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NEET 2018
Exam
NEET
Year
2018
Subject
Physics
Chapter
Thermal Properties of Matter

NEET 2018 · Physics · Thermal Properties of Matter

The power radiated by a black body is P and it radiates maximum...

The power radiated by a black body is P and it radiates maximum energy at wavelength, λ0{\lambda}_{0}. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength 34λ0\dfrac{3}{4}{\lambda}_{0}, the power radiated by it becomes nP. The value of n is

Options

  1. A25681\dfrac{256}{81}
  2. B43\dfrac{4}{3}
  3. C34\dfrac{3}{4}
  4. D81256\dfrac{81}{256}

Correct answer

A. 25681\dfrac{256}{81}

Step-by-step solution

By Stefan's law,
P=\sigmaAT4⇒P∝T4P = \sigmaA{T}^{4} ⇒ P ∝ {T}^{4}

According to Wien's law T∝1λmT ∝ \dfrac{1}{{\lambda}_{m}}

⇒P∝(1λm)4⇒P2P1=(λm1λm2)4⇒ P ∝ {\left(\dfrac{1}{{\lambda}_{m}}\right)}^{4} ⇒ \dfrac{{P}_{2}}{{P}_{1}} = {\left(\dfrac{{\lambda}_{{m}_{1}}}{{\lambda}_{{m}_{2}}}\right)}^{4}

⇒P2P1=(λ034λ0)4⇒nPP=25681⇒n=25681⇒ \dfrac{{P}_{2}}{{P}_{1}} = {\left(\dfrac{{\lambda}_{0}}{\dfrac{3}{4}{\lambda}_{0}}\right)}^{4} ⇒ \dfrac{nP}{P} = \dfrac{256}{81} ⇒ n = \dfrac{256}{81}

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