Talk:Basic Physics of Nuclear Medicine/Atomic & Nuclear Structure

In regards to question 17 on the quiz: $$A \propto Z^2$$ is incorrect; $$A \propto Z$$ is more correct and the best option available for your quiz. When you say something x is "directly proportional" to y ($$x \propto y$$) you mean $$x = k y$$ for some unknown constant k.  Or if you measure (x1 = 10, y1= 20) and you know they are linearly proportional, you can find out that if x2 = 20 that y2=40  (and if proportional to the square it would be y2=80).

Let's look at some randomly chosen stable forms:

Z = 1; A = 1

Z = 2; A = 4

Z = 3; A = 7

Z = 4; A = 9

Z = 5; A = 11

Z = 10; A = 20

Z = 20; A = 40

Z = 30; A = 64

Z = 40; A = 90

Z = 50; A = 120

The trend we see is that $$A \propto Z$$ fits much better than $$A \propto Z^2$$.

Take any two elements (besides H and He) say B (Z=5) where A(B)=11; and try using the a proportionality to Z and Z^2 to guess the A for Sn(Z=50). $$A \propto Z$$, would guess that since Z increased by a factor of 10, that A should increase by a factor of 10, to A(Sn) = 110. $$ A \propto Z^2$$ would guess that since Z increased by a factor of 10, A should increase by a factor of 100, and guess A(Sn)= 1100. The linear proportionality is much closer to the data than the squared proportionality, though there are better models then this. 67.244.53.200 (talk) 09:50, 3 September 2009 (UTC)

Thank you for your contribution marz (talk) 03:10, 4 September 2009 (UTC)