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\prob{26}
9277_26

A radioactive nucleus decays, with the activity shown in the graph above. The half-life of the nucleus is


  1. 2 min
  2. 7 min
  3. 11 min
  4. 18 min
  5. 23 min

Lab Methods}Log-Log graph

Since initially, the counts per minute is 6E3, the half-count amount would be 3E3. This occurs between 5 and 10 minutes. Choice (B) seems a good interpolation.

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
chrisfizzix
2008-10-06 11:48:03
The simple jump from intuition needed to understand Yosun's solution is as follows:

Each nucleus in the sample has the same probability of decaying in the next time dt. The number of nuclei that do decay in that time is just the number of nuclei in the sample times the probability of decay in time dt. This implies that the total decay rate at any given time is proportional to the number of nuclei in the sample. So, if the count rate falls by a factor of two over a finite time interval, then we know that the population of the sample must have also decreased by a factor of two.

Thus, to solve the problem, just find the count rate at t=0 and then find the later time at which the count rate is half that of t=0. Then the elapsed time in the half-life.

Initially, the count rate is 6 \times 10^4 . The count rate falls to 3 \times 10^4 at t=7 min. Thus, the half-life is 7 minutes.
Alternate Solution - Unverified
troy12345
2006-11-01 13:31:36
Though this is a more complicated way, it's the way they wanted you to do it in the problem:

The activity is given by log(dN/dt) = -L*t*log(e) + log(N0*l), where L is the inverse lifetime of the substance. So the slope of the graph is -L*log(e), which looks to be approximately -2/(40 minutes). Solving for L, we get L = -2/(40 * log(e) minutes). The halflife is given by T = ln(2)/L = log(e)*log(2)/(2/(40 * log(e) minutes)) = log(2)*20 minutes = 7 minutes.
Alternate Solution - Unverified
Comments
gabep
2009-09-26 03:52:54
This is a semi-log graph, not a log-log. ("Lab Methods}Log-Log graph")NEC
jarg
2009-03-16 15:28:09
Yosun you are one hell of a badass !
(no insult intented)
NEC
chrisfizzix
2008-10-06 11:48:03
The simple jump from intuition needed to understand Yosun's solution is as follows:

Each nucleus in the sample has the same probability of decaying in the next time dt. The number of nuclei that do decay in that time is just the number of nuclei in the sample times the probability of decay in time dt. This implies that the total decay rate at any given time is proportional to the number of nuclei in the sample. So, if the count rate falls by a factor of two over a finite time interval, then we know that the population of the sample must have also decreased by a factor of two.

Thus, to solve the problem, just find the count rate at t=0 and then find the later time at which the count rate is half that of t=0. Then the elapsed time in the half-life.

Initially, the count rate is 6 \times 10^4 . The count rate falls to 3 \times 10^4 at t=7 min. Thus, the half-life is 7 minutes.
Alternate Solution - Unverified
ma
2007-10-13 07:11:09
i also don't understand how you get 6000 as the inital count.
can anyone help?
Rune
2007-10-27 13:04:29
The way you read a log scale is for example, go where it says , that is 1x10^3. For every horizontal line after that, the coefficient increases by 1 so it goes 1x10^3,2x10^3, 3x10^3, etc etc until it gets to 10x10^3 which is the same as 1x10^4. You can see that it starts on the 6th line above , so that is 6x10^3
NEC
dumbguy
2007-10-10 18:37:12
How do we know that the initial counts is 6000? I believe my problem is in reading the log graph.
lambda
2007-10-20 16:57:08
dumbguy & ma: The initial count is the same as the "counts per minute" on the y-axis when time = 0. The line intersects the y-axis at "counts per minute" = 6*10^3 = 6000.
Answered Question!
hamood
2007-04-02 12:46:26
The way I went about this was by choosing two points on the graph..at t=0, A=6000, and at t=20, A=800.
Then we can plug this into the formula \alpha = \alpha^_0 e^(\lambda t)
\lambda is ofcourse the decay constant, related to half-life by
half life = ln 2/ \lambda
NEC
enrico
2007-01-04 15:53:42
The problem is modelled by

dN / dt = - k N.

And the half life is defined as 1 / k, how the hell you get this answer. The value of k is 122.5 min...

This is not physics...
andrumen1
2010-10-07 18:42:40
No, rememeber that \frac{1}{k} is called the characteristic time. Not the half-life, however, they are related.
NEC
enrico
2007-01-04 13:29:41
This problem is kind of weird, and I think the vertical axis should be just ln (N(t)) or the quantity of the material N(t), not the number of counts which would mean dN/dt.NEC
troy12345
2006-11-01 13:31:36
Though this is a more complicated way, it's the way they wanted you to do it in the problem:

The activity is given by log(dN/dt) = -L*t*log(e) + log(N0*l), where L is the inverse lifetime of the substance. So the slope of the graph is -L*log(e), which looks to be approximately -2/(40 minutes). Solving for L, we get L = -2/(40 * log(e) minutes). The halflife is given by T = ln(2)/L = log(e)*log(2)/(2/(40 * log(e) minutes)) = log(2)*20 minutes = 7 minutes.
Rune
2007-10-27 13:02:55
The way you read a log scale is for example, go where it says 10^3, that is 1x10^3. For every horizontal line after that, the coefficient increases by 1 so it goes 1x10^3,2x10^3, 3x10^3, etc etc until it gets to 10x10^3 which is the same as 1x10^4. You can see that it starts on the 6th line above 10^3, so that is 6x10^3
Rune
2007-10-27 13:04:52
oops, replied to the wrong person

apr2010
2010-04-09 02:33:31
There is never "the" way to do it. Somebody said, once u started calculated you lost already. Push yourself to see the obvious, it is good to know how u would do it though.
gravity
2010-11-07 23:51:46
I find often that they do intend on there being two solutions to a problem - one that takes mere seconds and another that will take much more time. It's still testing our intuition - physics isn't just about equations and proofs, we're not math majors.

Regardless, I'm realizing the last month I've spent studying physics is by far less useful than combing over these practice tests in fine detail. Good test-taking skills are a must on this test, and you have to be able to spot the "cheap" way to answer something right away.
Alternate Solution - Unverified
jax
2005-12-05 07:35:20
Replace the E4 with E3 in the solution...
yosun
2005-12-06 20:15:42
jax: thanks for the typo alert; it's been corrected.
Fixed Typos!

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This is a semi-log graph, not a log-log. ("Lab Methods}Log-Log graph")

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