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Standard XII

Physics

Blackbody

The plots of ...

Question

The plots of intensity vs wavelength for two black bodies having same area at temperatures $T_{1}$ and $T_{2}$ such that $T_{2} = 2 T_{1}$ is shown in the figure. Let the energy radiated per second by body 1 and body 2 be $E_{1}$ and $E_{2}$ respectively. Pick the correct statement.

E_{1} = 16 E_{2}

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E_{2} = 16 E_{1}

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E_{2} = 4 E_{1}

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None of the above

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Solution

The correct option is B $E_{2} = 16 E_{1}$
Since area of both the bodies is same, energy radiated per second $E \propto (T)^{4}$
$\Rightarrow \frac{E_{1}}{E_{2}} = {(\frac{T_{1}}{T_{2}})}^{4} = {(\frac{T_{1}}{2 T_{1}})}^{4}$
$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{1}{16}$

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Blackbody

Standard XII Physics

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The plots of intensity vs wavelength for two black bodies having same area at temperatures T1 and T2 such that T2=2T1 is shown in the figure. Let the energy radiated per second by body 1 and body 2 be E1 and E2 respectively. Pick the correct statement.

The correct option is B E2=16E1Since area of both the bodies is same, energy radiated per second E∝(T)4 ⇒E1E2=(T1T2)4=(T12T1)4 ⇒E1E2=116

The plots of intensity vs wavelength for two black bodies having same area at temperatures $T_{1}$ and $T_{2}$ such that $T_{2} = 2 T_{1}$ is shown in the figure. Let the energy radiated per second by body 1 and body 2 be $E_{1}$ and $E_{2}$ respectively. Pick the correct statement.

The correct option is B $E_{2} = 16 E_{1}$
Since area of both the bodies is same, energy radiated per second $E \propto (T)^{4}$
$\Rightarrow \frac{E_{1}}{E_{2}} = {(\frac{T_{1}}{T_{2}})}^{4} = {(\frac{T_{1}}{2 T_{1}})}^{4}$
$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{1}{16}$