# $\require{mhchem}$ $\ce{\beta^{+}}$ decay for $\ce{_9^18F}$, computing $\Delta m$

For the decay:

$$\require{mhchem}\ce{_9^18F\to_8^18O +e+ +{v}}$$

To compute $E$, I need $\Delta m$, the provided answer looks like:

$$m_i = 18.000938~u$$

$$m_f = 17.999159~u + 2~(5.49 \times 10^{-4}~u)$$

I believe $m_i$ is mass for $\ce{F}$ and $m_f$ mass for $\ce{O}$? Then where did the numbers $17.999159~u$ and $5.49 \times 10^{-4}~u$ come from?

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Related to OP's question: physics.stackexchange.com/q/61779/11062 – Waffle's Crazy Peanut Apr 21 '13 at 14:36
It is possible that you are worrying too much about the numbers in these problems to the exclusion of the physics. The physics is just the conservation of energy. Mass in the initial state = sum of masses in the final state plus sum of kinetic energies in the final state (all in the CoM frame of the initial particle). To get the geometry of the decay you also conserve momentum. – dmckee Apr 21 '13 at 14:48

The energy liberated from the reaction : $$\require{mhchem} \ce{_9^18F}\rightarrow\ce{_8^18O}+\ce{e^{+}}+\nu$$

$E=\Delta m\times c^2=(m_f-m_i)c^2=(m_{\ce{_^18O}}+m_{\ce{_^e^+}}+\ce{m_{\nu}}-m_{\ce{_^18F}})\times c^2$

## Edit:

$m_{^{18}O}= 17.9991603 u$
$m_{e^+}= 0.000548579909 u$
$m_{\nu}=$ neglected in some cases but you can find that in wikipedia
$m_{^{18}F}=18.0009380u$

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The reaction is simply $_9 F^{18} \to\ _8O^{18}+\ _{1}e^0+\nu_e$ (electron neutrino)

The mass of electron and positron are the same. The method for computing the mass defect still remains the same. Have a look at the question on finding $\Delta m$. That number $5.49\times 10^{-4}u$ is the mass of electron. Hence, $m_f$ is not just for oxygen, it's the total mass of products. Since the neutrino's mass is so small, we can just neglect it (or atleast for homework).

I don't think there's a $2$ coming around...

Edit based on comments: The reason it more mass is because in the link (you've provided), they've used oxygen of mass number $16$ which is frequent I think so. In case of this decay, the oxygen has 2 more neutrons. Since O-16, 17 and even 18 are stable, it's not a big problem. In case of the $2$ electrons issue, strictly speaking, no two electrons are formed in this reaction. But as dmckee says, the question creator should've taken both the released electron and the electron neutrino into account. So that, both the leptons may have been assumed to have the same mass.

Here's the Wiki article on F-18 decay.

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"I don't think there's a 2 coming around" It is likely that whoever wrote the example simply assumed the two leptons received the same distribution of energy from the decay. For the neutrino this would be expressed mostly as kinetic energy (it's mass being less than 1 eV). There is not particular reason for making such an assumption, but it is not horrible as decay neutrinos for many isotopes have energy spectra in the 0.1 to 1 MeV range making that split plausible. – dmckee Apr 21 '13 at 15:08
But where did values for $m_i$ and $m_f$ come about? I thought $m_i = 18.9984032 u$ according to chemicalelements.com/show/mass.html for example? Atomic mass of oxygen seems even more off $15.9994 u$ according to the link? – Jiew Meng Apr 22 '13 at 7:11
Oh I got that. I need to consider thats its F-18 and O-18, did some Googling to get that ... but I dont get how the 2 electrons come about, the answers says "To calculate the energy released, subtract the mass of the daughter products (including the mass of two electrons) from the parent mass", or maybe you still think it should not be 2? – Jiew Meng Apr 22 '13 at 7:30