A) The temperature of \[B\] is \[1934\ K\]
B) \[{{\lambda }_{B}}=1.5\mu m\]
C) The temperature of \[B\] is \[11604\ K\]
D) The temperature of \[B\] is \[2901\ K\]
Correct Answer: A
Solution :
According to Stefan?s law \[E=eA\sigma {{T}^{4}}\Rightarrow {{E}_{1}}={{e}_{1}}A\sigma T_{1}^{4}\] and \[{{E}_{2}}={{e}_{2}}A\sigma T_{2}^{4}\] \[\because \]\[{{E}_{1}}={{E}_{2}}\] \[\therefore \]\[{{e}_{1}}T_{1}^{4}={{e}_{2}}T_{2}^{4}\] \[\Rightarrow \]\[{{T}_{2}}={{\left( \frac{{{e}_{1}}}{{{e}_{2}}}T_{1}^{4} \right)}^{\frac{1}{4}}}={{\left( \frac{1}{81}\times {{(5802)}^{4}} \right)}^{\frac{1}{4}}}\]\[\Rightarrow \]\[{{T}_{B}}=1934\ K\] And, from Wein?s law \[{{\lambda }_{A}}\times {{T}_{A}}={{\lambda }_{B}}\times {{T}_{B}}\] \[\Rightarrow \frac{{{\lambda }_{A}}}{{{\lambda }_{B}}}=\frac{{{T}_{B}}}{{{T}_{A}}}\]\[\Rightarrow \]\[\frac{{{\lambda }_{B}}-{{\lambda }_{A}}}{{{\lambda }_{B}}}=\frac{{{T}_{A}}-{{T}_{B}}}{{{T}_{A}}}\] \[\Rightarrow \]\[\frac{1}{{{\lambda }_{B}}}=\frac{5802-1934}{5802}=\frac{3968}{5802}\Rightarrow {{\lambda }_{B}}=1.5\ \mu m\]
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