# Ratio of potassium atoms to calcium atoms in 1 hour

#### moenste

1. Homework Statement
Potassium 44 (4419K) has a half-life of 20 minutes and decays to form 1420Ca, a stable isotope of calcium.
(a) How many atoms would there be in a 10 mg sample of potassium 44?
(b) What would be the activity of the same?
(c) What would the activity be after one hour?
(d) What would the ratio of potassium atoms to calcium atoms be after one hour?
(NA = 6 * 1023 mol-1.)

Answers: (a) 1.4 * 1020, (b) 7.9 * 1016 Bq, (c) 9.8 * 1015 Bq, (d) 1:7

I got every part except (d).

2. The attempt at a solution
(a) 44 g of Potassium 44 contain 6 * 1023 atoms. (10 * 10-3) g of Potassium 44 contain ((10 * 10-3) / 44) * (6 * 1023) = 1.36 * 10 20 atoms.

(b) dN / dt = -λN = - ((ln 2) / (20 * 60)) * (1.36 * 1020) = -7.9 * 1016 s-1 -> 7.9 * 1016 Bq

(c) A = A0 e-λt = (7.9 * 1016) * e-((ln 2) / (20 * 60)) * (60 * 60) = 9.8 * 1015 Bq

(d) For this part I am lost. Any ideas please? I did the same calculations for Calcium as in part (a-c) and compared the numbers and they all got me around 0.3 and not the 1:7 (0.14).

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#### PWiz

The number of atoms of potassium left after time $t$ is given by $N=N_0 e^{-\lambda t}$ where $N_0$ is the number of atoms at time $t=0$. We also know that the number of calcium atoms after 1 hour is equal to the number of potassium atoms that decayed in 1 hour. Can you calculate the number of decayed atoms?

• moenste

#### epenguin

Homework Helper
Gold Member
d is the easiest part of the problem - you need only think about half-lives. What hapens by end of one half life, two,... how many half lives are there in an hour?

• moenste

#### moenste

The number of atoms of potassium left after time $t$ is given by $N=N_0 e^{-\lambda t}$ where $N_0$ is the number of atoms at time $t=0$.
Using formula: NK = 1.4 * 1020 * e-(ln 2 / 20 * 60) * 60 * 60 = 1.75 * 1019 atoms of K after 1 hr.

N0 Ca = ((10 * 10-3) / 14) * 6 * 1023) = 4.3 *1020 atoms of Ca at t = 0
NCa = 4.3 * 1020 * 0.125 = 5.357 * 1019 atoms of Ca at t = 1 hr.

Divide K by Ca gives 0.33 as I used to get before.

We also know that the number of calcium atoms after 1 hour is equal to the number of potassium atoms that decayed in 1 hour. Can you calculate the number of decayed atoms?
1.4 * 1020 - 1.75 * 1019 = 1.225 *1020. The answer doesn't fit the 5.4 * 1019 which I got in the previous calculation. Though if I 1.75 * 1019 / 1.225 * 1020 = 1 / 7 indeed. But why did I get 5.4 * 1019 in the first part?

d is the easiest part of the problem - you need only think about half-lives. What hapens by end of one half life, two,... how many half lives are there in an hour?
We have T1/2 = 20 min, so there are 3 of them in 1 hour.

#### PWiz

N0 Ca = ((10 * 10-3) / 14) * 6 * 1023) = 4.3 *1020 atoms of Ca at t = 0
No! The number of Ca atoms at t=0 is 0. No potassium atom from the sample has decayed at t=0.

• moenste

#### PWiz

By the way, did you notice something wrong with this? I've never seen an atom with a mass number less than its proton number • moenste

#### epenguin

Homework Helper
Gold Member
We have T1/2 = 20 min, so there are 3 of them in 1 hour.
OK there are three half-lives in an hour. So after an hour what fraction of the original 44K is left?

• moenste

#### moenste

No! The number of Ca atoms at t=0 is 0. No potassium atom from the sample has decayed at t=0.
I get it know, thank you.

