Radioactive decay: finding half-life, decay constant and initial atoms

[catkin]

[SOLVED] Decay equations: Given A0 & At, find half-life, lambda and N0

Homework Statement

This is from Advanced Physics by Adams & Allday, spread 8.13 Question 5.

The activity of a particular source falls from 5E8 Bq to 2E7 Bq in 20 minutes.
a) What is the half-life and decay constant for this nuclide?

Homework Equations

At = A0 / 2^^(t / t½)
A = A0e^^(-λt)
λt½ = ln2

The Attempt at a Solution

The half-life (t½) and decay constant (λ) can be found using either a) the first relevant equation to determine t½) and then the third relevant equation to get λ or, equivalently, b) the second relevant equation to determine λ and then the third to determine t½. The first is more elegant, giving the results in the order asked in the question.

My problem is with the maths of the first approach; I can do the second and get the answer given in the book.

Here's my foundered attempt at the first approach.

At = A0 / 2^^(t / t½)
At / A0 = 2^^(t / t½)
log2(At / A0) = t / t½
t½ = t / log2(A[SIZE="1"]t / A[SIZE="1"]0)
Trouble! A[SIZE="1"]t < A[SIZE="1"]0 so log[SIZE="1"]2(A[SIZE="1"]t / A[SIZE="1"]0) will be negative so t[SIZE="1"]½ will be negative and that's not in the problem domain. How can I re-organise the equations to get the positive solution for t[SIZE="1"]½?

There's an equivalent re-organisation in the equivalent solution method. I hoped it would give me enough of a clue but it hasn't . Here it is.

A = A[SIZE="1"]0e^^(-λt)
A / A[SIZE="1"]0 = e^^(-λt)
ln(A / A[SIZE="1"]0) = -λt
ln(A0 / A) = λt ← here's the re-organisation
and the rest is straightforward leading to
λ = 8E-4 s[SIZE="1"]-1 ct1sf

[B]Subsidiary question:[/B] the textbook gives 7.6E-4 s[SIZE="1"]-1. Is that correct? The question gives data to 1 significant figure (5E8 Bq to 2E7 Bq) so I don't believe the answer should be more precise.

[h2]Homework Statement [/h2]
b) How many atoms were there in the original source?

[h2]Homework Equations[/h2]
A = λN

[h2]The Attempt at a Solution[/h2]
My problem is that I'm ~10% and a factor of 10 out compared with the answer given in the book. This book does have a few typos so normally I'd assume the book was wrong but these 8.13 spread questions have been such a battle and working to 1 significant figure gives such big rounding discrepancies I don't have the confidence.

Here's my solution.

A = λN
N = A / λ
= 5E8 / 8E-4
= 6E11 ct1sf (book gives 6.6E10. This calculation gives 6.3E11 ct2sf)

 

[nrqed]

Homework Statement

This is from Advanced Physics by Adams & Allday, spread 8.13 Question 5.

The activity of a particular source falls from 5E8 Bq to 2E7 Bq in 20 minutes.
a) What is the half-life and decay constant for this nuclide?

Homework Equations

At = A0 / 2^^(t / t½)
A = A0e^^(-λt)
λt½ = ln2

The Attempt at a Solution

The half-life (t½) and decay constant (λ) can be found using either a) the first relevant equation to determine t½) and then the third relevant equation to get λ or, equivalently, b) the second relevant equation to determine λ and then the third to determine t½. The first is more elegant, giving the results in the order asked in the question.

My problem is with the maths of the first approach; I can do the second and get the answer given in the book.

Here's my foundered attempt at the first approach.

At = A0 / 2^^(t / t½)
At / A0 = 2^^(t / t½)

You made a mistake here. Your second line should read
$$\frac{A_0}{A_t} = 2^{t/t_{1/2}}$$ This is why you end up with a log of a negative number.

 

[Sourabh N]

I - significant digits error in book- answer coming out to be 7.6...
II - typo error in book- answer coming out to be 6.6E11.

 

[nrqed]

I - significant digits error in book- answer coming out to be 7.6...
II - typo error in book- answer coming out to be 6.6E11.

