10 points to whoever figures this out?
Thursday, December 24th, 2009 at
7:25 AM
The percent abundance of 40K (t ½ = 1.27 x 109 years) in naturally occurring potassium is 1.18 x 10-2. Calculate the total 40K activity, in units of disintegrations per hour, in a person with a mass of 77.3 kg of which 0.339% is potassium. Use 39.96 amu as the atomic mass of 40K.
Yes it is hard and I’m not sure how to do it.
If it helps any, activity = hn, where n is degenerations and h is 0.693/the half life (t1/2)
fractional abundance = # mol of radioactive nuclei/ # mol of sample
oops h is decay constant.
Filed under: Potassium Diet
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mass of 40K from the person
77.3 kg*0.339 = 26.2047 kg
total moles of abundant 40K
26.2047 kg(1000g/1kg)(1mole/39.96g)*0.0118
= 7.74 moles 40K
t1/2 = 1.27 x10^9years(12months/1year)(30days/1month)*
(24hour/1day) = 1.097 x 10^13 hour
t1/2 = ln2 / k
k = ln2 / t1/2
= ln2/ 1.097 x 10^13 hour
= 3.52 x 10-14 disintegration per hour
best answer?
wow that is really hard
ok give me the10 points anyway cause I sat here and read that whole thing.