Explanation for a series of numbers (radioactive Decay)

corey0422622443 · #1 $Old$ September 13th, 2009, 11:22 PM

Hello Again!
Thank you to all who helped with my last problem! i sincerely appreciate it!
I have done all the maths to calculate the radioactive decay of the three substances (a,b & c), but i have been asked by my tutor to describe the results over time (paying particular attention to limits). Can anybody help explain what he might mean by this or perhaps share some opinions?

thanking you all in advance!
-corey

chisigma · #2 $Old$ September 14th, 2009, 12:17 AM

That is a typical nuclear beta-decay reaction of the form...

$A \rightarrow B \rightarrow C$

If we indicate with $m_{a} (t)$ , $m_{b} (t)$ and $m_{c}(t)$ the mass of the elements as a function of time and $t_{a}$ , $t_{b}$ and $t_{c}$ their 'half life time' and consider that C is 'stable' [i.e. $t_{c} = \infty$ ...] , we obtain...

$m_{a} (t) = m_{a} (0) \cdot 2^{-\frac{t}{t_{a}}}$

$m_{b} (t) = \{m_{b} (0) + m_{a} (0) (1-2^ {-\frac{t}{t_{a}}}) \} \cdot 2^ {-\frac{t}{t_{b}}}$

$m_{c} (t) = m_{c} (0) + \{m_{b} (0) + m_{a} (0) (1-2^ {-\frac{t}{t_{a}}}) \} \cdot (1- 2^ {-\frac{t}{t_{b}}})$

From the diagrams it seems to be...

$m_{a} (0)= 800 g$

$m_{b} (0) = 100 g$

$m_{c} (0) = 0 g$

The numerical value of $t_{a}$ and $t_{b}$ can be extrapolated from the diagrams...

A very interesting problem!...

Kind regards

$\chi$ $\sigma$

corey0422622443 · #3 $Old$ September 14th, 2009, 12:31 AM

thanks alot! I think i am missing something however, how does this relate to limits? thats the piece i am struggling with

(+rep to you btw)

chisigma · #4 $Old$ September 14th, 2009, 01:38 AM

Very easy!... at the time $t=0$ the global mass is $m_{a} (0) + m_{b} (0) + m_{c} (0) = 900 g$ . At the time $t=\infty$ the global mass is the mass of C alone and for the law of canservation must be $m_{c} (\infty) = 900 g$ ...

Kind regards

$\chi$ $\sigma$