A) \[3.33\mu m\]
B) \[0.66\mu m\]
C) \[1\mu m\]
D) \[1m\]
Correct Answer: A
Solution :
For emitted radiation, according to Wein's displacement law \[{{\lambda }_{m}}\propto \frac{1}{T}\] \[\therefore \] \[{{\lambda }_{m}}T\]= constant \[{{\lambda }_{1}}{{T}_{1}}={{\lambda }_{2}}{{T}_{2}}\] Given, \[{{\lambda }_{1}}=4\mu m,\] \[{{T}_{1}}=2000K\] and \[{{T}_{2}}=2400K\] \[\therefore \] \[4\times 2000={{\lambda }_{2}}\times 2400\] \[{{\lambda }_{2}}=\frac{20}{6}=3.33\mu m\]
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