question_answer

A black body is at a temperature of \[2880\ K\]. The energy of radiation emitted by this object with wavelength between \[499\ nm\] and \[500\ nm\] is \[{{U}_{1}}\], between \[999\ nm\] and \[1000\ nm\] is \[{{U}_{2}}\] and between \[1499\ nm\] and \[1500\ nm\] is \[{{U}_{3}}\]. The Wein's constant\[b=2.88\times {{10}^{6}}\ nm\,K\]. Then [IIT 1998]

A)            \[{{U}_{1}}=0\]                   

B)            \[{{U}_{3}}=0\]

C)            \[{{U}_{1}}>{{U}_{2}}\]   

D)            \[{{U}_{2}}>{{U}_{1}}\]

Correct Answer: D

Solution :

                   Wein's displacement law is \[{{\lambda }_{m}}T=b\] Þ \[{{\lambda }_{m}}=\frac{b}{T}=\frac{2.88\times {{10}^{6}}}{2880}=1000\,nm.\] Energy distribution with wavelength will be as follows From the graph it is clear that U2 > U1.