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Radioactive decay question

  • #1
Krushnaraj Pandya
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697
71

Homework Statement


Consider nucleus A decaying to B with decay constant D1, B decays to either X or Y (decay constants D2 and D3). at t=0, number of nuclei of A,B,X and Y are J,J,0 and 0. and N1,N2,N3 and N4 are the number of nuclei of A,B,X and Y at any instant.
My question is, what is the relation between D1,D2 and D3 for B to first increase, peak and then decrease?

Homework Equations


rate of decay = DN

The Attempt at a Solution


I figured the rate of accumulation of B is N1D1- N2D2- N3D3, since this also involves N, I don't know how to figure out a relation between D1,D2 and D3 alone for the specified conditions. I also figured out that no matter what D1,D2,D3 are- after a long time B will have zero nuclei (so will A).
If D1 is greater it will cause a peak but I'm not sure how the values will change with subsequent change in N
 

Answers and Replies

  • #2
Charles Link
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Your equation should read ## \frac{dN_2}{dt}= N_1D_1-N_2D_2-N_2D_3 ##, with an ## N_2 ## in the 3rd term. ## \\ ## When ## N_1D_1=N_2(D_2+D_3) ## that will be when ## N_2 ## peaks, with ## \frac{dN_2}{dt}=0 ##. ## \\ ## You can actually solve the equation ## \frac{dN_1}{dt}=-D_1 N_1 ## for ## N_1(t) ## , and plug into the ## \frac{dN_2}{dt} =N_1 D_1-N_2(D_2+D_3) ## equation, and solve this equation for ## N_2 =N_2(t) ##. (It's been a while since I did this particular calculation, but I believe it uses a technique of integrating factors). ## \\ ## As I remember it, when you have ## \frac{dX}{dt}+AX =f(t) ##, (where ## X=X(t) ##), you can multiply both sides by ## e^{At} ##, (called an integrating factor), and the left side is then ## \frac{d(Xe^{At})}{dt} ##. The right side has ## f(t)e^{At} ##, and you can then integrate to get ## Xe^{At} =\int f(t) e^{At} \, dt ##.
 
Last edited:
  • #3
Krushnaraj Pandya
Gold Member
697
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Your equation should read ## \frac{dN_2}{dt}= N_1D_1-N_2D_2-N_2D_3 ##, with an ## N_2 ## in the 3rd term. ## \\ ## When ## N_1D_1=N_2(D_2+D_3) ## that will be when ## N_2 ## peaks, with ## \frac{dN_2}{dt}=0 ##. ## \\ ## You can actually solve the equation ## \frac{dN_1}{dt}=-D_1 N_1 ## for ## N_1(t) ## , and plug into the ## \frac{dN_2}{dt} =N_1 D_1-N_2(D_2+D_3) ## equation, and solve this equation for ## N_2 =N_2(t) ##. (It's been a while since I did this particular calculation, but I believe it uses a technique of integrating factors). ## \\ ## As I remember it, when you have ## \frac{dX}{dt}+AX =f(t) ##, (where ## X=X(t) ##), you can multiply both sides by ## e^{At} ##, (called an integrating factor), and the left side is then ## \frac{d(Xe^{At})}{dt} ##. The right side has ## f(t)e^{At} ##, and you can then integrate to get ## Xe^{At} =\int f(t) e^{At} \, dt ##.
Interesting, I understand this for the most part but I doubt my textbook wants me to derive the answer using the calculus you mentioned since only basic parallel disintegration is in our course- the answer given is simply D1>D2+D3 for the given conditions to occur, is there a simple intuitive way to know that?
 
  • #4
Charles Link
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Interesting, I understand this for the most part but I doubt my textbook wants me to derive the answer using the calculus you mentioned since only basic parallel disintegration is in our course- the answer given is simply D1>D2+D3 for the given conditions to occur, is there a simple intuitive way to know that?
I believe ## N_2 ## will always experience a peak. If ## D_2 ## and ## D_3 ## are very large, that will occur when ## N_2 ## is very close to zero and ## N_1 ## very close to its initial value. In this case, I believe your book has an incorrect answer. The equation that I posted is the correct equation: ## N_1 D_1=N_2(D_2+D_3) ##
 
  • #5
Krushnaraj Pandya
Gold Member
697
71
I believe ## N_2 ## will always experience a peak. If ## D_2 ## and ## D_3 ## are very large, that will occur when ## N_2 ## is very close to zero and ## N_1 ## very close to its initial value. In this case, I believe your book has an incorrect answer. The equation that I posted is the correct equation: ## N_1 D_1=N_2(D_2+D_3) ##
I understand, thank you very much :D
 

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