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) $2000\text{K}$	(1) $\text{The radiation at peak wavelength can lead to emission of photoelectrons from a metal of work function}4\text{eV.}$
(Q) $3000\text{K}$	(2) $\text{The radiation at peak wavelength is visible to human eye.}$
(R) $5000\text{K}$	(3) $\text{The radiation at peak emission wavelength will result in the widest central maximum of a single slit diffraction.}$
(S) $10000\text{K}$	(4) $\text{The power emitted per unit area is}\frac{1}{16}\text{of that emitted by a blackbody at temperature}6000\text{K.}$
	(5) $\text{The radiation at peak emission wavelength can be used to image human bones.}$

The correct option is P → 3, Q → 4, R → 2, S → 1.

To match each black body temperature in List-I with the corresponding statement in List-II, let us analyze the physical laws involved step-by-step:

1. Wien's Displacement Law:
The peak wavelength ${\lambda }_m$ of radiation emitted by a black body at absolute temperature $T$ is given by:

${\lambda }_m=\frac{b}{T}$

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

2. Stefan-Boltzmann Law:
The total power emitted per unit area by a black body is proportional to the fourth power of its absolute temperature:

$E=\sigma T^4$

3. Single Slit Diffraction Central Maximum:
The angular width of the central maximum in single slit diffraction is given by:

$\beta =\frac{2\lambda }{a}\propto \lambda$

Thus, a larger wavelength results in a wider central maximum.

Now, let us evaluate each item in List-I:

Analysis of (P) $T=2000\text{K}$:
The peak wavelength is calculated as:

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

Among all given temperatures, $T=2000\text{K}$ produces the longest peak wavelength ($1450\text{nm}$). Since diffraction width is directly proportional to wavelength, this peak wavelength results in the widest central maximum.
P matches with (3).

Analysis of (Q) $T=3000\text{K}$:
Using Stefan-Boltzmann law, compare the power per unit area emitted at $3000\text{K}$ to that at $6000\text{K}$:

$\frac{E_{3000}}{E_{6000}}={\left(\frac{3000}{6000}\right)}^4={\left(\frac{1}{2}\right)}^4=\frac{1}{16}$

The power emitted per unit area is indeed $\frac{1}{16}$ of that emitted at $6000\text{K}$.
Q matches with (4).

Analysis of (R) $T=5000\text{K}$:
The peak wavelength is calculated as:

${\lambda }_m=\frac{2.9\times {10}^{-3}}{5000}=0.58\times {10}^{-6}\text{m}=580\text{nm}$

A wavelength of $580\text{nm}$ lies in the yellow region of the visible light spectrum ($400\text{nm}$ to $700\text{nm}$). Therefore, the radiation at peak wavelength is visible to the human eye.
R matches with (2).

Analysis of (S) $T=10000\text{K}$:
The peak wavelength is calculated as:

${\lambda }_m=\frac{2.9\times {10}^{-3}}{10000}=0.29\times {10}^{-6}\text{m}=290\text{nm}$

The energy of a photon corresponding to this peak wavelength is:

$E=\frac{hc}{{\lambda }_m}=\frac{1.24\times {10}^{-6}\text{V-m}}{0.29\times {10}^{-6}\text{m}}\approx 4.28\text{eV}$

Since photon energy $4.28\text{eV}\ge ;4\text{eV}$ (work function), this radiation can cause the photoelectric emission of electrons from the metal surface.
S matches with (1).

Hence, the correct matching is: P → 3, Q → 4, R → 2, S → 1.