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Q.

In beta decay, the emitted beta particles have a range of values of their kinetic energy. The maximum energy Emax  of beta particles emitted in the decay of C14 is 0.165 MeV. Assuming number of beta particles (dN) with energy between E and E + dE have the form dN E×(EmaxE)2×dE. Let a sample of C14 is decaying into N14 and activity is kept constant at rate 1 millicurie. All beta particles emitted are being absorbed by a target and causing evolution of heat. If the heat evolution rate is x×106MeV/s, find x (rounded off to nearest integer)

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answer is 2.

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Detailed Solution

Average value of  E=0EmaxdN×E0EmaxdN=Emax3
So heat evolution rate  =Emax3×3.7×107Meνs   =2.03×106Meνs
 
 

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