Ratio of potassium atoms to calcium atoms in 1 hour

[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?
(N_A = 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).

 

[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?

 

[epenguin]

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]

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: N_K = 1.4 * 10²⁰ * e^{-(ln 2 / 20 * 60) * 60 * 60} = 1.75 * 10¹⁹ atoms of K after 1 hr.

N_{0 Ca} = ((10 * 10^-3) / 14) * 6 * 10²³) = 4.3 *10²⁰ atoms of Ca at t = 0
N_Ca = 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 T_1/2 = 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.

 

[PWiz]

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

 

[epenguin]

We have T_1/2 = 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]

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]

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

 

[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]

you meant 0.125 right (not 0.0125)? :)

Yes, corrected.

 

[F1Dan]

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?
(N_A = 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).

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?
(N_A = 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