- #1

exequor

- 393

- 0

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.