Beta - and Beta + Decay (Q-Value)

In summary: And it's ok for the present exercise.In summary, the Q-value for the electron emission beta decay of Co-60 is 2.82380 MeV (correct to 6 significant figures) and the Q-value for the positron emission beta decay of Mg-23 is 3.03480 MeV (correct to 6 significant figures). These values are calculated using the atomic masses of the elements, subtracting the atomic mass of the electron, and then multiplying by the speed of light squared. The Q-value represents the energy released in the reaction and is positive in both cases.
  • #1
says
594
12

Homework Statement


1) Calculate the Q-value for the electron emission beta decay of Co-60 *

2) Calculate the Q-value for the positron emission beta decay of Mg-23 *

* (both correct to 6 significant figures)

Atomic Masses (u)
electron = 0.0005485803
Co-60 = 59.9338222
Ni-60 = 59.9307906
Mg-23 = 22.9941249
Na-23 = 22.98976967

Atomic Number
Co-27
Ni-28
Mg-12
Na-11

Homework Equations



Q = Σmic2-Σmfc2
Beta decay (electron emission) = X ---> Y(Z+1) + -1e + v_
where v_ = anti-neutrino
Beta decay (positron emission) = X ---> Y(Z-1) + +1e + v
where v = neutrino

The Attempt at a Solution


1 (Co-60)

Q = (59.9338222 - (59.9307906+0.0005485803))*c2
= 59.9338222 - 59.9313392
= 0.0024830 * 931.5020
= 2.3129
= 2.31290 MeV (correct to 6 significant figures)

2 (Mg-23)

Q = (22.9941249 - (22.98976967+0.0005485803))*c2
= 22.9941249-22.99031825
= 0.0038066*931.5020
= 3.54580 MeV (correct to 6 significant figures)

Q is positive in both cases. Q is the energy that is liberated in the reaction. Am i correct in neglecting and mass that the anti-neutrino and neutrino have?
 
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  • #2
Neglecting the neutrino mass is fine.
But e.g. here I find a different decay energy. Can you spot the difference and understand why ?
 
  • #3
The Co-60 value looks similar, but the Mg-23 does not. Your reference says that Mg-23 decays via electron capture and not positron emission beta decay.

Other than that I'm not really sure what the difference is, because Q = Σmic2-Σmfc2 and I've got all the masses in the reaction.
 
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  • #4
Co60:
Did I give the right link ? It's on table4. To me it does not look similar at all: 2.824 versus 2.313 ?

Are you sure you have to deduct the electron mass ? That means you consider it as being created from the reaction energy [edit] which is correct. But how about your statement "have all the masses"?
Checking here might set you thinking about what the given data stand for: the nuclear masses or the atomic masses ?
 
  • #5
I've been using the atomic masses to calculate the Q-value. The second link you sent suggests I should use the nuclear mass. I'm a bit confused now
 
  • #6
Q = Σmic2-Σmfc

Q = Σmic2 - Σ mfc2

mi = atomic mass of element - number of electrons = mass of nucleus

mf = (atomic mass of element - number of electrons) + mass of one electron = mass of nucleus + one electron

Is these the correct masses I should be using? I thought m was the atomic mass, but from what I gather in what you've sent me it's the mass of the nucleus (nuclear mass).
 
  • #7
Q = ( 59.9338222 - 0.0148116681 ) - ( 59.9307906 - 0.0153602484 + 0.0005485803 ) * c2
= 0.0030315 * 931.5020
= 2.8238
= 2.82380 MeV (correct to 6 significant figures) :smile:

I'll have to do the same for the other one too!

Just a quick question. I've been asked to calculate this to 6 sig figs. 0.0030315 * 931.5020 = 2.823848313, however because we are multiplying we have to round to the .0000 because of the 931.5020. This makes the answer 2.8238, and because the answer is wanted to 6 sig figs it = 2.82380 MeV. Is my reasoning correct here?
 
