Physics: radioactivity question?

by exequor
Tags: physics, radioactivity
exequor is offline
May2-05, 04:09 PM
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P: 393
You are told that potassium 44 has a half life of 20 minutes and decays to form calcium 14.

Q1. how many atoms would there be in a 10mg sample of potassium?
Answer: I used avogadro's constant to find this by: (6E23/44)*10E-3 = 1.4E20 atoms

Q2. what would be the activity of the sample?
Answer: I used -dN/dt = lamda*N ... -dN/dt = (0.693/20*60)*1.4E20 = 8E16 Bq

Q3. what would the activity be after 1 hour?
Answer: I found the number of atoms after one hour (N=No*e^(-lamba*t)) then I found the activity for that number of atoms

lamda*t = (0.6931/20)*60 = 2.1
N = No*e^(-lamba*t) = 1.4E20*e^-2.1 = 1.7E19

hence -dN/dt = (0.693/20*60)*1.7E19 = 9.8E15 Bq

Q4. what would the ratio of potassium atoms to calcium atoms be after one hour?
answer: i know what the number of potassium atoms would be after one hour (1.7E19) but since the question stated that potassium 44 decays to form calcium 14 i don't know that I have to do to obtain the answer for this?

any help would be greatly appreciated.
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OlderDan is offline
May2-05, 05:50 PM
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You need to check your isotopes and get the numbers correct. If the calcium product of the potassium decay is stable, then all the potassium atoms that decay remain as calcium atoms. If it is not stable, then you would need to know the halflife of the calcium to do the problem.

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