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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Thermal Properties of Matter

NTA Abhyas 2020 · Physics · Thermal Properties of Matter

The power radiated by a black-body is P _ 0 and it radiates maximum...

The power radiated by a black-body is P0{P}_{0} and it radiates maximum energy around the wavelength λ0.{\lambda}_{0} . If the temperature of the black-body is now changed so that it radiates maximum energy around a wavelength 3λ04\dfrac{3{λ}_{0}}{4}, then the power radiated by it will be

Options

  1. A43P0\dfrac{4}{3}{P}_{0}
  2. B169P0\dfrac{16}{9}{P}_{0}
  3. C6427P0\dfrac{64}{27}{P}_{0}
  4. D25681P0\dfrac{256}{81}{P}_{0}

Correct answer

D. 25681P0\dfrac{256}{81}{P}_{0}

Step-by-step solution

Let T0{T}_{0} be the initial temperature of the black body

∴λ0T0=b∴ {\lambda}_{0}{T}_{0} = b(Wien's law)

Power radiated, P0=CT04,{P}_{0} = C{T}_{0}^{4}, where, CC is constant.

If TT is new temperature of black body, then

3λ04T=b=λ0T0orT=43T0\dfrac{3{\lambda}_{0}}{4}T = b = {\lambda}_{0}{T}_{0}orT = \dfrac{4}{3}{T}_{0}

Power radiated, P=CT4=CT04(43)4P = CT^{4} = C{T}_{0}^{4}{\left(\dfrac{4}{3}\right)}^{4}

P=25681P0P = \dfrac{256}{81}{P}_{0}
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