MHT-CET Physics and Chemistry Full Test-6 PHY-Que-40 Solution

Q.40. A black body radiates maximum energy at wavelength ‘λ’ and its emissive power is E. Now due to change in temperature of that body, it radiates maximum energy at wavelength  \frac{2\lambda}{3} . At that temperature emissive power is

A. \frac{51E}{8}

B. \frac{81E}{16}

C. \frac{61E}{27}

D. \frac{71E}{19}

Answer :- B. \frac{81E}{16}

Explanation :-

From Wien’s Displacement Law,

\lambda_{\max } = \frac{b}{T} \Rightarrow T = \frac{b}{\lambda_{\max }}

From Stefan-Boltzmann Law,

E = \sigma T^4 = \sigma \left( \frac{b}{\lambda_{\max }} \right)^4

Let the new emissive power be \mathrm{E}^{\prime} .
\therefore \quad E^{\prime} = \sigma \left( \frac{b}{\frac{2 \lambda_{\max }}{3}} \right)^4

E^{\prime} = \frac{81}{16} E