Finding unlabeled vitamin B12 using radioactive tracer dilution

[darren t]

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 B12. The sample was then extracted and the Vitamin B12 purified. The final product contained 49ug of Vitamin B12 and 1.58x10^5 c.p.m of radioactivity. Calculate the amount of unlabelled Vitamin B12 in the sample

I have no idea at all were to start! Could some one please help
Thank you in advance! All the best and happy new year. Darren

 

[malawi_glenn]

First you need to find out the half lives of the isotopes under consideration. Do you know the relation between activity and halv life?

 

[darren t]

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 cpm/56 per microgram of labelled B12

Second, when the 56 ug is added to an unknown weight w of unlabelled B12, the ratio of unlabelled to labelled B12 is w/56.

The 49 ug of the mixture gives a count rate of 1.58 x 10^5 cpm and therefore contains [1.58 x 10^5/7.36 x 10^5] x 56 = 12.02 ug of labelled B12. The ratio of unlabelled to labelled B12 for this sample is (49 - 12.02)/12.02 and we must assume that this figure has not been altered by the processing of the materials. That is

w/56 = (49 - 12.02)/12.02 giving w = 173 ug

 

[Borek]

This is a simple dilution problem, and 173 ug seems to be the correct answer.

In the real life you should check half life of 60Co, and compare it with the timescale of the expriment - if the difference is large enough you may safely assume that activity of the sample remains constant during experiment.