# Chemical Forums

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#### quyenp

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« on: October 20, 2015, 08:15:24 PM »

I have 2 two problems realated to a radioactive exercise.
228Th is a member of the thorium series (ending with the stable 208Pb). What volume in cm3 of helium at 00C and 1atm collected when 1.00 gram of 228Th (t1/2=1.91 years) is stored in a cointainer for 20.0 years. The half-lives of all intermadiaste nuclides are short compared to the half-life of 228Th.
The first solution that I've made is calculating:
-the mole of 228Th remained after 20 years
-the mole of 228Th has decayed
-the mole of Helium based on the equation: 228Th 208Pb+54He
-the volume of helium
and the result is 0,49 L
But the result is 3.56 L. The key is calculating the activity of Thorium, then they have the number of Helium and finally have the volume of Helium
So what is the difference between two solutions?
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#### Borek

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« Reply #1 on: October 20, 2015, 09:10:39 PM »

Looks to me like 0.49 L (actually 0.483 L) is a correct answer.
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#### quyenp

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« Reply #2 on: October 20, 2015, 09:36:29 PM »

Hi Borek! The answer is 3.56 L. This is a question from IChO taken place in 1999. Some exercise may calculate directly from the radioactive equation,for example we can calculate the number of He based on the equation from U-238 to Pb-206Therefore, I am really confused.
.
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#### Borek

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« Reply #3 on: October 20, 2015, 10:23:20 PM »

3.56 L is listed as a correct answer, which doesn't make it automatically correct. I see nothing wrong in your approach and in the numbers you got, actually I reproduced them using the original data from your post.

This is well past 10 half times, less than 1/1024 of the original Th is left - so for this problem we can safely assume it decayed completely. There were around 4.31×10-3 moles* of Th, and your equation - about 5 moles of He being produced for each mole of 228Th is spot on. That yields 0.0215 moles of He, or 0.483 L.

Do they show how they got their answer?

*actually a bit more, as molar mass of 228Th is slightly lower than the molar mass of Th listed in tables.
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#### Enthalpy

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« Reply #4 on: October 22, 2015, 04:27:27 AM »

I too get 0.49L at 0°C and 1atm with 5 He per 228Th.

I wonder whether 1g of 228Th has ever been isolated.
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#### quyenp

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« Reply #5 on: October 22, 2015, 01:22:11 PM »

3.56 L is listed as a correct answer, which doesn't make it automatically correct. I see nothing wrong in your approach and in the numbers you got, actually I reproduced them using the original data from your post.

This is well past 10 half times, less than 1/1024 of the original Th is left - so for this problem we can safely assume it decayed completely. There were around 4.31×10-3 moles* of Th, and your equation - about 5 moles of He being produced for each mole of 228Th is spot on. That yields 0.0215 moles of He, or 0.483 L.

Do they show how they got their answer?

*actually a bit more, as molar mass of 228Th is slightly lower than the molar mass of Th listed in tables.
First, they calculate the activity of Thorium at the t = 0 (A =λN = 0.693*1*6.022.1023 / (228*1.91) = 9.58*1022 , then calculate the number of particles of He during 20 years (N = 9.58*1020*20*5 = 9.58.1022 ) and finally they get the volume of He at standard condition (V = 9.58*1022*22.4/ (6.022*1023) = 3.56*103 cm3)
I just think maybe the difference between the half-life of various radioactive elements makes the exercise have another approach. To be honest, I am not sure about this! What do you think Borek?
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#### gippgig

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« Reply #6 on: October 22, 2015, 01:57:41 PM »

The activity is not constant (it decreases as the Th decays) so multiplying the current activity by the number of years gives a wrong answer. That approach would only work if the half-life was much longer than the time interval.
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#### Borek

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« Reply #7 on: October 22, 2015, 09:04:18 PM »

The activity is not constant (it decreases as the Th decays) so multiplying the current activity by the number of years gives a wrong answer. That approach would only work if the half-life was much longer than the time interval.

That.

Plus, there is a simple reality check. 224Th is a "limiting reagent" here. One can easily calculate mass of the helium produced using their approach, subtract it from the mass of thorium initially present and see what is the average mass of an atom left. Is there enough mass for 208Pb (note that number atoms didn't change!)?
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#### AWK

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