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_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!

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

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

^{2}= 25.

No, if multiply the distance by 8, you divide the countrate by 8[tex]40 = 80[/tex] Times distance by 8, and so divide countrate by 32.

Count rate = 40

^{2}= 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= 8

^{2}.

_Mayday_

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