OK there are three half-lives in an hour. So after an hour what fraction of the original 44K is left?
N / N0 = e-λt = 0.125 is remaining.
1.4 * 1020 * 0.125 = 1.75 *1019 atoms are left
1.4 * 1020 - 1.75 * 1019 = 1.225 *1020 Ca atoms in 1 hour
1.75 * 1019 / 1.225 * 1020 = 1 / 7 ratio of K atoms to Ca atoms in 1 hour

#### epenguin

Homework Helper
Gold Member
You don't need formulae with exponentials or logs to work this out.
After 20 min. half remains, after 40min half of that, after 60min half of that again - so ½×½×½ = 1/8 , and eighth. Yes 0.125. Ratio to product (43Ca I suppose it is) 1:7

Last edited:
• moenste

#### moenste

You don't need formulae with exponentials or logs to work this out.
After 20 min. half remains, after 40min half of that, after 60min half of that again - so ½×½×½ = 1/8 , and eighth. Yes 0.0125. Ratio to product (43Ca I suppose it is) 1:7
Now I get it. Thank you. And you meant 0.125 right (not 0.0125)? :)

#### epenguin

Homework Helper
Gold Member
• moenste

#### F1Dan

1. Homework Statement
Potassium 44 (4419K) has a half-life of 20 minutes and decays to form 1420Ca, a stable isotope of calcium.
(a) How many atoms would there be in a 10 mg sample of potassium 44?
(b) What would be the activity of the same?
(c) What would the activity be after one hour?
(d) What would the ratio of potassium atoms to calcium atoms be after one hour?
(NA = 6 * 1023 mol-1.)

Answers: (a) 1.4 * 1020, (b) 7.9 * 1016 Bq, (c) 9.8 * 1015 Bq, (d) 1:7

I got every part except (d).

2. The attempt at a solution
(a) 44 g of Potassium 44 contain 6 * 1023 atoms. (10 * 10-3) g of Potassium 44 contain ((10 * 10-3) / 44) * (6 * 1023) = 1.36 * 10 20 atoms.

(b) dN / dt = -λN = - ((ln 2) / (20 * 60)) * (1.36 * 1020) = -7.9 * 1016 s-1 -> 7.9 * 1016 Bq

(c) A = A0 e-λt = (7.9 * 1016) * e-((ln 2) / (20 * 60)) * (60 * 60) = 9.8 * 1015 Bq

(d) For this part I am lost. Any ideas please? I did the same calculations for Calcium as in part (a-c) and compared the numbers and they all got me around 0.3 and not the 1:7 (0.14).
1. Homework Statement
Potassium 44 (4419K) has a half-life of 20 minutes and decays to form 1420Ca, a stable isotope of calcium.
(a) How many atoms would there be in a 10 mg sample of potassium 44?
(b) What would be the activity of the same?
(c) What would the activity be after one hour?
(d) What would the ratio of potassium atoms to calcium atoms be after one hour?
(NA = 6 * 1023 mol-1.)

Answers: (a) 1.4 * 1020, (b) 7.9 * 1016 Bq, (c) 9.8 * 1015 Bq, (d) 1:7

I got every part except (d).

2. The attempt at a solution
(a) 44 g of Potassium 44 contain 6 * 1023 atoms. (10 * 10-3) g of Potassium 44 contain ((10 * 10-3) / 44) * (6 * 1023) = 1.36 * 10 20 atoms.

(b) dN / dt = -λN = - ((ln 2) / (20 * 60)) * (1.36 * 1020) = -7.9 * 1016 s-1 -> 7.9 * 1016 Bq

(c) A = A0 e-λt = (7.9 * 1016) * e-((ln 2) / (20 * 60)) * (60 * 60) = 9.8 * 1015 Bq

(d) For this part I am lost. Any ideas please? I did the same calculations for Calcium as in part (a-c) and compared the numbers and they all got me around 0.3 and not the 1:7 (0.14).

d) is solved as the following:
Find how many K atoms have decayed in 1 hrs (i.e. N initial minus N final ) for which you should get 12.55 x 10^ 19 which equals to # Ca atoms formed after 1 hrs elapsed. Now ( # K/#Ca)after one hour = 1.75 x10^19/12.55 x10^19 = 0.14