Animated Solution for Physics - Properties of Solids and Liquids: Match the temperature of a black body given in List-I with an appropriate statement in List-II, and choose the correct option. [Given: Wien's constant as $2.9\times 10^{-3}\text{ m-K}$ and $\frac{hc}{e}=1.24\times 10^{-6}\text{ V-m}$]

This problem is a beautiful masterpiece that seamlessly weaves together concepts from Thermodynamics, Wave Optics, and Modern Physics. To conquer it, we must rely on two fundamental pillars of blackbody radiation:

1. Wien's Displacement Law: The peak emission wavelength ${\lambda }_m$ is inversely proportional to the absolute temperature $T$. Mathematically, ${\lambda }_{m}T=b$, where $b=2.9\times 10^{-3}\text{ m-K}$. 2. Stefan-Boltzmann Law: The total power emitted per unit area is proportional to the fourth power of the absolute temperature, $P\propto T^4$.

Armed with these laws, let's decode the mysteries hidden within each temperature.

The Coldest Star and the Widest Wave

Let's begin with the lowest temperature, $T=2000\text{ K}$.

Using Wien's Displacement Law, we can calculate the peak wavelength:

Because $2000\text{ K}$ is the lowest temperature among the choices, it naturally produces the longest peak wavelength. Now, let's bridge this to Wave Optics. In single-slit diffraction, the angular spread of the central maximum is governed by $\sin \theta =\frac{\lambda }{a}$. A larger wavelength $\lambda$ directly results in a wider central maximum. Therefore, the radiation at $2000\text{ K}$ will produce the widest central maximum.

Conclusion: (P) matches with statement (3).

The Power of the Fourth Power

Next, we examine $T=3000\text{ K}$.

While we could calculate its peak wavelength (which turns out to be $966\text{ nm}$, deep in the infrared region), none of the statements in List-II correspond to an infrared wave. Instead, let's look at the power emitted. According to the Stefan-Boltzmann Law, $P=\sigma T^4$.

Let's compare the power emitted at $3000\text{ K}$ to the power emitted at $6000\text{ K}$:

This is an exact match for statement (4), proving that the power emitted per unit area is indeed $1/16$ of that emitted by a blackbody at $6000\text{ K}$.

Conclusion: (Q) matches with statement (4).

The Light We Can See

Moving up the temperature scale, we reach $T=5000\text{ K}$.

Let's find its peak wavelength using Wien's Law:

Does $580\text{ nm}$ sound familiar? The human eye has evolved to detect a very specific band of electromagnetic radiation, roughly between $400\text{ nm}$ and $700\text{ nm}$. Since $580\text{ nm}$ falls perfectly within this visible spectrum (appearing as a bright yellow-green), this radiation is clearly visible to the human eye.

Conclusion: (R) matches with statement (2).

The Ultraviolet Photoelectric Kick

Finally, we arrive at the hottest body, $T=10000\text{ K}$.

Its peak wavelength is the shortest:

This wavelength lies in the ultraviolet (UV) region. To see if it can trigger the photoelectric effect, we must calculate the energy of a single photon at this wavelength. Using the given constant $\frac{hc}{e}=1.24\times 10^{-6}\text{ V-m}$:

The energy of the incident photon ($4.27\text{ eV}$) is strictly greater than the metal's work function ($4\text{ eV}$). Because $E>\varphi$, these highly energetic UV photons will successfully knock out photoelectrons from the metal surface.

Conclusion: (S) matches with statement (1).

By trusting the fundamental laws of physics, we have systematically decoded the entire matrix. The correct mapping is P $\to$ 3, Q $\to$ 4, R $\to$ 2, S $\to$ 1.