Half-Life Problem: Solve w/o Given Amount of Sample

In summary: No. The problems at that link only deal with the simplest situation: a decay from one species to a much longer... something else.In summary, you use a formula such as $$n\frac{t}{T}$$ but are unsure you've never done one without a given amount of the sample.
  • #1
amazingphysics2255
60
4
Homework Statement
Yttrium 90 is also radioactive and is a beta particle emitter with a half-life of 2.67 days.

Discuss the amount of Yttrium that would be present in the above example after the
Strontium had been decaying for the first 8 days.
Relevant Equations
N/a
I believe I use a formula such as $$n\frac{t}{T}$$ but am unsure I've never done one without a given amount of the sampleSo if I was to use this I would get 8/(divided by)2.67 = 2.99625468 rounded to either 3 or 2.997 But i don't think this is write because you need to use 1/2 in this formula.

edit: So 8/2.67 is going to equal the number of half-lives has gone by so 2.997 halflives have gone by

(1/2)^2.997= 0.125 rounded so is the answer 0.125g?
 
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  • #2
Find a relevant equation ... your belief isn't good enough .

As for the intial amount, pick something and call it ##n_0##. The exercise asks what fraction of that is left over after 8 days.

[edit] o:) misread the (partial ?) problem statement. Sorry ...
 
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  • #3
So my answer is incorrect?...I don't believe I'm allowed to make up a random value to solve this either.
 
  • #4
Your question seems rather incomplete.
Are you starting out with 100 % strontium 90, and then you need to find out how much Yttrium there is after 8 days? Eventually you get an equilibrium where the $$ \frac {N_Y} {N_{Sr}} = \frac {T_Y} {T_{Sr}} $$
(N is number of atoms, T is half-life)
Using the equilibrium value isn't too accurate, since there are only 3 half-lifes of yttrium gone, so you'll need to solve a differential equation for the amount of yttrium that's left.
 
  • #5
@willem2 The half-life of Strontium 90 is about 29 years and it decays into Yttrium 90. Yttrium 90 is also radioactive and is a beta particle emitter with a half-life of 2.67 days. Discuss the amount of Yttrium that would be present in the above example after the Strontium had been decaying for the first 8 days.

Perhaps this bit at the start will help?
 
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  • #6
8/2.67= 2.997 (1/2)^2.997=0.12526 = 12.526% percent of the sample remaining?

or If you round 2.996254682 to = 3 rounded it = 0.125 decimal or 1/8 as a fraction or 12.5%
 
  • #7
amazingphysics2255 said:
8/2.67= 2.997 (1/2)^2.997=0.12526 = 12.526% percent of the sample remaining?

or If you round 2.996254682 to = 3 rounded it = 0.125 decimal or 1/8 as a fraction or 12.5%
As @willem2 posted, you need to start with a differential equation or two.
Create variables for the number of atoms of each species that are present at time t. Then write equations for their rates of change.
 
  • #8
amazingphysics2255 said:
8/2.67= 2.997 (1/2)^2.997=0.12526 = 12.526% percent of the sample remaining?

or If you round 2.996254682 to = 3 rounded it = 0.125 decimal or 1/8 as a fraction or 12.5%
Yes. That is correct.
 
  • #9
Chestermiller said:
Yes. That is correct.
But it doesn’t help in answering the question. There is not a specific initial quantity of Y90. It is being created and destroyed simultaneously.
 
  • #10
haruspex said:
But it doesn’t help in answering the question. There is not a specific initial quantity of Y90. It is being created and destroyed simultaneously.
Thanks. I misunderstood the question.
 
  • #11
haruspex said:
As @willem2 posted, you need to start with a differential equation or two.
Create variables for the number of atoms of each species that are present at time t. Then write equations for their rates of change.
I think you need to give the OP a more detailed idea of how to proceed.
 
  • #12
haruspex said:
As @willem2 posted, you need to start with a differential equation or two.
Create variables for the number of atoms of each species that are present at time t. Then write equations for their rates of change.
Chestermiller said:
I think you need to give the OP a more detailed idea of how to proceed.
amazingphysics2255 said:
I didn't make the problem, so I solved it with what I was given.
Let ##N_Y(t)## and ##N_S(t)## be the number of atoms of Yttrium and Strontium respectively at time t. You know the values at time 0.
If the decay rates are ##\lambda_Y## and ##\lambda_S##, what is the rate of change of each count at time t?
 
