Animated Solution for Physics - Properties of Solids and Liquids: A black body is at a temperature of $2880\text{ K}$. The energy of radiation emitted by this body with wavelength between $499\text{ nm}$ and $500\text{ nm}$ is $U_1$, between $999\text{ nm}$ and $1000\text{ nm}$ is $U_2$ and between $1499\text{ nm}$ and $1500\text{ nm}$ is $U_3$. The Wien constant, $b=2.88\times 10^6\text{ nm-K}$. Then,

Imagine a black body glowing intensely at a very high temperature of $2880\text{ K}$. As it glows, it doesn't just emit light of a single color; instead, it emits radiation across a continuous spectrum of wavelengths. However, the energy is not distributed evenly. There is a specific "sweet spot"—a peak wavelength where the emitted energy is at its absolute maximum. To visualize this, we can plot the spectral emissive power ($E_{\lambda }$) against the wavelength ($\lambda$).

To find exactly where this peak occurs, we rely on a beautiful piece of physics known as Wien's Displacement Law. This law states that the wavelength corresponding to maximum emission (${\lambda }_m$) is inversely proportional to the absolute temperature ($T$) of the black body. Mathematically, it is expressed as:

Here, $b$ is Wien's constant, which is given to us as $2.88\times 10^6\text{ nm-K}$.

Let's substitute our known values into Wien's formula to find the peak wavelength for our specific black body:

Calculating this gives us a clean, exact value:

This means the peak of our energy distribution curve lies exactly at $1000\text{ nm}$.

Now, let's analyze the energy bands provided in the question. We are given three specific intervals, each with a width of exactly $1\text{ nm}$:

$U_1$: Energy between $499\text{ nm}$ and $500\text{ nm}$ $U_2$: Energy between $999\text{ nm}$ and $1000\text{ nm}$ $U_3$*: Energy between $1499\text{ nm}$ and $1500\text{ nm}$

Geometrically, the energy emitted in a specific wavelength interval corresponds to the area under the $E_{\lambda }$ vs $\lambda$ curve for that interval. Since all three intervals have the exact same width ($1\text{ nm}$), the area is directly proportional to the average height of the curve in that region.

Because we established that the curve reaches its absolute maximum peak at ${\lambda }_m=1000\text{ nm}$, the interval for $U_2$ (which sits right at this peak) will naturally have the greatest height, and consequently, the largest area.

Therefore, the energy $U_2$ is strictly greater than both $U_1$ and $U_3$.