The plots of intensity versus wavelength for three black bodies at temperatures $T_{1}, T_{2}$ and $T_{3}$ respectively are shown in fig. Their temperatures are such that:

T_{1} > T_{2} > T_{3}

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T_{1} > T_{3} > T_{2}

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T_{2} > T_{3} > T_{1}

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T_{3} > T_{2} > T_{1}

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Solution

The correct option is B $T_{1} > T_{3} > T_{2}$
As we know that from emissive power $E \propto T^{4}$ and from wein's displacement law $λ_{m a x} \times T = c o n s t a n t$ . So if emissivity will be maximum them simultaneously it's temperature will be maximum, but from the wein's law we can say that the particle which has maximum temperature will have lower wavelength.
So, $E \propto T \propto \frac{1}{λ}$
So, we can say:
$T_{1} > T_{3} > T_{2}$
$E_{1} > E_{3} > E_{2}$
But, $λ_{1} < λ_{3} < λ_{2}$

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