Q.Earth's mean temperature can be assumed to be 280 K. How will the curve of blackbody radiation look for this temperature? Find out $\lambda_{max}$. In which part of the electromagnetic spectrum does this value lie? (Take Wien's constant $b = 2.897\times10^{-3}$ m K)

Wien's Displacement Law: From Intuition to Formula

You already know that hot objects glow. A piece of iron heated in a forge first turns dull red, then bright orange, then yellow-white, and finally blue-white as it gets hotter. That colour change is not random — it follows a strict physical rule.

The Core Intuition

Think of temperature as a measure of how violently the atoms in an object are jiggling. When atoms vibrate faster (higher temperature), they emit electromagnetic waves with shorter wavelengths. Shorter wavelength means higher frequency, which means more energy per photon.

So as temperature rises, the "peak" of the emitted radiation shifts toward the blue end of the spectrum. As temperature falls, the peak shifts toward the red end. This is why a cool star looks red and a very hot star looks blue-white.

The Precise Statement

Wien's Displacement Law gives you the exact relationship between the temperature of a blackbody (an ideal thermal emitter) and the wavelength at which its emission is strongest:

where $b$ is Wien's displacement constant:

In most exam problems, you use:

What This Tells You

If you know the temperature of a star, a filament, or any hot object, you can immediately find the wavelength where it shines brightest. Conversely, if you measure the peak wavelength of a star's light, you can calculate its surface temperature.

Example: The Sun's surface temperature is about 5800 K. Its peak wavelength is:

That is green light — right in the middle of the visible spectrum. The Sun actually peaks in green, but our eyes perceive the combined light as white. …