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Help! How to calculate the remaining activity of a radioactive substance

 
 
YRX
 
Reply Sun 29 Nov, 2020 09:09 pm
I've got a question on a book about the half-life of a radioactive substance.
The given graph stated that the activity of the radioactive material is 1000 counts per minute at t=0.
It was given that the activity of the particular substance reduced to 800 counts per minute at 20th minute. At the 35th minute, the activity of the substance was further reduced to 400 counts per minute.
The question asked for the remaining activity after an hour.
Given selections are 50, 100, 200 and 400.

My solution is as below,
First, the initial activity of the substance is 1000 counts per minute (cpm). After a half-life, the activity should be reduced to about 500 which appears to be in between the time frame 20 to 35 minutes. Thus, I can deduce that the half-life of the substance should be in between 20 to 35 minutes.
20<Half-life<35
Divided by an hour
3>No. of half-life in an hour>1.7143
Calculating the approximate range of remaining activity should be:
1000 x (1/2)^3< Remaining activity< 1000 x (1/2)^1.7143
125< Remaining activity<304.7504
And thus, the only suitable answer should be 200 cpm as it was the only number in the calculated range.

My teacher's solution is as below,
According to the given value, the half-time of the radioactive substance is 15 minutes as the radioactive activity reduced by half from 20th to 35th minute (800 to 400 as stated above). And thus, starting from t=0, when t=60, 4 half-lives have occurred and the calculation should be:
1000 x (1/2)^4 = 62.5 cpm
The answer is 50 as it is the nearest value to the calculation.
(My reasoning for accusing the solution is incorrect are as below
If the half-life of the substance is 15 minutes, at 20th minute, the activity should be smaller than 500cpm.)

The answer given by the book was 50. There were no solutions or explanation

PS* Can anyone teach me how to attach the image cus it would be much more easier for me to ask questions lol.
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maxdancona
 
  1  
Reply Sun 29 Nov, 2020 10:22 pm
@YRX,
I agree with your logic. If the decay is exponential you would think that the count would be below half of the original value at t=20.

Did you ask your teacher?

It doesn't sound like this is an exponential function.

maxdancona
 
  1  
Reply Sun 29 Nov, 2020 10:26 pm
@YRX,
To post an image, you need to put it on an image hosting service. The. You put the URL in an
Code:[img]http://hosting.com/pathyoyourimag[/img]


I use imagebb. There are others. If you post an image to one of these sites, give me the URL and I can help you get it into a post.
0 Replies
 
YRX
 
  1  
Reply Sun 29 Nov, 2020 10:28 pm
@maxdancona,
I've asked but my teacher doesn't seem to really accept the solution XD
btw can you teach me in detail how to use the image hosting service?
YRX
 
  1  
Reply Sun 29 Nov, 2020 11:21 pm
@YRX,
[img][https://ibb.co/rdCbjQF/img]
0 Replies
 
maxdancona
 
  1  
Reply Mon 30 Nov, 2020 09:02 am
@YRX,
OK... I see the graph. I see the problem.

That "0" on the x-axis of the graph is clearly incorrect... the graph does not start at T=0. In fact, it looks like the graph starts at T=15 (I didn't do the math, but assuming it is an exponential curve it wouldn't be that difficult to calculate).

0 Replies
 
maxdancona
 
  2  
Reply Mon 30 Nov, 2020 09:05 am
@YRX,
https://i.ibb.co/kmR37G5/Phys-49.png

To do this I typed the following

Code:[img]https://i.ibb.co/kmR37G5/Phys-49.png[/img]


I got this URL by going to your link.. and then right clicking on the image and "open in new tab". That opened just the image on a new tab with a URL I could cut and paste.

To skip this step, there is a button that says something like "link to HTML"... I just copy the URL from there.
0 Replies
 
maxdancona
 
  1  
Reply Mon 30 Nov, 2020 09:16 am
@YRX,
Let's do the math.

Your teacher is correct that the half-life is clearly 15 minutes.

We are looking for a function in the form of C = A(1/2)^(t/15).

We can calculate the value of A with any one of these points.

800 = A (1/2)^(20/15)

I get the value of A as 2015.87. Since A corresponds to the value of the count at T=0... looking at the graph makes me feel good (it makes sense).

So then I plug in 60 for t.

C = 2015.87*(1/2)^(60/15)

I get 126 (I am doing this quickly... I could be making a mistake).

I believe 126 makes sense.... because it is a little less than 2 half-lives from 400 (and should be a little more than a quarter of this value).


50--- is clearly NOT the correct answer. Unless maybe they mean an hour after the origin... in which case they would really mean t = 75?

This is a horribly written problem.

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