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 $$2.9 \times 10^{-3}$$ m-K and $$\frac{hc}{e} = 1.24 \times 10^{-6}$$ 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.

The peak wavelength of the radiation from a black body is given by Wien’s displacement law
$$\lambda_{\text{max}} = \frac{b}{T}, \quad b = 2.9 \times 10^{-3}\ \text{m·K}$$

To interpret each statement of List-II we first work out either $$\lambda_{\text{max}}$$ (for visibility, diffraction etc.) or the emitted power (for Stefan’s law).

$$\lambda_{\text{max}} = \frac{2.9 \times 10^{-3}}{2000} = 1.45 \times 10^{-6}\ \text{m} = 1450\ \text{nm}$$
This lies in the infrared region and has the longest wavelength among the four temperatures. For a single-slit, the angular width of the central maximum $$\propto \lambda$$, so the longest wavelength gives the widest central maximum. Hence statement (3) matches.

By Stefan-Boltzmann law $$P \propto T^{4}$$. If $$P_{3000} / P_{6000} = 1/16$$, then
$$\left(\frac{T_{3000}}{6000}\right)^{4} = \frac{1}{16} \implies T_{3000} = 3000\ \text{K}$$
Thus statement (4) matches.

$$\lambda_{\text{max}} = \frac{2.9 \times 10^{-3}}{5000} = 5.8 \times 10^{-7}\ \text{m} = 580\ \text{nm}$$
This wavelength lies well inside the visible band (400 - 700 nm). Hence statement (2) matches.

Photon energy corresponding to $$\lambda_{\text{max}}$$ must exceed the work function 4 eV.
Using $$E\ (\text{eV}) = \dfrac{hc}{e\lambda}$$ with $$\dfrac{hc}{e} = 1.24 \times 10^{-6}\ \text{V·m}$$,
the threshold wavelength for 4 eV is
$$\lambda_{\text{threshold}} = \frac{1.24 \times 10^{-6}}{4} = 3.1 \times 10^{-7}\ \text{m} = 310\ \text{nm}$$
For $$T = 10000\ \text{K}$$,
$$\lambda_{\text{max}} = \frac{2.9 \times 10^{-3}}{10000} = 2.9 \times 10^{-7}\ \text{m} = 290\ \text{nm}$$
Since $$290\ \text{nm} \lt 310\ \text{nm}$$, these photons have energy $$\gt 4\ \text{eV}$$ and can eject photoelectrons. Hence statement (1) matches.

Collecting the results:
P → 3, Q → 4, R → 2, S → 1

Therefore the correct choice is
Option C which is: P → 3, Q → 4, R → 2, S → 1.