# Ratio of potassium atoms to calcium atoms in 1 hour

• moenste
In summary: 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).In summary, potassium 44 (4419K) has a half-life of 20 minutes and decays to form 1420Ca, a stable isotope of calcium. A 10 mg sample of potassium 44 would contain 1.4 * 1020 atoms. The activity of the sample would be 7.9 * 1016 Bq and after one hour, the activity would be
moenste

## 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).

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
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
PWiz said:
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?

epenguin said:
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.

moenste said:
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
moenste said:
1420Ca
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
moenste said:
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
PWiz said:
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.

epenguin said:
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

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
epenguin said:
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)? :)

moenste said:
you meant 0.125 right (not 0.0125)? :)

Yes, corrected.

moenste
moenste said:

## 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).
moenste said:

## 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

## 1. What is the significance of the ratio of potassium atoms to calcium atoms in 1 hour?

The ratio of potassium atoms to calcium atoms in 1 hour can provide important information about the rate of decay of these elements and their relative abundance in a sample.

## 2. How is the ratio of potassium atoms to calcium atoms in 1 hour measured?

The ratio of potassium atoms to calcium atoms in 1 hour is typically measured using a technique called mass spectrometry, which separates and analyzes the isotopes of these elements.

## 3. What factors can affect the ratio of potassium atoms to calcium atoms in 1 hour?

The ratio of potassium atoms to calcium atoms in 1 hour can be influenced by various factors such as temperature, chemical reactions, and the presence of other elements in the sample.

## 4. Is the ratio of potassium atoms to calcium atoms in 1 hour constant?

No, the ratio of potassium atoms to calcium atoms in 1 hour can vary depending on the conditions of the sample. It can also change over time due to the natural process of radioactive decay.

## 5. What can the ratio of potassium atoms to calcium atoms in 1 hour tell us about a sample?

The ratio of potassium atoms to calcium atoms in 1 hour can provide insight into the age and composition of a sample, as well as potential environmental factors that may have affected it.

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