JEE Advanced 2023 — Paper 1 — PHY-IV solution

JEE Advanced2023Paper 1PHY-IV

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 $2.9 \times 1 0^{- 3}$ m-K and $\frac{h c}{e} = 1.24 \times 1 0^{- 6}$ 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 $\to$ 3, Q $\to$ 5, R $\to$ 2, S $\to$ 3
B
P $\to$ 3, Q $\to$ 2, R $\to$ 4, S $\to$ 1
C
P $\to$ 3, Q $\to$ 4, R $\to$ 2, S $\to$ 1
D
P $\to$ 1, Q $\to$ 2, R $\to$ 5, S $\to$ 3

Solution

Peak wavelengths from Wien's law. $λ_{m} T = 2.9 \times 1 0^{- 3}$ m-K.

(P) $T = 2000$ K $\Rightarrow λ_{m} = 1450$ nm (infrared, largest among the four). The longest wavelength produces the widest diffraction maximum since the half-angle scales as $λ / a$ . Match $\to$ (3).

(Q) $T = 3000$ K $\Rightarrow λ_{m} = 967$ nm. The Stefan-Boltzmann law gives power per unit area $\propto T^{4}$ , so $\frac{P _{3000}}{P _{6000}} = (\frac{1}{2})^{4} = \frac{1}{16}$ . Match $\to$ (4).

(R) $T = 5000$ K $\Rightarrow λ_{m} = 580$ nm, which lies in the visible spectrum. Match $\to$ (2).

(S) $T = 10000$ K $\Rightarrow λ_{m} = 290$ nm. The corresponding photon energy is $\frac{h c}{λ _{m}} = \frac{1.24 \times 1 0 ^{- 6}}{290 \times 1 0 ^{- 9}} \approx 4.28$ eV, which exceeds a $4$ eV work function and ejects photoelectrons. Match $\to$ (1).

Mapping: P $\to$ 3, Q $\to$ 4, R $\to$ 2, S $\to$ 1. Answer: (C).

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