Radioactive Decay Rates and the Relationship between Decay Constants

[Krushnaraj Pandya]

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

 

[Charles Link]

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 $Xe^{At} =\int f(t) e^{At} \, dt$.

 

[Krushnaraj Pandya]

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?

 

[Charles Link]

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: $N_1 D_1=N_2(D_2+D_3)$

 

[Krushnaraj Pandya]

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