_{m}}T=b(constant)$

$\displaystyle \Rightarrow {{\left( {{\lambda }_{m}} \right)}_{1}}{{T}_{1}}={{\left( {{\lambda }_{m}} \right)}_{2}}{{T}_{2}}$

$\displaystyle \Rightarrow {{T}_{2}}=\frac{{{\left( {{\lambda }_{m}} \right)}_{1}}}{{{\left( {{\lambda }_{m}} \right)}_{2}}}\times {{T}_{1}}=\frac{4000{\AA}}{2000\text{ }{\AA}}\times (2000+273)K$

$\displaystyle \Rightarrow {{T}_{2}}=4546K$

$\displaystyle \therefore {{T}_{2}}=(4546-273)=4273{}^\circ C$

Example 3.

A hot black body emits the energy at the rate of 16 Jm^–2s^–1 and its most intense radiation corresponds to 20,000 Å. When the temperature of this body is further increased and its most intense radiation corresponds to 10,000 Å, then find the value of energy radiated in Jm^–2s^–1 .

Solution :

From Wein’s Displacement Law,

$\displaystyle T\propto \frac{1}{{{\lambda }_{m}}}$

Here, becomes half ⇒ Temperature doubles.

Now from Stefan’s Law,

$\displaystyle E=\sigma {{T}^{4}}$

$\displaystyle \Rightarrow \frac{{{E}_{1}}}{{{E}_{2}}}={{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{4}}$

$\displaystyle \Rightarrow {{E}_{2}}={{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{4}}\times {{E}_{1}}={{\left( \frac{2{{T}_{1}}}{{{T}_{1}}} \right)}^{4}}\times 16$

$\displaystyle \therefore {{E}_{2}}=256J{{m}^{-2}}{{s}^{-1}}$

Wein’s Displacement Law

Statement of Wien’s Displacement Law

Wein’s Displacement Law in terms of frequency and temperature

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