Modern Physics 3 Question 38

37. A radioactive nucleus X decays to a nucleus Y with a decay constant λx=0.1s1,Y further decays to a stable nucleus Z with a decay constant λy=1/30s1. Initially, there are only X nuclei and their number is N0=1020. Set-up the rate equations for the populations of X,Y and Z. The population of Y nucleus as a function of time is given by Ny(t)=N0λx/(λxλy)[exp(λyt)exp(λxt)]. Find the time at which NY is maximum and determine the populations X and Z at that instant.

(2001, 5M)

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Solution:

  1. (a) Let at time t=t, number of nuclei of Y and Z are NY and NZ. Then,

Rate equations of the populations of X,Y and Z are

dNXdt=λXNXdNYdt=λXNXλYNY and dNZdt=λYNY

(b) Given NY(t)=N0λXλXλY[eλYteλXt]

For NY to be maximum

dNY(t)dt=0 i.e λXNX=λYNY (iv)  [from Eq. (ii)]  or λX(N0eλXt)=λYN0λXλXλY[eλYteλXt] or λXλYλY=eλYteλXt1λXλY=e(λXλY)t or (λXλy)tln(e)=lnλXλY or t=1λXλYlnλXλY

Substituting the values of λX and λY, we have

t=1(0.11/30)ln0.11/30=15ln(3)t=16.48s

 or 

(c) The population of X at this moment,

NX=N0eλXt=(1020)e(0.1)(16.48)NX=1.92×1019NY=NXλXλY [From Eq. (iv) ]=(1.92×1019)(0.1)(1/30)=5.76×1019NZ=N0NXNY=10201.92×10195.76×1019 or NZ=2.32×1019



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