GR9277 #26



Alternate Solutions 
chrisfizzix 20081006 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 and then find the later time at which the count rate is half that of . Then the elapsed time in the halflife.
Initially, the count rate is . The count rate falls to at min. Thus, the halflife is 7 minutes.   troy12345 20061101 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.  

Comments 
gabep 20090926 03:52:54  This is a semilog graph, not a loglog. ("Lab Methods}LogLog graph")   jarg 20090316 15:28:09  Yosun you are one hell of a badass !
(no insult intented)   chrisfizzix 20081006 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 and then find the later time at which the count rate is half that of . Then the elapsed time in the halflife.
Initially, the count rate is . The count rate falls to at min. Thus, the halflife is 7 minutes.   ma 20071013 07:11:09  i also don't understand how you get 6000 as the inital count.
can anyone help?
Rune 20071027 13:04:29 
The way you read a log scale is for example, go where it says , that is 1x. For every horizontal line after that, the coefficient increases by 1 so it goes 1x,2x, 3x, etc etc until it gets to 10x which is the same as 1x. You can see that it starts on the 6th line above , so that is 6x

  dumbguy 20071010 18:37:12  How do we know that the initial counts is 6000? I believe my problem is in reading the log graph.
lambda 20071020 16:57:08 
dumbguy & ma: The initial count is the same as the "counts per minute" on the yaxis when time = 0. The line intersects the yaxis at "counts per minute" = 6*10^3 = 6000.

  hamood 20070402 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 = e^( t)
is ofcourse the decay constant, related to halflife by
half life = ln 2/   enrico 20070104 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 20101007 18:42:40 
No, rememeber that is called the characteristic time. Not the halflife, however, they are related.

  enrico 20070104 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.   troy12345 20061101 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 20071027 13:02:55 
The way you read a log scale is for example, go where it says , that is 1x. For every horizontal line after that, the coefficient increases by 1 so it goes 1x,2x, 3x, etc etc until it gets to 10x which is the same as 1x. You can see that it starts on the 6th line above , so that is 6x

Rune 20071027 13:04:52 
oops, replied to the wrong person

apr2010 20100409 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 20101107 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 testtaking skills are a must on this test, and you have to be able to spot the "cheap" way to answer something right away.

  jax 20051205 07:35:20  Replace the E4 with E3 in the solution...
yosun 20051206 20:15:42 
jax: thanks for the typo alert; it's been corrected.

 

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