The energy spectrum of a black body exhibits a maximum around a wavelength {lambda _o}. The temperature of the black body is now changed such

Q: The energy spectrum of a black body exhibits a maximum around a wavelength ${\lambda _o}.$ The temperature of the black body is now changed such that the energy is maximum around a wavelength $\frac{{3{\lambda _o}}}{4}$.The power radiated by the black body will now increase by a factor of

a (a) According to Wein's law $\lambda_mT$ = constant ==> ${\lambda _{{m_1}}}{T_1} = {\lambda _{{m_2}}}{T_2} \Rightarrow {T_2} = \frac{{{\lambda _{{m_1}}}}}{{{\lambda _{{m_2}}}}}{T_1} = \frac{{{\lambda _0}}}{{3{\lambda _0}/4}} \times {T_1} = \frac{4}{3}{T_1}$ Now $P \propto {T^4} \Rightarrow \frac{{{P_2}}}{{{P_1}}} = {\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^4}$ ==> $\frac{{{P_2}}}{{{P_1}}} = {\left( {\frac{{4/3\;{T_1}}}{{{T_1}}}} \right)^4} = \frac{{256}}{{81}}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

The energy spectrum of a black body exhibits a maximum around a wavelength ${\lambda _o}.$ The temperature of the black body is now changed such that the energy is maximum around a wavelength $\frac{{3{\lambda _o}}}{4}$.The power radiated by the black body will now increase by a factor of

A$256/81$
B$64/27$
C$16/9$
D$4/3$

Medium

STD 11 Science
NEET
STD 11 - 10.2. transmission of heat
Physics

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