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JEE Advanced · Physics · 12. Thermal Properties

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 hce=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 116 of that emitted by a blackbody at temperature 6000 K
    5 The radiation at peak emission wavelength can be used to image human bones

  1. A P3, Q5, R2, S3
  2. B P3, Q2, R4, S1
  3. C P3, Q4, R2, S1
  4. D P1, Q2, R5, S3
Verified Solution

Answer & Solution

Correct Answer

(C) P3, Q4, R2, S1

Step-by-step Solution

Detailed explanation

Maximum wavelength for each case:
(P) 2000 K\(\begin{array}{l}\lambda_m T=b \\ \Rightarrow \lambda_m=\frac{b}{T}=\frac{2.9 \times 10^{-3}}{2000} \\ =1.45 \times 10^{-6} m \\ =1450 nm\end{array}\)
(Q) 3000 K\(\begin{array}{l}\lambda_m T=b \\ \Rightarrow \lambda_m=\frac{2.9 \times 10^{-3}}{3000} \\ =966.66 nm\end{array}\)
(R) 5000 K\(\begin{array}{l}\lambda_m T=b \\ \Rightarrow \lambda_m=\frac{2.9 \times 10^{-3}}{5000} \\ =580 nm\end{array}\)
(S) 10000 K\(\begin{array}{l}\lambda_m T=b \\ \Rightarrow \lambda_m=\frac{2.9 \times 10^{-3}}{10,000} \\ =290 nm\end{array}\)
Now for List-II: 1 λth=hcϕ=1.24×10-64=0.31×10-6 m=310 nm
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