Nuclear decay of a mixture of P and S. (2009 USAPhO Question A2)

[vthem]

The answer is not provided by AAPT.

Homework Statement

A mixture of ^{32}P and ^{35}S (two beta emitters widely used in biochemical research) is placed next to a detector and allowed to decay, resulting in the data (attached) below. The detector has equal sensitivity to the beta particles emitted by each isotope, and both isotopes decay into stable daughters.
You should analyze the data graphically. Error estimates are not required.a.Determine the half-life of each isotope.
^{35}S has a signicantly longer half-life than ^{32}P.
b.Determine the ratio of the number of ^{32}P atoms to the number of ^{35}S atoms in the original sample.

Homework Equations

\frac{dN}{dt}=-λΝ

The Attempt at a Solution

|\frac{dN}{dt}|=N_P λ_P+N_S λ_S (1)
T_{\frac{1}{2}S} \gg T_{\frac{1}{2}P}⇔λ_S \ll λ_P(2)
N=N_0 e^{-λt} (3)

But I can't figure out how to use them in order to get a result.

 

[BvU]

What did you do with the instruction to analyze the data graphically ?

What would you do to get the half life of ³²P if the exercise only had ³²P and no S ?

 

[vthem]

What did you do with the instruction to analyze the data graphically ?

What would you do to get the half life of ³²P if the exercise only had ³²P and no S ?

|\frac{dN}{dt}|=λ\cdot N_{0,P}\cdot e^{-λt}⇔\ln{|\frac{dN}{dt}}|=\ln(λ\cdot N_{0,P})-λ\cdot t

Then on a \ln{|\frac{dN}{dt}}| - t diagram the slope would be -λ and it would "cut" the \ln{|\frac{dN}{dt}}| axis on \ln(λ \cdot N_{0,P}). Since T_{1/2}=\frac{\ln2}{λ} I have both a and b.

What should I do in this case?

 

[FermiAged]

Plot the data and recognize where one isotope is the dominant presence and where both isotopes are present.

 

[vthem]

Plot the data and recognize where one isotope is the dominant presence and where both isotopes are present.

In a \ln{|\frac{dN}{dt}}| - t diagram and see where the slope is constant or you mean something else?

 

[vthem]

I'm asking which plot...

 

[BvU]

The one you have mentioned a number of times already

 

[BvU]

a \ln{|\frac{dN}{dt}}| - t diagram

...

 

[BvU]

The only thing you can plot is the activity versus time. If you make the y-axis logarithmic, there would be one straight line if there were only one contribuant, e.g. P. Now there are two. After a while, the P isn't dominating any more and you would have seen a straight line depicting the S activity. Determine N0 and t 1/2 and subtract from the total activity. What's left is the P.

If you only would have followed the instruction: graphically determine...

 

[berkeman]

Thread closed for Moderation...

 

[berkeman]

Thread re-opened after some cleanup.

@vthem -- You have been receiving very good help in this thread. Your original post (OP, post #1) has a table. The helpers are asking you to *plot* that data on a 2-D plot, and have even given you some very good hints for ways to analyze that data to answer the question that you are asked in this problem. Please post your graph in the next post that you make here in this thead. Thank you.

 

[vthem]

I got λ=-0.00856 and b=8,68. This is only for S though. Excel file

 

[FermiAged]

Assuming your values are for the latter part of the data, a half-life of 81 days shows good agreement with my Chart of the Nuclides (no nerd should be without it) value of 87.2 days for S-35. Remember that this will include a small amount of the shorter lived isotope which will bias the result downward a bit. This method works better with increasing differences between the decay constants of the isotopes present.

Work backwards to separate this isotope from the other.

 

[dauto]

Note how your plot has an almost constant slope at first but than curves and has a different almost constant slope later. At first the long lived isotope contributes little because its decay is so slow. Later the short lived isotope contributes little because it is almost all gone.

 

[vthem]

Note how your plot has an almost constant slope at first but than curves and has a different almost constant slope later. At first the long lived isotope contributes little because its decay is so slow. Later the short lived isotope contributes little because it is almost all gone.

Thanks a lot for the reply! (I also hit the Thanks button :P )

So because the λ of the long lived isotope is too little the slope will give me the other λ. Right?