TS EAMCET · Physics · Thermal Properties of Matter
The wavelength of the radiation emitted by a black body is \(1 \mathrm{~mm}\) and Wien's constant is \(3 \times 10^{-3} \mathrm{mK}\). Then the temperature of the black body will be
- A \(3 \mathrm{~K}\)
- B \(30 \mathrm{~K}\)
- C \(300 \mathrm{~K}\)
- D \(3000 \mathrm{~K}\)
Answer & Solution
Correct Answer
(A) \(3 \mathrm{~K}\)
Step-by-step Solution
Detailed explanation
As we know temperature of a black body is given by, \[ \lambda T=b \quad \text { (Wien's displacement law) } \] Where, wavelength of the radiation, \[ \lambda=1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m} \] Wien's constant, \(\mathrm{b}=3 \times 10^{-3} \mathrm{mK}\)…
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