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Topic: Radioactive decay energy  (Read 6638 times)

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Offline Rutherford

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Radioactive decay energy
« on: August 30, 2012, 01:54:42 PM »
The formula is:
E=(mi-mf)c2
mi=initial mass, mf=final mass
I only found problems where those masses are expressed in unified atomic mass unit, this should be applied only for one atom, but when I have, for example, 100kg of uranium 239 and by beta decay it decays to 239Np, let's say 50kg was made, how to calculate the energy released?

Offline sjb

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Re: Radioactive decay energy
« Reply #1 on: August 30, 2012, 02:02:40 PM »
The formula is:
E=(mi-mf)c2
mi=initial mass, mf=final mass
I only found problems where those masses are expressed in unified atomic mass unit, this should be applied only for one atom, but when I have, for example, 100kg of uranium 239 and by beta decay it decays to 239Np, let's say 50kg was made, how to calculate the energy released?

What is the exact mass of 239U?

Offline Rutherford

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Re: Radioactive decay energy
« Reply #2 on: August 30, 2012, 02:36:09 PM »
The atomic mass of uranium is 239.05429.
Of neptunium 239.05293.
« Last Edit: August 30, 2012, 02:57:02 PM by Raderford »

Offline Hunter2

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Re: Radioactive decay energy
« Reply #3 on: August 31, 2012, 01:13:51 AM »
I think no needed. If from 100 Kg U 50 kg converted to Np then 50 kg has to be left

E = 50 kg * (300.000.000 m/s)2 = 4,5 x 1018 J

Offline sjb

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Re: Radioactive decay energy
« Reply #4 on: August 31, 2012, 03:22:18 AM »
I think no needed. If from 100 Kg U 50 kg converted to Np then 50 kg has to be left

E = 50 kg * (300.000.000 m/s)2 = 4,5 x 1018 J

Possibly, given the sig fig issues, yes. But does 50 kg exactly of 239U yield 50 kg exactly of 239Np ?

Offline Hunter2

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Re: Radioactive decay energy
« Reply #5 on: August 31, 2012, 03:33:23 AM »
Quote
Possibly, given the sig fig issues, yes. But does 50 kg exactly of 239U yield 50 kg exactly of 239Np ?

Probably not. But compared will be  E=(mi-mf)c2
mi=initial mass, mf=final mass

So it doesn't matter what is the yield.

The mass difference of a neutron and a proton is the difference. In U 239 one Neutron will be converted to a Proton to get Np 239.

50 Kg U means 49.999972 kg Np

Offline Rutherford

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Re: Radioactive decay energy
« Reply #6 on: August 31, 2012, 04:12:16 AM »
I tried another way. I used first the unified atomic masses converted to kg, and I got that E=2.04*10-13J, this is the energy released when 1 atom of uranium gets decayed. In the example (50000/239)*6*1023=1.255*1026 atoms decayed. So the energy released is E*N=2.56*1013J. Why our results difference so much?

Offline Hunter2

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Re: Radioactive decay energy
« Reply #7 on: August 31, 2012, 04:33:06 AM »
How do you calculate E = 2.04* 10-13 J?

U MG = 239 g/mol  contain 92 Protones and 147 Neutrons

Offline Rutherford

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Re: Radioactive decay energy
« Reply #8 on: August 31, 2012, 05:23:47 AM »
Mass difference is 239.05429u-239.05293u=0.00136u=2.2583*10-30kg, so E=2.0325*10-13J (small difference because I used numbers with more digits now).

Offline Hunter2

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Re: Radioactive decay energy
« Reply #9 on: August 31, 2012, 06:10:23 AM »
No that is not the mass. You calculated the mass defect.

The mass is 92 x mass of proton + 147 x mass of neutron = mass of 1 atom.

Offline Rutherford

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Re: Radioactive decay energy
« Reply #10 on: August 31, 2012, 06:46:17 AM »
But the mass defect is caused by the released energy.

Offline Hunter2

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Re: Radioactive decay energy
« Reply #11 on: August 31, 2012, 06:51:58 AM »
The question is 50 kg Uranium disappeared and 50 kg or a little bit less Neptunium is created.

23992U => 23993Np + ß

I am wrong you are right its the mass of the defect to be taken.

http://en.wikipedia.org/wiki/Decay_energy
« Last Edit: August 31, 2012, 07:08:32 AM by Hunter2 »

Offline Rutherford

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Re: Radioactive decay energy
« Reply #12 on: August 31, 2012, 07:12:36 AM »
Okay, I got confused because I read some text about the mass defect. 50kg of Np was made so a little more than 50kg of U decayed (because mass of n0 is bigger than the mass of p+), now I use that mass as Δm to calculate the energy released.
Where does the released energy then come from?

Offline Hunter2

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Offline Rutherford

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Re: Radioactive decay energy
« Reply #14 on: August 31, 2012, 07:30:33 AM »
Then it should be: Δm=the mass that decayed(not 100kg) - the mass of Np.
Δm=2.8446*10-4g
E=2.56*10-13J again.

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