  • #8
I wouldn't subtract 27 me and 28 me on each side and then add one me on the Ni again,...
Besides the neutrino we also ignore the difference in total binding energy (check the wikipedia article -- looks like that's ok to do).

Subtracting the atomic masses as give leaves only 5 digits. Were the atomic masses given in the exercise ? Then the exercise itself leaves you a digit short !
If you quote 0.0028238 GeV/c^2 you could argue that's seven digits.

[edit] the 931.502 is 6 digits, so that's ok. -- Note that it's about relative errors when multiplying !
But I found 931.4941 ?!

(have to run!)
 
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  • #9
Yeah, the atomic masses were given as:

electron = 0.0005485803
Co-60 = 59.9338222
Ni-60 = 59.9307906

Conversions were given as well
1 eV = 1.602189E-19 Joules
c = 2.99792458E+08 m/s
1 amu = 1.660566E-27 kg
1 amu = 931.5020 MeV

There's no way I'm converting amu to kg, calculating joules, converting to MeV. No thank you!

59.9338222 - 0.0148116681 = 59.9190105 ( 9 sig figs )
59.9307906 - 0.0153602484 + 0.0005485803 = 59.9159790 ( 9 sig figs )

59.9190105 - 59.9159790 = 0.0030315 (5 sig figs)

0.0030315 (5 sig figs) * 931.5020 (7 sig figs) = 2.8238 (5 sig figs) = 2.82380 (6 sig figs)
 
  • #10
(Mg-23)

Q = [ (22.9941249 - 0.0065829636) - (22.98976967 - 0.0060343833 + 0.0005485803) ]
= 0.0032580 * 931.5020
= 3.0348
= 3.03480 MeV (correct to 6 sig figs)

Your link suggests Mg-23 decays via electron capture and the decay energy = 4.057 MeV (http://atom.kaeri.re.kr/ton/nuc4.html)
 
  • #11
This link, at the bottom of the page, says that B+ decay of Mg-23 = 3.0348 MeV

http://crete.homeip.net/show_nuclide/120023/
 
Last edited by a moderator:
  • #12
A difference of 2 me .

Again, the wikipedia link explains why the 3.0348 MeV is the right answer -- in 5 digits. Really
2.8238 (5 sig figs) = 2.82380 (6 sig figs)
is bluffing: you don't know it's a zero at the end. But as I stated: not your fault.
 

Related to Beta - and Beta + Decay (Q-Value)

1. What is beta decay?

Beta decay is a type of nuclear decay in which an unstable nucleus emits either a beta particle (a high-energy electron) or a positron (a positively charged electron). This process occurs in order to stabilize the nucleus and reduce the overall energy of the atom.

2. What is the difference between beta- and beta+ decay?

Beta- decay occurs when a neutron in the nucleus is converted into a proton, emitting a beta particle (electron) and an anti-neutrino. Beta+ decay, on the other hand, occurs when a proton in the nucleus is converted into a neutron, emitting a positron (positive electron) and a neutrino. Both types of decay result in a more stable nucleus.

3. What is the Q-value of beta decay?

The Q-value of beta decay is the difference in energy between the initial and final states of the nucleus. It is a measure of the energy released or absorbed during the decay process and is typically given in units of electron volts (eV).

4. How is the Q-value of beta decay calculated?

The Q-value of beta decay can be calculated using the formula: Q = (Mi - Mf)c^2, where Mi is the initial mass of the nucleus, Mf is the final mass of the nucleus, and c is the speed of light. The Q-value can also be determined experimentally by measuring the energy of the emitted particles.

5. What factors influence the Q-value of beta decay?

The Q-value of beta decay is influenced by the difference in mass between the initial and final states of the nucleus, as well as the energy levels of the particles involved in the decay. Additionally, the type of beta decay (beta- or beta+) and the atomic number of the element also play a role in determining the Q-value.

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