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.

No, if the distance is divided by 5 (multiplied by 1/5) then the times are multiplied by 5^{2}= 25.

No, if multiply the distance by 8, you divide the countrate by 8^{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}.

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!

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!