Physics: radioactivity question?

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Potassium-44 has a half-life of 20 minutes and decays into calcium-14. A 10 mg sample contains approximately 1.4E20 atoms of potassium. The initial activity of the sample is calculated to be about 8E16 Bq, which decreases to 9.8E15 Bq after one hour. The decay process indicates that for every potassium atom that decays, one calcium atom is formed, leading to a ratio of potassium to calcium atoms of 1:8.2 after one hour. Understanding the stability of the decay products is crucial for accurate calculations.
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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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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.
 


To find the ratio of potassium atoms to calcium atoms after one hour, we can use the equation N(K)/N(Ca) = e^-lambda*t, where N(K) is the number of potassium atoms and N(Ca) is the number of calcium atoms. We already have the value for N(K) after one hour (1.7E19), so we just need to find the value for N(Ca).

To do this, we can use the fact that potassium 44 decays to form calcium 14. This means that for every one potassium 44 atom that decays, one calcium 14 atom is formed. Therefore, the ratio of N(K)/N(Ca) is 1:1.

Plugging this into the equation, we get 1.7E19/N(Ca) = e^-2.1, which can be rearranged to solve for N(Ca):

N(Ca) = 1.7E19/e^-2.1

N(Ca) = 1.7E19/0.122

N(Ca) = 1.4E20

Therefore, after one hour, there would be 1.7E19 potassium atoms and 1.4E20 calcium atoms, giving a ratio of 1:8.2.
 
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