The power radiated by a black body is \(P\) and it radiates maximum energy at wavelength \(\lambda_0\). Temperature of the black body is now changed so that it radiates maximum energy at the wavelength \(\frac{3}{4}\lambda_0\). The power radiated by it now becomes \(nP\). The value of \(n\) is:
1. \( \frac{3}{4} \)
2. \( \frac{4}{3} \)
3. \( \frac{256}{81} \)
4. \( \frac{81}{256}\)
Subtopic: Wien's Displacement Law |
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NEET - 2018
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A black body is at a temperature of \(5760~\mathrm{K}\). The energy of radiation emitted by the body at wavelength \(250~\mathrm{nm}\) is \(U_1\), at wavelength \(500~\mathrm{nm}\) is \(U_2\) and that at \(1000~\mathrm{nm}\) is \(U_3\). Wien’s constant, \(\mathrm{b}=2.88 \times 10^6 \mathrm{~nm}-\mathrm{K}\). Which of the following is correct?
1. \(
\mathrm{U}_3 =0
\)
2. \(\mathrm{U}_1 >\mathrm{U}_2
\)
3. \(\mathrm{U}_2 >\mathrm{U}_1
\)
4. \(\mathrm{U}_1 =0\)
Subtopic: Wien's Displacement Law |
64%
From NCERT
NEET - 2016
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