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Q.

A black body at 1227°emits radiation with maximum intensity at wave length of 5000 angstroms. The temperature of a body is increased by 1000°C . The maximum intensity will be observe at  

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a

4000 angstroms

b

5000 angstroms 

c

6000 angstroms 

d

3000 angstroms 

answer is D.

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Detailed Solution

 Given, T=1227+273=1500K

T′=1227+1000+273=2500K

λm=5000Å

λm′λm=TT′

 Hence, λm'=15002500×5000=3000Å. 

 

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