  • #13
amazingphysics2255 said:
Is this not 90
90 is the number of nucleons in each atom of the Strontium. It is nothing to do with the number of atoms.
At the start there is no Yttrium, only Strontium
amazingphysics2255 said:
Firstly, isn't it beta decay?
The type of decay is irrelevant.
amazingphysics2255 said:
do you mean t time as in 8 days?
I mean at some arbitrary time t. A rate of decay ##\lambda## means that for each atom there is a probability ##\lambda dt## that it will decay in a short period dt.
If there are N(t) such atoms, how many on average will decay in time dt?
amazingphysics2255 said:
is this problem like mine?
No. The problems at that link only deal with the simplest situation: a decay from one species to a much longer lived one. Your problem involves decay to and from an intermediate species.
 
  • #14
haruspex said:
90 is the number of nucleons in each atom of the Strontium. It is nothing to do with the number of atoms.
At the start there is no Yttrium, only Strontium
So there is only 1 atom Strontium. Which has 52 neutrons and 38 Protons
 
  • #15
amazingphysics2255 said:
So there is only 1 atom Strontium. Which has 52 neutrons and 38 Protons
There is an unknown quantity of Strontium. Unless there is further information on that, interpret the question as asking what fraction of the total number of atoms will be Yttrium.
 
  • #16
All I know is. The half-life of Strontium 90 is about 29 years and it decays into Yttrium 90. Yttrium is a beta particle emitter with a half-life of 2.67 days.

I am then told to
Discuss the amount of Yttrium that would be present in the above example after the Strontium had been decaying for the first 8 days.

So I need to find out the amount of Yttrium that is in the strontium after decaying for 8 days?
 
  • #17
amazingphysics2255 said:
All I know is. The half-life of Strontium 90 is about 29 years and it decays into Yttrium 90. Yttrium is a beta particle emitter with a half-life of 2.67 days.

I am then told to
Discuss the amount of Yttrium that would be present in the above example after the Strontium had been decaying for the first 8 days.

So I need to find out the amount of Yttrium that is in the strontium after decaying for 8 days?
Right, and I can lead you to the answer. Start by trying to answer the question I asked in post #14.
 
  • #18
haruspex said:
I mean at some arbitrary time t. A rate of decay λλ means that for each atom there is a probability λdtλdt that it will decay in a short period dt.
If there are N(t) such atoms, how many on average will decay in time dt?
So I need to find how many on average will decay in the dt? and dt is not 8 days is it?
 
  • #19
amazingphysics2255 said:
So I need to find how many on average will decay in the dt? and dt is not 8 days is it?
dt is some very brief period.
If an event happens at rate x, on average, how many events would you expect in time dt?
 
  • #20
I don't know if this help but I figured
For strontium you have the function

$$f(t)=100e^{ \frac{\ln(0.5)}{29 \times 365}}$$

as for your question, I'm not going to lie I don't fully understand it as to why I haven't answered it
 
  • #21
amazingphysics2255 said:
I don't know if this help but I figured
For strontium you have the function

$$f(t)=100e^{ \frac{\ln(0.5)}{29 \times 365}}$$
That's a useful start, but the 100 is only needed if you want the answer as a percentage. That will get confusing here, so let's drop that.
The exponent is missing the factor t; what you have written is not actually a function of time.
The constant in the exponent is (minus) the rate of decay, ##\lambda_S##, in events per day.
So we can write your equation more generally as
##N_S(t)=N_S(0)e^{-\lambda_S t}##
This can be obtained from the differential equation ##\frac{dN_S}{dt}=-\lambda_SN_S##. That is, the rate of loss of Strontium atoms is proportional to the number remaining and the decay rate ##\lambda_S##.

The next step is to obtain the differential equation for the number of Yttrium atoms. It is similar to the equation above, but we now need an extra term to represent the Yttrium atoms being created by decay of the Strontium atoms. Have a go at that.
 
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  • #22
If you are like me, the idea of writing down and solving differential equations may be off-putting. There is an approach that dodges that difficulty.

We have a formula for the rate at which Yttrium-90 that is being added to the sample as a function of time.

We can easily derive a formula for the fraction of the Yttrium-90 added at a particular time that will survive until the 8 day interval expires.

We could simply integrate the product of the two to figure out how much Yttrium-90 is added and will survive.
 
