Calculation of a half-life how?

Anyway, glad we could help. In summary, the conversation discusses the general form of growth/decay models and how to solve for the growth rate (k) using the equation A(t) = A_i * e^(kt), where A_i is the initial amount, t is time, and A(t) is the amount after time t. The conversation also explains how to use this equation to find the remaining amount of a substance after a given time, using the example of a radioactive substance with a half-life of 5 days and an initial mass of 12kg. The conversation concludes with a discussion on the difference between arithmetic and geometric sequences in relation to growth/decay models.
  • #1
LoBun
17
0
"A radioactive substance has a half-life of 5 days, initial mass of 12kg. How much the original isotope will remain after 10 days?"

This is just an example, the question itself is not important, i just want to know how to solve this type of problems generally.. thank you!
 
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  • #2
The general form for growth/decay is the model of:

[tex]A(t) = A_{i}e^{kt}[/tex]

where Ai is the initial amount, k the growth rate, and t is time.

You need to solve for k... so. Set it up like this.

[tex]6 = 12e^{5k}[/tex]

Solve for k and use the new equation to find any other questions pertaining to this model.
 
  • #3
What's the definition of half-life? In this example, what happens after 5 days?
 
  • #4
If I am not mistaken..

Half life is the time taken for half the number of nuclides to decay. So, as far as I understand, if the half life is of 5 days, after this time the substance will have half its initial mass.. in this case.. 6kg?

Still somewhat confused.. :confused:
 
  • #5
Look at my post... that's exactly what I have. Can you please indicate what is confusing?

EDIT: The followings posts are probably an easier way to look at the problem. Since 10 days is conveniently twice the half-life, you don't need to use a growth/decay model.
 
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  • #6
That is right. The half life is also independent of how many particles you have (within the scale we're discussing here - [itex]\mbox{kg}[/itex]). So how much will you have left after 10 days?

Of course, that's just because this particular question is set up such that the answer is obvious (once you understand it). In general you will need to solve an equation to figure out how much remains.
 
  • #7
LoBun said:
Half life is the time taken for half the number of nuclides to decay. So, as far as I understand, if the half life is of 5 days, after this time the substance will have half its initial mass.. in this case.. 6kg?
Right. And what about after another 5 days (for a total of 10 days)?
 
  • #8
Maybe because..

..you've confused a few numbers..

Is it supposed to be:


10 = 12e^(5k) ?
 
  • #9
May want to check out the "radioactive decay" post in the Differential Equations section. Remember to use enough decimal digits to get an accurate answer (Naeem, are you reading this?).
 
  • #10
No... A(t) is the amount of the substance. The initial amount was 12 kg and after 5 days (t), you have 6 left. I'm fairly certain my method is correct.

This method isn't necessary... if you look at the above posts you'll see why.
 
  • #11
Hum..

Doc Al said:
Right. And what about after another 5 days (for a total of 10 days)?

Would it be.. 1/4 of the initial mass?
 
  • #12
Yes, that's correct. :)

Just to show you the method I used, I'll solve this problem for you.

[tex]6 = 12e^{5k}[/tex]
[tex]k = \frac{-\ln(2)}{5}[/tex]

Now you have the function
[tex]A(t) = 12e^{\frac{-t\ln{2}}{5}[/tex]

Plugging in t = 10 you get

[tex]A(t) = 12e^{\frac{-10\ln{2}}{5}[/tex]
[tex] A(t) = 3[/tex]

Which is 1/4 of the original amount.
 
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  • #13
The equation for an exponentially decaying quantity [itex]I(t)[/itex] is

[tex] I(t) = I_0e^{-kt}[/tex]

where [itex]I_0 = I(0)[/itex]. If the half-life is [itex]T[/itex], what it means is that

[tex]I(T) = \frac{I_0}{2}[/tex]

In your case, you are given the quantity at time 0 as [itex]12 \mbox{kg}[/itex], so [itex]I_0 = 12 \mbox{kg}[/itex]. The half-life is 5 days, so (omitting units for brevity)

[tex]I(5) = \frac{I_0}{2} = 6 = I_0 e^{-5k} = 12e^{-5k}[/tex]

ie. you're solving

[tex] 6 = 12e^{-5k}[/tex]

for [itex]k[/itex]. Once you do that, you need to find [itex]I(10)[/itex]. All you need to do to do that is to evaluate the formula for [itex]I(t)[/itex] with [itex]t=10[/itex]:

[tex]I(10) = 12e^{-10k}[/tex]

which isn't a problem now, since you found [itex]k[/itex] in the last step!
 
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  • #14
So in the case of this question it is pretty much an..

..arithmetic sequence.. :]

Thank you all very much, now I am sure i understood the idea and its definitely not as hard as i thought..
 
  • #15
No problem.
 
  • #16
Geometric sequence actually. An exponentially decaying quantity is one that is decaying continuously in analogy to the discrete ratio for successive terms of a geometric sequence.
 

Related to Calculation of a half-life how?

What is the half-life of a substance?

The half-life of a substance is the amount of time it takes for half of the initial amount of the substance to decay or disintegrate.

How is the half-life calculated?

The half-life is calculated using the formula t1/2 = ln(2)/λ, where t1/2 is the half-life, ln(2) is the natural logarithm of 2, and λ is the decay constant.

What is the importance of calculating the half-life?

Calculating the half-life is important for understanding the rate of decay of a substance, determining its stability, and predicting its behavior in various environments.

Can the half-life of a substance change?

The half-life of a substance is a constant value for a specific substance under specific conditions. It cannot change unless the conditions or the substance itself changes.

What factors can affect the half-life of a substance?

The half-life of a substance can be affected by temperature, pressure, and the presence of other substances that can interfere with its decay process.

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