The graph shows the blackbody radiation curve | vedclass

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(A) According to Wien's displacement law,the wavelength $\lambda_m$ corresponding to the maximum intensity of radiation is inversely proportional to t

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(A) According to Wien's displacement law,the wavelength $\lambda_m$ corresponding to the maximum intensity of radiation is inversely proportional to the absolute temperature $T$,i.e.,$\lambda_m \propto \frac{1}{T}$.Given $T_2 > T_1$,it follows that $\lambda_{m2} This means the peak of the radiation curve for $T_2$ must shift towards a shorter wavelength (left) compared to the peak for $T_1$.Additionally,according to the Stefan-Boltzmann law,the total intensity (area under the curve) is proportional to $T^4$. Since $T_2 > T_1$,the curve for $T_2$ must lie entirely above the curve for $T_1$.Comparing the given options,the graph in option $A$ correctly shows the peak for $T_2$ shifted to the left (smaller $\lambda_m$) and the entire curve for $T_2$ lying above the curve for $T_1$.

The graph shows the blackbody radiation curves at temperatures $T_1$ and $T_2$. Which of the following is correct? $(T_2 > T_1)$

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