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Count Rate (Quick Clarification)

  • Thread starter _Mayday_
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[SOLVED] Count Rate (Quick Clarification)

Problem

The count rate of a source is 64 counts a second when 50mm away from the source. Assuming the inverse square law is obeyed, what would be the expected count rate when the source is 80mm away.

Solution

[tex]64 = 50[/tex][/quote]
I know what you mean but I cringe at seeing that. 64 is NOT equal to 50!

Distance doubles, count rate quarters.
Yes, that is true.

[tex]1280 = 10[/tex] Divide distance by 5, and so times countrate by 20
No, if the distance is divided by 5 (multiplied by 1/5) then the times are multiplied by 52= 25.

[tex]40 = 80[/tex] Times distance by 8, and so divide countrate by 32.

Count rate = 40
No, if multiply the distance by 8, you divide the countrate by 82= 64.
Oh, I see, you grabbed that "4" from "distance doubles" and are multiplying everything by it. It doesn't work that way. If you divide the distance by 2, then, as you say, the countrate is multiplied by 4. But if you divide by 2 again (to get 1/4 of the original distance) you multiply by 4 again- so you actually multiply by 16. If you divide by 4 yet again (to get 1/8 of the original distance) you multiply by 4 again: 4(16)= 64= 82.


_Mayday_
 
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Answers and Replies

Hootenanny
Staff Emeritus
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I'm afraid your solution isn't correct. Try writing down an equation representing the relationship between distance and count rate.
 
795
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As the distance doubles, the count rate falls by a factor of 4.

Count rate = 64 Distance = 50mm

Count rate = 65536 Distance = 10mm

Count rate = 1 Distance = 80mm
 
795
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That isn't correct. I must be going wrong. Am I correct in saying that if the distance doubles, then the count rate will quarter?
 
140
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That isn't the definition of the inverse square law.
 
795
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What is the definition of the inverse square law?
 
140
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[tex]A = \frac{constant}{B^2}[/tex]

Determine what A and B are in this case and find the value of the constant first
 
795
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[tex]Activity = \frac{Constant}{Distance^2}[/tex]
 
Hootenanny
Staff Emeritus
Science Advisor
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[tex]Activity = \frac{Constant}{Distance^2}[/tex]
Correct, now all you need to do is determine the value of the constant.
 
795
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That's funny we have studied the inverse square law, but the teacher has somehow managed to miss that! It looks to me to be the inverse square law in it's simplest form and yet he missed it? On a 12 question paper this is the only one I appear to have gotten wrong, which is suprising as it is only question 2! I'll have a word with my teacher!

[itex]Countrate = \frac{Constant}{Distance^2}[/itex]

[itex]64 = \frac{Constant}{50^2}[/itex]

[itex]64 = \frac{Constant}{2500}[/itex]

[itex]64 \times 2500 = Constant[/itex]

[itex]Constant = 160000[/itex]

[itex]Countrate = \frac{160000}{6400}[/itex]

Countrate = 25 Counts.

Okay, now using common sense I can say that this sounds about right. If I double 50mm to 100mm, and then quater 64 I get 16. But now I know that it must be greater than 16, as the distance doesn't quite double. Though this may sound stupid, atleast it is reassuring in my mind. If my answer is incorrect, then I probably sound like an idiot now! :haha:

Thanks to both Hoot and Astrorob for knocking some sense into me!

_Mayday_
 
Last edited:
140
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No problem, glad to be of help.
 

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