Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers and students. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Reading Wikipedia on "Beta Decay" they list the equation of Carbon-14 decaying into N-14. see equation https://en.wikipedia.org/wiki/Beta_decay

I presume the 14 refers to total protons and neutrons. The 6 in C, I believe means six protons. In N, I believe the 7 refers to protons. Where did the seventh proton come from the decay of C-14 into N-14? I understand neutrons can decay into protons, but the math doesn't work for me.

share|improve this question

For all radioactive decay (or other nuclear reaction) of a nuclide into other nuclides, the atomic number $Z$ and mass number $A$ need to be conserved.

$$\ce{_{Z_1}^{A_1}X -> _{Z_2}^{A_2}Q + _{Z_3}^{A_3}R }$$ $$Z_1 = Z_2 + Z_3$$ $$A_1 = A_2 + A_3$$

Additionally, the charges must be conserved. If $\ce{_{Z_3}^{A_3}R }$ is an alpha particle $\ce{_2^2\alpha^{2+}}$ or $\ce{_2^2He^{2+}}$, then:

$$\ce{_{Z}^{A}X -> _{Z-2}^{A-2}Q^{2-} +_2^2He^{2+}}$$

Neutrons decay into protons and electrons (beta particles). Both particles are needed for the math to work. And in order for the math to work, you need to know that the atomic number $Z$ is $0$ for a neutron (no protons), and $1$ for a proton (1 proton, or the $\ce{H+}$ ion). The electron does not have an atomic number. In fact, it is not that $Z$ and charges must be conserved separately. The sum of atomic number and charge is conserved.

$$\ce{_0^1n -> _1^1p+ + _{0}^0e-}$$

For the decay of $\ce{^{14}_6C}$ into $\ce{_7^{14}N}$, we first have a neutron decay into a proton and an electron, which is captured by the newly formed $\ce{_7^{14}N^+}$. The nitrogen nuclide is initially a cation because carbon only had six electrons, and nitrogen needs seven.

$$\ce{_6^{14}C -> _7^{14}N+ +e- -> _7^{14}N}$$

share|improve this answer
3  
If the emitted electron is "captured by the newly formed nitrogen", what does a Geiger counter detect? – DJohnM May 6 '14 at 21:41
    
Something is really wrong here. I mean wikipedia is talking about N and e- and not N+ and e-. To me your solution is logical because the charges are conserved, but everybody else appear to have the "wrong" equation... :D – inf3rno Jan 12 at 5:14
    
Asked it again from physicists: physics.stackexchange.com/questions/228929/… – inf3rno Jan 12 at 6:03
    
I counted the energy of the radiocarbon decay here: biology.stackexchange.com/questions/42239/… in kJ/mol, which is in the $10^7-10^8$ range, while chemical bonds have energy usually in the $10^2-10^3$ range. The ionization energy of nitrogen is around 1400 kJ/mol, so it is in the $10^4-10^5$ range, thus it certainly cannot recapture the leaving electron assuming it carries a relevant part of the decay energy. I edited the answer. – inf3rno Jan 13 at 7:00
    
@inf3rno Please don't make such drastic changes to an answer. You can always post a new answer with additions to the already given answer. – Martin - マーチン Jan 13 at 7:36

By beta decay in the nucleus of the atom a neutron decays into a proton, an electron, and an electron antineutrino and a lot of energy. It will lose a little mass, since $E = mc^2$, so the decay energy is coming from the mass we lose. The electron (or beta particle) has a high energy and leaves the atom.

$$\ce{_0^1n -> _1^1p+ + _{0}^0e- + \overline{\nu_{e}}}$$

By the decay of radiocarbon $8(n) + 6(p) = 14$ the product will be a nitrogen $7(n) + 7(p) = 14$, which means we will have one less neutron and one more proton in the nucleus.

$$\ce{_6^{14}C -> _7^{14}N+ + _{0}^0e- + \overline{\nu_{e}}}$$

The total charge remains zero, since we will have one more electron (beta particle) and one more proton in the system.

According to wikipedia the decay energy of radiocarbon is $E_\mathrm{d} = 0.156476~\mathrm{MeV} = 2.50702189 \cdot 10^{-17}\rm~kJ$. For a mole atoms this means $E_\mathrm{d} \cdot N_\mathrm{A} = 1.509764 \cdot 10^7\rm~kJ/mol$. The first ionization energy of nitrogen is $1.402 \cdot 10^3\rm~kJ/mol$, so the nitrogen cation won't be able to recapture the beta particle, since it will carry too much energy. I am not certain about how big fraction of the decay energy the electron will carry, but certainly much more than the first ionization energy. Since the nitrogen has a relative high electronegativity (3.1), it is likely, that it will get an electron from the environment from an electron donor, for example hydrogen. Radiocarbon labelled biomolecules are likely to fall apart by nuclear transmutations, since the bond energies in these molecules are around $10^2-10^3\rm~kJ/mol$, so a single decay can break $10^4-10^5$ bonds. That's why beta radiations are dangerous.

share|improve this answer

Charge and energy are always conserved. Mass is a form of energy, and what we call 'mass number' - the sum of neutrons + protons - is a close approximation to mass (because electrons are so light, having just over a two-thousandth of the mass of a proton or a neutron).

The conversion of a neutron into a proton when an electron is released takes care of charge conservation, because the charge on the proton is exactly the opposite of the charge on the electron.

The energy needed to create the electron in beta decay comes from the reduced binding energy of the nucleons in N-14 compared with those in C-14.

Does that make sense of the maths for you?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.