[quyenp]

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 (t_1/2=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?

[Borek]

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

[quyenp]

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.
.

[Borek]

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.

[Enthalpy]

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.

[quyenp]

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?

[gippgig]

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.

[Borek]

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!)?

[AWK]

Assuming complete decomposition of Th-228 one can obtain
(1/228)x5x22.4(or 22.7)=0.491(or 0.498) L