The percentage error in the Radioactive substance Population

Click For Summary
The discussion revolves around calculating the percentage error in the population of a radioactive substance after 9 days, given an underestimated half-life of 2.25 days, which is actually 2.5 days due to a 10% error. The original poster is struggling to relate this corrected half-life to the population after 9 days, seeking clarification on how to proceed. Other participants emphasize the need to show detailed calculations to identify any errors in the approach. The conversation highlights the importance of using the correct half-life to determine the population decay accurately. Ultimately, the original poster is looking for guidance on applying the correct equations to find the desired percentage error in population.
curious_mind
Messages
48
Reaction score
9
Homework Statement
Initially the half-life is measured 2.25 days, but later it was found underestimated by 10%. It is required to find the percentage error in "Population" of the substance after 9 days.
Relevant Equations
## N = N_0 e^{-\lambda t} ##
Please check the question below as given originally. Answer given is 25%. I am unable to proceed.

It is given that the half-life is underestimated by 10% therefore it must be larger than originally estimated.
What I can find using the percentage error formula is ##\left( \dfrac{Actual-Estimated}{Actual} \right) \times 100% = \left( \dfrac{Actual-2.25}{Actual} \right) \times 100%=10%##

So, ##Actual = 2.5 ~days##
Now, I am unable to make relation of this with the population after 9 days, which is required to find in the question. The answer given is ##25%##. How it is obtained?

Thanks.
 

Attachments

  • Capture1.PNG
    Capture1.PNG
    23.8 KB · Views: 117
Physics news on Phys.org
What answer did you get, and how did you get there ?
 
hmmm27 said:
What answer did you get, and how did you get there ?
I am not getting required answer that's why posted question here. For The original question, I attached a picture. Answer given is 25%.

What I obtained is the actual half-life given the estimated half life and the underestimation percentage error of measuring half life.

It is Required to find the percentage error in the population ##N## after 9 days. I am unable to proceed for it.
 
curious_mind said:
I am not getting required answer that's why posted question here. For The original question, I attached a picture. Answer given is 25%.

What I obtained is the actual half-life given the estimated half life and the underestimation percentage error of measuring half life.

It is Required to find the percentage error in the population ##N## after 9 days. I am unable to proceed for it.
As @hmmm27 already said, please show your work. That you obtained a result that doesn't match a known answer is insufficient information for us to figure out what you did wrong, if anything.
 
kuruman said:
As @hmmm27 already said, please show your work. That you obtained a result that doesn't match a known answer is insufficient information for us to figure out what you did wrong, if anything.
I felt I already showed my work and even posted original question in the thread. But elaborating more..

We are given percentage error in half life measurement of radioactive substance as 10%. It is given as underestimated, which means the actual half-life should be more, ok?

Now percentage error of any physical quantity is given by (Actual-Estimated)/Actual * 100% which is given 10%

So (Actual-2.25)/Actual * 100% =10%

This way I obtained actual half-life to be = 2.5 days.

Now it is asked to find the percentage error in population of substance after 9 days.

Which I am unable to tell what to do next
 
You need to find the population with the correct half-life after 9 days. How will you do that? You have posted a "relevant equation." Do you think it could be a useful equation?
 
If have close pipe system with water inside pressurized at P1= 200 000Pa absolute, density 1000kg/m3, wider pipe diameter=2cm, contraction pipe diameter=1.49cm, that is contraction area ratio A1/A2=1.8 a) If water is stationary(pump OFF) and if I drill a hole anywhere at pipe, water will leak out, because pressure(200kPa) inside is higher than atmospheric pressure (101 325Pa). b)If I turn on pump and water start flowing with with v1=10m/s in A1 wider section, from Bernoulli equation I...

Similar threads

  • · Replies 21 ·
Replies
21
Views
1K
Replies
4
Views
3K
Replies
4
Views
2K
Replies
23
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
0
Views
1K