Asked by Andrew
The average human body has 1.4x10^2 g of total K within it. Calculate the activity in disintegration/year of the potassium-40 in the average human body using your value of t1/2 (or K) of this substance
Answers
Answered by
bobpursley
Potassium-40 (40K) is a radioactive isotope of potassium which has a very long half-life of 1.251×109 years. It makes up 0.012% (120 ppm) of the total amount of potassium found in nature. Potassium-40 is a rare example of an isotope that undergoes both types of beta decay.so if 140 grams of K are in the body, then .00012x140 grams is the initial K-40 (.0168grams). Now, that relates then to how many atoms?
.0168grams/40 * avagrado[s number or 2.53E20 atoms.
But the half life is 1.251×109 years, so the disintegrations per year
N=No*e^-kt
dN/dt=-kNoe^-kt
at t=0, where k=.693/1.25E9
dN/dt=.693*2.53E20/1.25E9=1.4E11 dis/year
check my work.
.0168grams/40 * avagrado[s number or 2.53E20 atoms.
But the half life is 1.251×109 years, so the disintegrations per year
N=No*e^-kt
dN/dt=-kNoe^-kt
at t=0, where k=.693/1.25E9
dN/dt=.693*2.53E20/1.25E9=1.4E11 dis/year
check my work.
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