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(C) $1$. Wien's Displacement Law states $\lambda_m T = b$, where $b = 2.9 	imes 10^{-3} \, m-K$. Thus, $\lambda_m \propto 1/T$.$2$. For $(P)$ $2000 \

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$P \rightarrow 3, Q \rightarrow 4, R \rightarrow 2, S \rightarrow 1$

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(C) $1$. Wien's Displacement Law states $\lambda_m T = b$, where $b = 2.9 	imes 10^{-3} \, m-K$. Thus, $\lambda_m \propto 1/T$.$2$. For $(P)$ $2000 \, K$: $\lambda_m = (2.9 	imes 10^{-3}) / 2000 = 1.45 	imes 10^{-6} \, m = 1450 \, nm$. Since $\lambda_m$ is inversely proportional to $T$, the lowest temperature $(2000 \, K)$ gives the largest $\lambda_m$. The width of the central maximum in single-slit diffraction is proportional to $\lambda$, so $(P ightarrow 3)$.$3$. For $(Q)$ $3000 \, K$: Power emitted per unit area $E = \sigma T^4$. Ratio $E_{3000}/E_{6000} = (3000/6000)^4 = (1/2)^4 = 1/16$. Thus, $(Q ightarrow 4)$.$4$. For $(R)$ $5000 \, K$: $\lambda_m = (2.9 	imes 10^{-3}) / 5000 = 0.58 	imes 10^{-6} \, m = 580 \, nm$. This wavelength lies in the visible spectrum. Thus, $(R ightarrow 2)$.$5$. For $(S)$ $10000 \, K$: $\lambda_m = (2.9 	imes 10^{-3}) / 10000 = 0.29 	imes 10^{-6} \, m = 290 \, nm$. The energy of a photon $E = hc/\lambda_m = (1.24 	imes 10^{-6}) / (0.29 	imes 10^{-6}) \approx 4.27 \, eV$. Since $4.27 \, eV > 4 \, eV$, it can cause photoelectric emission. Thus, $(S ightarrow 1)$.$6$. Matching: $P ightarrow 3, Q ightarrow 4, R ightarrow 2, S ightarrow 1$. Correct option is $(C)$.

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.

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