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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Rutherford on August 30, 2012, 01:54:42 PM

Title: Radioactive decay energy
Post by: Rutherford on August 30, 2012, 01:54:42 PM: The formula is:
E=(m_i-m_f)c²
m_i=initial mass, m_f=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?
Title: Re: Radioactive decay energy
Post by: sjb on August 30, 2012, 02:02:40 PM: Quote from: Raderford on August 30, 2012, 01:54:42 PM
The formula is:
E=(m_i-m_f)c²
m_i=initial mass, m_f=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?
Title: Re: Radioactive decay energy
Post by: Rutherford on August 30, 2012, 02:36:09 PM: The atomic mass of uranium is 239.05429.
Of neptunium 239.05293.
Title: Re: Radioactive decay energy
Post by: Hunter2 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
Title: Re: Radioactive decay energy
Post by: sjb on August 31, 2012, 03:22:18 AM: Quote from: Hunter2 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

Possibly, given the sig fig issues, yes. But does 50 kg exactly of ²³⁹U yield 50 kg exactly of ²³⁹Np ?
Title: Re: Radioactive decay energy
Post by: Hunter2 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=(m_i-m_f)c²
m_i=initial mass, m_f=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
Title: Re: Radioactive decay energy
Post by: Rutherford 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?
Title: Re: Radioactive decay energy
Post by: Hunter2 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
Title: Re: Radioactive decay energy
Post by: Rutherford 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).
Title: Re: Radioactive decay energy
Post by: Hunter2 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.
Title: Re: Radioactive decay energy
Post by: Rutherford on August 31, 2012, 06:46:17 AM: But the mass defect is caused by the released energy.
Title: Re: Radioactive decay energy
Post by: Hunter2 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
Title: Re: Radioactive decay energy
Post by: Rutherford 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?
Title: Re: Radioactive decay energy
Post by: Hunter2 on August 31, 2012, 07:18:01 AM: http://www.science.uwaterloo.ca/~cchieh/cact/nuctek/decayenergy.html
Title: Re: Radioactive decay energy
Post by: Rutherford 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.
Title: Re: Radioactive decay energy
Post by: Hunter2 on August 31, 2012, 08:32:00 AM: correct