So everything works now?

 

[catkin]

You made a mistake here. Your second line should read
$$\frac{A_0}{A_t} = 2^{t/t_{1/2}}$$ This is why you end up with a log of a negative number.

Wow! That was quick! Thanks.

Isn't it the same thing, though? If the above is inverted it becomes
At / A0 = 1 / 2^^(t / t½)
and then
At = A0 / 2^^(t / t½)
which is where I started. Or have I been battling with this stuff too long and become punch drunk? It's been a long day!

 

[Sourabh N]

At = A0 / 2^^(t / t½)
At / A0 = 2^^(t / t½) - wrong, it should be 2^^(-t...

 

[catkin]

At = A0 / 2^^(t / t½)
At / A0 = 2^^(t / t½) - wrong, it should be 2^^(-t...

Thanks Maybe I should leave this until another day but ...

I didn't write
At / A0 = 2^^(t / t½)
I wrote
At / A0 = 1 / 2^^(t / t½)
which is (?) equivalent to
At / A0 = 2^^(-t / t½)
which is the correction you just gave.

Must go to bed now (Physics first lesson tomorrow AM), so please don't think me rude when I go offline. :zzz:

 

[Sourabh N]

I just copied and pasted it from your #1 post.

 

[catkin]

I'm so sorry; both of you identified what I'd done wrong and I still couldn't see it. Thanks for your patience.

Following your advice I can now solve the problem and, sketching a graph, the answer looks right but it's not the same as the answer given in the book. Here is my new solution.

$$A_{t}= \frac {A_{0}} {2^{(t / t_{1 / 2 } )}}$$

$$\frac{A_{t}} {A_{0}} = \frac {1} {2^{(- t / t_{1 / 2})}}$$

$$log_{2}( \frac {A_{t}} {A_{0}} ) = \frac {-t } {t_{1 / 2}}$$

$$t_{1 / 2} = \frac {- log(2) \ t } {log( \frac {A_{t}} {A_{0}} )}$$

Using minutes as the time unit

$$t_{1 / 2} = \frac {- 20 log(2) } {log( \frac {2E7} {5E8})}$$

giving = 4 minutes ct1sf (4.3 ct2sf)

The answer given in the book is 15 minutes. Which is correct?

If 4 minutes is correct then I would like to know what is wrong with my alternate solution which solves for λ first (8E-4 s-1 ct1sf) and then converts λ to t½ giving approximately 15, the book's answer! I'll wait to see if 4 minutes is correct before posting my alternate solution.

 

[Sourabh N]

I think I did some calc error earlier. Correct me if I'm wrong :

$$\lambda$$ = ln(A$$_{0}$$ / A) / t

$$\lambda$$ = ln(5e8 / 2e7) / 20*60

$$\lambda$$ = ln(25) / 1200

$$\lambda$$ = 2.7e-3

 

[Sourabh N]

posted twice, sorry.

 

[catkin]

Thanks, Sourabh N

Yes -- I think that's right. It's what I got at first and thought was wrong because it differed from the answer given in the book. Somehow I managed to get the answer given in the book but now I can't find how. I've tried swapping At and A0, "forgetting" to convert minutes to seconds, using log() instead of ln().

Thanks again for your help. I'm confident you have the right solution. "Grrr" to textbooks with the wrong answers and typos!

Best

C

 

[Sourabh N]

Its a very interesting situation, one doesn't know how one got the wrong answer
I also "struggled" for some time to find out how I got the wrong answer, here's my way :
ln (A0/A) = ln (5e8/2e7) = ln (0.4) {supposed to be ln (25) }

 

[catkin]

That's it! That's how I got it wrong and got the same answer as the book -- by swapping At and A0.

 

[catkin]

Apologies to Adams and Allday!

I copied the question wrongly; the initial activity was 5E7 Bq, not 5E8 Bq. Strange that I got that wrong because I scanned and OCRed the question from the textbook! Improbable that, having got the question mangled, I could get the same answer as the book by transposing two values in the working!

Who said the Universe is a random, deterministic mechanism without guiding intelligence?!