Black Body Radiation

To solve this problem, we need to match the temperature of a black body from List-I with the appropriate statement from List-II using Wien's Law and other given properties of blackbody radiation. Wien's Law is given by:

, where .

Let's calculate for each temperature in List-I:

1. 2000 K:
   (Infrared region, used in wide diffraction patterns)
2. 3000 K:
   (Near visible region, corresponds to work function explanations)
3. 5000 K:
   (Visible light, human eye sensitivity peak)
4. 10000 K:
   (UV region, photoelectron emission possible)
   The power per unit area comparison with 6000 K is necessary for context.

Next, let's match these with the statements in List-II:

1. (P) 2000 K (3) The radiation at peak emission wavelength will result in the widest central maximum of a single slit diffraction.
2. (Q) 3000 K (4) The power emitted per unit area is of that emitted by a blackbody at temperature 6000 K.
3. (R) 5000 K (2) The radiation at peak wavelength is visible to the human eye.
4. (S) 10000 K (1) The radiation at peak wavelength can lead to emission of photoelectrons from a metal of work function 4 eV.

Correct Matches: P 3, Q 4, R 2, S 1