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 frac hc e =1.24 × 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

P rightarrow 3, Q rightarrow 5, R rightarrow 2, S rightarrow 3

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

P rightarrow 1, Q rightarrow 2, R rightarrow 5, S rightarrow 3

Physics · Heat Transfer

JEE Advanced 2023 — Paper 1 — Question 31

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 $2.9 \times 10^{-3} \mathrm{~m}-\mathrm{K}$ and $\frac{\mathrm{hc}}{\mathrm{e}}=1.24 \times 10^{-6} \mathrm{~V}-\mathrm{m}$ ]

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

Option A:
$\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 5, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 3$
Option B:
$\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 2, \mathrm{R} \rightarrow 4, \mathrm{~S} \rightarrow 1$
Option C:
$\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 1$
Correct
Option D:
$\mathrm{P} \rightarrow 1, \mathrm{Q} \rightarrow 2, \mathrm{R} \rightarrow 5, \mathrm{~S} \rightarrow 3$

Answer: C

Step-by-step solution

(1) radiation at peak $\lambda=\frac{\mathrm{hc}}{4 \mathrm{eV}}=\frac{1.24 \times 10^{-6}}{4}=0.31 \times 10^{-6}=3100 \mathrm{~A}^{\circ}$

\lambda_{\mathrm{m}} \mathrm{~T}=2.9 \times 10^{-3}

\lambda_{\mathrm{m}}=\frac{2.9 \times 10^{-3}}{\mathrm{~T}}=3100 \times 10^{-10}

\mathrm{T}=\frac{2.9 \times 10^{7}}{3100}=9354 \mathrm{~K} \rightarrow 10000 \mathrm{~K}

(2) $\lambda_{m}$ visible to human eye (violet to red) (For 700 nm ) $\mathrm{T}=\frac{2.9 \times 10^{-3}}{7000 \times 10^{-10}}=\frac{29000}{7}=4142 \rightarrow 5000 \mathrm{~K}$ (For 400 nm ) $\mathrm{T}=\frac{2.9 \times 10^{-3}}{4000 \times 10^{-10}}=7250$ (3) widest central maximum $\Rightarrow \lambda_{\max } \Rightarrow \mathrm{T}_{\min } \Rightarrow 2000 \mathrm{~K}$ (4) power per unit area $=\frac{1}{16}$ (power by block body at $\mathrm{T}=6000 \mathrm{~K}$ )

=\frac{1}{16} \sigma(6000)^{4}=\sigma \mathrm{T}^{4} \Rightarrow \mathrm{~T}=3000 \mathrm{~K}

(5) $\lambda=1 \stackrel{0}{\mathrm{~A}}$

\mathrm{T}=\frac{2.9 \times 10^{-3}}{10^{-10}}=2.9 \times 10^{7} \mathrm{~K}

(p) $\rightarrow 3$ , (q) $\rightarrow 4$ , (r) $\rightarrow 2$ , (s) $\rightarrow 1$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2023
Paper: Paper 1
Subject: Physics
Chapter: Heat Transfer
Topic: Convection and Radiation