Recent content by darren t

hey: So i did my calculation again t = 5715 / 0.693 * ln (100 / 75) t = 8246.75 * ln (1.33) t = 8246.75 * 0.28 t = 2309.09 years Its still does not seem right! I think I might go with your way HallsofIvy :)

Thanks for getting back to me A is the amount of radioactivity it has when it is new and Ao refers to how much radioactivity is has now! Normally this value would be in Curies ect,. but no other data was given! Your right the value should be 57 not 570, that was a typo on my end, Sorry!

[Hey I have also been asked this question, A mouse dropping was found in a house in Ireland and had a 14C:12C ratio that was 57% of a modern day standard. Use this value to estimate the aprox date when the mouse visited the house? The half life of 14C is 5715 year I used the following...

Hi thanks for getting back to me Thats all the information i was given. I have tried to calculate it. What do you think?? Note first that the measured count rate for any sample can be converted to the weight of labelled B12 from the first line of information count rate = 7.36 x 10^5...

Hi Guys, Im just having a little trouble with this question, I was just wondering whether some one could point me in the right direction. Fifty six micrograms of 60Co - labeled Vitamin B12 containing 7.36x10^5 c.p.m were added to a sample containing an unknown amount of unlabeled Vitamin...