JEE Advance - Physics (2010 - Paper 1 Offline - No. 8)
Two spherical bodies A (radius 6 cm ) and B (radius 18 cm ) are at temperature T1 and T2, respectively.
The maximum intensity in the emission spectrum of A is at 500 nm and in that of B is at 1500 nm.
Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that
of B?
Answer
9
Explanation
According to Wien's displacement law, $$\lambda$$mT = constant
$$\therefore$$ $${({\lambda _m})_A}{T_A} = {({\lambda _m})_B}{T_B}$$
or, $${{{T_A}} \over {{T_B}}} = {{{{({\lambda _m})}_B}} \over {{{({\lambda _m})}_A}}} = {{1500\,nm} \over {500\,nm}}$$ or $${{{T_A}} \over {{T_B}}} = 3$$ ...... (i)
According to Stefan Boltzmann law, rate of energy radiated by a black body
$$E = \sigma A{T^4} = \sigma 4\pi {R^2}{T^4}$$ [Here, $$A = 4\pi {R^2}$$]
$$\therefore$$ $${{{E_A}} \over {{E_B}}} = {\left( {{{{R_A}} \over {{R_B}}}} \right)^2}{\left( {{{{T_A}} \over {{T_B}}}} \right)^4} = {\left( {{{6\,cm} \over {18\,cm}}} \right)^2}{(3)^4}$$ (Using (i))
$$ = 9$$
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