Saturday, November 20, 2010

Answer to query by JM from N99/2/7bii and c

After balancing the equation you end up with a particle which has 0 mass and +1 charge. Hence it is called a positron. Not frequently encountered . Mainly particles like alpha , beta, protons and neutrons. Above all you have gamma rays emitted.

c) The rock is all potassium in the beginning. Today it is of the ratio of 1 for every 7 Argon found. Argon is the product of Potassium decay. Therefore i can deduced that in the beginning there were 8 potassium and no argon but today 1 potassium and 7 argon. Hence the ratio of potassium today to that initially is 1:8.
1/8 = e-(^t) . t can be found. The age of the rock.

cii) If Argon managed to escape the rock. The number of argon should be more than 7. Hence using the ratio again 1/N where N is more than 8 , t is greater. Hence the above t is an underestimate.

N06/II/7b and c
The ratio is misleading so you have to change it to the same denominator in order to know how many radioactive 235 is present. So the factor to multiply is 7 . Hence I have 1/140 today and 7/140 long time ago . So how long is that.
The number I am concerned is just 1 today and 7 long time ago. Hence the ratio 1/7 = e-(^t) should give you the time elapse.