Nuclear equation of the decay, decay constant, # of atoms

[moenste]

Homework Statement

A sample of iodine contains 1 atom of the radioactive isotope iodine 131 (¹³¹I) for every 5 * 10⁷ atoms of the stable isotope iodine-127. Iodine has a proton number of 52 and the radioactive isotope decays into xenon 131 (¹³¹Xe) with the emission of a single negatively charged particle.

(a) What particle is emitted when iodine 131 decays? Write the nuclear equation which represents the decay.

(b) The diagram shows how the activity of a freshly prepared sample of the iodine varies with time. Use the graph to determine the decay constant of iodine 131. Give your answer in s^-1.

(c) Determine the number of iodine atoms in the original sample.

Answers: (b) 1.0 * 10^-6 s^-1, (c) 1.0 * 10¹⁸

2. The attempt at a solution
(a) Is it something like: ¹³¹₅₂I → ¹²⁷₅₂I → ¹³¹_?Xe?

(b) N = N₀ e^{-λ t}, where the decay consant is λ, N = remaining number of atoms, N₀ = original number of atoms, t = time.

We have N₀ = 5 * 10⁷ atoms, how do we find N?

 

[jbriggs444]

(b) N = N₀ e^{-λ t}, where the decay consant is λ, N = remaining number of atoms, N₀ = original number of atoms, t = time.

We have N₀ = 5 * 10⁷ atoms, how do we find N?

You do not have that N₀ = 5 x 10⁷ atoms. You are asking how to find N₀.

The graph has some useful information. It is expressed in units of Bequerels. If you are a dummy like myself you could Google that unit.

If you have a rate of decay in decays per second and you have already calculated a rate of decay in decays per nucleon per second, the next step seems obvious.

 

[moenste]

You do not have that N₀ = 5 x 10⁷ atoms. You are asking how to find N₀.

The graph has some useful information. It is expressed in units of Bequerels. If you are a dummy like myself you could Google that unit.

If you have a rate of decay in decays per second and you have already calculated a rate of decay in decays per nucleon per second, the next step seems obvious.

So: ln N / N₀ = - λ t. I take N = 0.5 * 10⁴ Bq, N₀ = 5 * 10⁷ atoms, t = 16 days = 1 382 400 s.

So: ln 0.5 * 10⁴ / 5 * 10⁷ = - λ * 1 382 400 → λ = 6.66 * 10^-6 s^-1.

 

[lychette]

an important relationship you should know...text book stuff is that
decay constant λ = (ln2)/ t_1/2 ... t_{1/2 = half life}
or, if you prefer it t_1/2 = (ln2)/λ

 

[moenste]

an important relationship you should know...text book stuff is that
decay constant λ = (ln2)/ t_1/2 ... t_{1/2 = half life}
or, if you prefer it t_1/2 = (ln2)/λ

But we don't know half-life.

 

[lychette]

But we don't know half-life.

you have a graph !

 

[moenste]

you have a graph !

It shows activity per time, not half-life.

 

[lychette]

the activity is decreasing with time...do you understand what 'half life' means in radioactive decay ?

 

[moenste]

the activity is decreasing with time...do you understand what 'half life' means in radioactive decay ?

The time required to lower the activity of the sample to half of the original value?

 

[lychette]

yes

 

[moenste]

yes

Activity 1 * 10⁴ Bq with time 8 days → = 691 200 s.

λ = ln 2 / 691 200 = 1 * 10^-6 s^-1. Should be correct?

 

[lychette]

Activity 1 * 10⁴ Bq with time 8 days → = 691 200 s.

λ = ln 2 / 691 200 = 1 * 10^-6 s^-1. Should be correct?

ideally you should take a few values of half life and get an average
you have gone from 2 to 1 x 10⁴Bq
you could go ,from 1 to 0.5 0r any other combination...as long as it is 1/2
you have changed days into seconds...this is good !

 

[moenste]

ideally you should take a few values of half life and get an average
you have gone from 2 to 1 x 10⁴Bq
you could go ,from 1 to 0.5 0r any other combination...as long as it is 1/2
you have changed days into seconds...this is good !

Could you suggest what to do in (a)?

(a) What particle is emitted when iodine 131 decays? Write the nuclear equation which represents the decay.

Is it something like: ¹³¹₅₂I → ¹²⁷₅₂I → ¹³¹_?Xe?

I know α, β and γ particles, is this the situation when they are applicable? Like ¹³¹₅₂I → ¹²⁷₅₀I + ⁴₂α? But how do I approach ¹³¹_?Xe then?

And if it's not these particles, what should be done?

 

[lychette]

¹³¹

Could you suggest what to do in (a)?

Is it something like: ¹³¹₅₂I → ¹²⁷₅₂I → ¹³¹_?Xe?

I know α, β and γ particles, is this the situation when they are applicable? Like ¹³¹₅₂I → ¹²⁷₅₀I + ⁴₂α? But how do I approach ¹³¹_?Xe then?

And if it's not these particles, what should be done?

It is the isotope Iodine131 that decays to Xenon131 so the nucleus contains the same number of nucleons ( protons + neutrons)
A negative particle was emitted...any ideas what the negative particle is and how this changes atomic number and nucleon number (mass number) of the nucleus?

 

[moenste]

¹³¹

It is the isotope Iodine131 that decays to Xenon131 so the nucleus contains the same number of nucleons ( protons + neutrons)
A negative particle was emitted...any ideas what the negative particle is and how this changes atomic number and nucleon number (mass number) of the nucleus?

So it is: ¹³¹₅₂I → ¹³¹₅₃Xe + ⁰_-1β. Correct?

