JEE Advanced 2023 Paper 1 · Q15 · Heat Radiation / Stefan-Boltzmann

Question

Match the temperature of a black body given in List-I with an appropriate statement in List-II. [Given: Wien's constant $b = 2.9 	imes 10^{-3}$ m·K and $hc/e = 1.24 	imes 10^{-6}$ V·m]
List-I:
(P) $2000$ K
(Q) $3000$ K
(R) $5000$ K
(S) $10000$ K
List-II:
(1) The radiation at peak wavelength can lead to emission of photoelectrons from a metal of work function $4$ eV
(2) The radiation at peak wavelength is visible to human eye
(3) The radiation at peak emission wavelength will result in the widest central maximum of a single slit diffraction
(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

SolveKar AI · Accepted Answer

Peak wavelength $\lambda_m T = b$, photon energy $E = hc/\lambda_m = (hc/e)T/b$ eV $= 0.428	imes 10^{-3} T$ eV. (P) $T=2000$ K: $E=0.86$ eV — long wavelength gives widest diffraction maximum — (3). (Q) $T=3000$ K: $E=1.28$ eV, $\lambda_m = 0.97\,\mu$m; Stefan: $(3000/6000)^4 = 1/16$ — (4). (R) $T=5000$ K: $\lambda_m = 580$ nm — visible — (2). (S) $T=10000$ K: $E=4.28$ eV $> 4$ eV — photoelectrons emitted — (1).

JEE Advanced 2023 Paper 1 · Q15 · Heat Radiation / Stefan-Boltzmann

Solution