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 × 10−3 m-K and ℎ𝑐 𝑒 = 1.24 × 10−6 V-m] List-I List-II (P) 2000 K (1) The radiation at peak wavelength can lead to emission of photoelectrons from a metal of work function 4 eV. (Q) 3000 K (2) The radiation at peak wavelength is visible to human eye. (R) 5000 K (3) The radiation at peak emission wavelength will result in the widest central maximum of a single slit diffraction. (S) 10000 K (4) The power emitted per unit area is 1/16 of that emitted by a blackbody at temperature 6000 K. (5) The radiation at peak emission wavelength can be used to image human bones. (A) 𝑃 → 3,𝑄 → 5, 𝑅 → 2, 𝑆 → 3 (B) 𝑃 → 3,𝑄 → 2, 𝑅 → 4, 𝑆 → 1 (C) 𝑃 → 3,𝑄 → 4, 𝑅 → 2, 𝑆 → 1 (D) 𝑃 → 1,𝑄 → 2, 𝑅 → 5, 𝑆 → 3

 To solve this problem, we need to match the temperature of a black body given in List-I with the appropriate statement in List-II using Wien's displacement law and the given constants.

Step 1: Understand Wien's Displacement Law

Wien's displacement law states that the wavelength at which a black body emits the most radiation (peak wavelength, ${\lambda }_{\text{max}}$) is inversely proportional to its temperature $T$:

$${\lambda }_{\text{max}}=\frac{b}{T}$$

where $b=2.9\times 10^{-3}\text{m-K}$ is Wien's constant.

Step 2: Calculate ${\lambda }_{\text{max}}$ for Each Temperature

1. For $T=2000\text{K}$:

   $${\lambda }_{\text{max}}=\frac{2.9\times 10^{-3}}{2000}=1.45\times 10^{-6}\text{m}=1450\text{nm}$$

   This is in the infrared region.

2. For $T=3000\text{K}$:

   $${\lambda }_{\text{max}}=\frac{2.9\times 10^{-3}}{3000}=9.67\times 10^{-7}\text{m}=967\text{nm}$$

   This is near the visible region but still mostly infrared.

3. For $T=5000\text{K}$:

   $${\lambda }_{\text{max}}=\frac{2.9\times 10^{-3}}{5000}=5.8\times 10^{-7}\text{m}=580\text{nm}$$

   This is in the visible region (yellow light).

4. For $T=10000\text{K}$:

   $${\lambda }_{\text{max}}=\frac{2.9\times 10^{-3}}{10000}=2.9\times 10^{-7}\text{m}=290\text{nm}$$

   This is in the ultraviolet region.

Step 3: Match Each Temperature with the Correct Statement

1. $P=2000\text{K}$:

   • The peak wavelength is 1450 nm (infrared). This does not match any of the statements directly, but it is closest to (3) because infrared radiation can result in a wide central maximum in diffraction.

2. $Q=3000\text{K}$:

   • The peak wavelength is 967 nm (near infrared). This does not match any statement directly, but it is closest to (4) because the power emitted per unit area at 3000 K is less than at 6000 K.

3. $R=5000\text{K}$:

   • The peak wavelength is 580 nm (visible). This matches (2) because the radiation is visible to the human eye.

4. $S=10000\text{K}$:

   • The peak wavelength is 290 nm (ultraviolet). This matches (1) because ultraviolet radiation can emit photoelectrons from a metal with a work function of 4 eV.

Step 4: Choose the Correct Option

Based on the above analysis:

• $P\to 3$

• $Q\to 4$

• $R\to 2$

• $S\to 1$

The correct option is (C):

$$\boxed{{\displaystyle C}}$$