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  • #23
The half-life of the Strontium is very large compared to the half-life of the Yttrium (29 years vs 8 days). So during the 8 days, the amount of strontium hardly changes at all, and the rate of production of Yttrium is essentially constant.

Your initial calculation assumed that all the Yttrium that was produced from the Strontium decay suddenly appeared at time zero. In that case, there would be 12.5% of the Yttrium that was produced remaining after 8 days. But, since the Yttrium is being produced at a constant rate (rather than all at once), some of the Yttrium produced does not decay for 8 days. Most of it decays for less than 8 days. For example, the Yttrium produced on day 5 has only 3 days to decay. Based on this, do you think that the fraction of all the Yttrium produced during the 8 days and that still remains after reaching day 8 will be greater of less than 12.5%.
 
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  • #24
amazingphysics2255 said:
@haruspex When I wrote my function for strontium I also wrote this for Yttrium. So for the yttrium, you have to start with the decayed parts of the strontium.
$$f_{2}(t)= (100-f(t))e^{ \frac{\ln(0.5)}{2.7}}$$ But you've since advised me this is not the case.

@Chestermiller I believe it will be less than 12.5%

@jbriggs444 Yes I'd like to see that method?
It will be substantially greater than 12.5 %. I calculate about 42%. This is because all the yttrium produced has been around for less than 8 days. Do you know anything about writing and solving simple ordinary differential equations?
 
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  • #25
I have been wondering what is the course from which this question was taken? and what is the actual full
statement of the problem?
 
  • #26
I just asked it's not meant to be solved. Like this, they just want me to discuss.
 
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  • #27
amazingphysics2255 said:
you have to start with the decayed parts of the strontium.
That would work if first the Strontium decayed for 8 days and then the Yttrium produced decayed for eight days. As everybody is trying to explain to you it is complicated by the two processes occurring simultaneously.

The simplest approach is to combine two suggestions that have been made:
Take the quantity of Strontium to be constant, since very little will have decayed in eight days.
Consider the quantity of Yttrium that was created in time dt at time t before the end of the eight days.
Multiply that by the fraction that will remain at the end (i.e. at time t later).
Perform the integral.
Chestermiller said:
I calculate about 42%.
To clarify, the context for that is the fraction of all the Yttrium that was produced that is still around after eight days.
 
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  • #28
amazingphysics2255 said:
I just asked it'snot meant to be solved

This is very curious problem. So no diff eq need be solved yet there substantial changes occurring over the time period begging for a diff eq. I'm not sure what answer is being looked for.
 

1. What is the "Half-Life Problem?"

The "Half-Life Problem" is a mathematical concept that involves determining the amount of time it takes for a substance to decrease by half through radioactive decay or chemical reactions. It is commonly used in fields such as physics, chemistry, and biology to determine the rate of decay or growth of a substance.

2. How do you solve the "Half-Life Problem" without a given amount of sample?

To solve the "Half-Life Problem" without a given amount of sample, you need to know the initial amount of the substance, the half-life of the substance, and the elapsed time. Then, you can use the formula A=A0(1/2)t/h, where A is the final amount, A0 is the initial amount, t is the elapsed time, and h is the half-life. This formula can be used to calculate the final amount of the substance after a certain amount of time has passed.

3. What are some common applications of the "Half-Life Problem"?

The "Half-Life Problem" has many applications in science and technology. It is used to determine the age of fossils and archaeological artifacts through radioactive dating. It is also used in nuclear physics to determine the decay rate of radioactive elements. In medicine, it is used to calculate the rate at which drugs are metabolized in the body. Additionally, it is used in environmental studies to track the decay of pollutants in the environment.

4. Can the "Half-Life Problem" be applied to all substances?

No, the "Half-Life Problem" can only be applied to substances that undergo radioactive decay or chemical reactions that follow the principle of exponential decay. These substances have a constant half-life, meaning that the amount of substance decreases by the same percentage over a specific period of time.

5. How is the "Half-Life Problem" used in nuclear energy production?

The "Half-Life Problem" is used in nuclear energy production to determine the rate of nuclear reactions and the amount of radioactive waste produced. It is also used in the design and maintenance of nuclear reactors to ensure safe and efficient operation. Understanding the principles of the "Half-Life Problem" is crucial in the management and disposal of nuclear waste to prevent harm to the environment and human health.

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