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Question

Physics Question on Thermodynamics

Match the temperature of a black body given in L-I with an appropriate statement in L-II and choose the correct option. (given: wien is constant as 2.9×1032.9\times10^{-3} m.k. hce=1.24×106v.m\frac{hc}{e}=1.24\times10^{-6}\,v.m

| ** List-I**| | **List-II **
---|---|---|---
(P)| 2000 K| (1)| The radiation at peek wavelength can lead to the emission of photo e^- from a metal of work function 4 eV
(Q)| 3000 K| (2)| The radiation at peek wavelength is visible to the human eye
(R)| 5000K| (3)| The radiation at peek emission wavelength will result in the widest central maximum of a single slit diffraction
(S)| 10000K| (4)| The power emitted per unit area is 116\frac{1}{16} of that emitted by a black body at a temperature of 6000K.
| | (5)| The radiation at peak emission wavelength can be used to image human bone.

Answer

The correct answer is : P\rightarrow3; Q\rightarrow4; R\rightarrow2; S\rightarrow1
λ×T=b\lambda\times T=b
λ=bT\lambda=\frac{b}{T}
E=hcλ=hcTbE=\frac{hc}{\lambda}=\frac{hcT}{b}
E=(hceb)×TebE=(\frac{hc}{eb})\times T\,eb
E=1.24×1062.9×103×TevE=\frac{1.24\times10^{-6}}{2.9\times10^{-3}}\times T\,ev
E=(0.428×103×T)eVE=(0.428\times 10^{-3}\times T)\,\,eV
(P) T = 2000K \Rightarrow E = 0.856eV
(Q) T = 3000K \Rightarrow E = 1.284eV
(R) T = 5000K \Rightarrow E=2.14eV
(S) T = 10000K \Rightarrow E=4.28eV