But why do they say something about

A sample of iodine contains 1 atom of the radioactive isotope iodine 131 (131I) for every 5 * 107 atoms of the stable isotope iodine-127. Iodine has a proton number of 52

some "isotope iodine-127" and etc. I thought that the 131 sample decays to 127 sample and then decays to 131 Xe.

And also what to do in (c)?

(c) Determine the number of iodine atoms in the original sample.

N = N₀ e^{-λ t}, where the decay consant is λ, N = remaining number of atoms, N₀ = original number of atoms, t = time. We need to find N₀, but what is the current number of atoms N?

I also tried to use dN / dt = - λ N to find N, where dN / dt = 1 * 10⁴ Bq activity and λ = 1 * 10^-6 s^-1, but that gives 10¹⁰, not 10¹⁸ as in the answer.

 

[lychette]

So it is: ¹³¹₅₂I → ¹³¹₅₃Xe + ⁰_-1β. Correct?

But why do they say something about

some "isotope iodine-127" and etc. I thought that the 131 sample decays to 127 sample and then decays to 131 Xe.

And also what to do in (c)?

N = N₀ e^{-λ t}, where the decay consant is λ, N = remaining number of atoms, N₀ = original number of atoms, t = time. We need to find N₀, but what is the current number of atoms N?

I also tried to use dN / dt = - λ N to find N, where dN / dt = 1 * 10⁴ Bq activity and λ = 1 * 10^-6 s^-1, but that gives 10¹⁰, not 10¹⁸ as in the answer.

You have it correct. The information is needed for part (c)
They want the total number of iodine atoms...not just the radioactive ones. They have told you the proportion of atoms that are radioactive

 

[gneill]

Just a note on the question accuracy -

Iodine has a proton number of 52 and the radioactive isotope decays into xenon 131 (131Xe)

Looks like a typo in the question. According to my periodic table Iodine has a proton number of 53, not 52. Xenon has a proton number of 54. So the transition from Iodine to Xenon involves the number of protons going from 53 to 54.

This revelation won't change the type of particle emitted, but it will make your transition equation more realistic.

So it is: ¹³¹₅₂I → ¹³¹₅₃Xe + ⁰_-1β. Correct?

Just bump the proton numbers up by one on the atoms.

 

[moenste]

You have it correct. The information is needed for part (c)
They want the total number of iodine atoms...not just the radioactive ones. They have told you the proportion of atoms that are radioactive

Just a note on the question accuracy -

Looks like a typo in the question. According to my periodic table Iodine has a proton number of 53, not 52. Xenon has a proton number of 54. So the transition from Iodine to Xenon involves the number of protons going from 53 to 54.

This revelation won't change the type of particle emitted, but it will make your transition equation more realistic.Just bump the proton numbers up by one on the atoms.

So, for (c):
We have 1 atom of radioactive iodine 131 for every 5 * 10⁷ atoms of stable isotope. But it's just one atom, we don't know how much atoms are there in total.

In N = N₀ e^{-λ t} -- how to find N so it would be possible to calculate N₀?

 

[gneill]

So, for (c):
We have 1 atom of radioactive iodine 131 for every 5 * 10⁷ atoms of stable isotope. But it's just one atom, we don't know how much atoms are there in total.

In N = N₀ e^{-λ t} -- how to find N so it would be possible to calculate N₀?

Your approach using the derivative in post #15 was fine. You found the number of radioactive atoms in the original sample. Each of those atoms will be accompanied by 5 * 10⁷ atoms of stable isotope.

 

[moenste]

Your approach using the derivative in post #15 was fine. You found the number of radioactive atoms in the original sample. Each of those atoms will be accompanied by 5 * 10⁷ atoms of stable isotope.

But 10¹⁰ * 5 * 10⁷ = 5 * 10¹⁷, not 1 * 10¹⁸.

 

[gneill]

But 10¹⁰ * 5 * 10⁷ = 5 * 10¹⁷, not 1 * 10¹⁸.

Why did you choose to use 1.0 x 10⁴ Bq as the activity at time zero when you were calculating N_o? What does the graph say?

 

[moenste]

Why did you choose to use 1.0 x 10⁴ Bq as the activity at time zero when you were calculating N_o? What does the graph say?

It's when the activity falls to 50 % of the original value and we can use our decay constant found in (b).

 

[gneill]

It's when the activity falls to 50 % of the original value and we can use our decay constant found in (b).

Well that's wrong . The initial sample is the sample that existed at time zero, at the start of the chart.

You used the half activity time to find the decay constant, which is fine, but that does not make the half activity point the initial sample.

 

[moenste]

Well that's wrong . The initial sample is the sample that existed at time zero, at the start of the chart.

You used the half activity time to find the decay constant, which is fine, but that does not make the half activity point the initial sample.

Hm, so in that case: λ = 1 * 10^-6 s^-1, dN / dt = - λ N, where N = ? and dN / dt = 2 * 10⁴ Bq (initial activity).

N = (2 * 10⁴) / (1 * 10^-6) = 2 * 10¹⁰ atoms of the radioactive isotope iodine 131. We need to determine the number of iodine atoms in the original sample of the stable isotope iodine 127, so: 2 * 10¹⁰ * 5 * 10⁷ = 1 * 10¹⁸ atoms.

 

[lychette]

Just a note on the question accuracy -

Looks like a typo in the question. According to my periodic table Iodine has a proton number of 53, not 52. Xenon has a proton number of 54. So the transition from Iodine to Xenon involves the number of protons going from 53 to 54.

This revelation won't change the type of particle emitted, but it will make your transition equation more realistic.Just bump the proton numbers up by one on the atoms.

Well spotted, I also checked atomic numbers