At first, consider two particles decay:

$A\rightarrow B + e^-$ Where A is initially rest.
So $\vec{p_B}+\vec{p_{e^-}}=0$

now
\begin{align}
\frac{p_B^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}&=E_{released}
\\
\frac{p_{e^-}^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}&=E_{released}
\tag{1}
\end{align}

see here you have **uncoupled equation** (equ.1) for $p_{e^-}$ .. So, solving above (equ.1) you will get a **fixed** $p_{e^-}$. Hence the energy($\sqrt{p_{e^-}^2+m_{e^-}^2}$) of the $\beta$ particle is always fixed in the two body $\beta$ decay. (you can find **$p_{B}$** too using the momentum conservation formula)

At first, consider three particles decay:

$A\rightarrow B + e^-+\nu_e$ Where A is initially rest. So $\vec{p_B}+\vec{p_{e^-}}+\vec{\nu_e}=0$

so
\begin{align}
\frac{p_B^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}+p_{\nu_e}&=E_{released}
\\
\frac{(\vec{p_{e^-}}+\vec {p_{\nu_e}})^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}+p_{\nu_e}&=E_{released}
\tag{2}
\end{align}

Now see in contrast to the two particles decay equation, here we have **five unknowns** $\big{(}p_{e^-},p_{\nu},p_{B},\theta\ (\rm the\ angle\ between\ \vec{p_{e^-}},\vec{p_{\nu}}),\phi\ (\rm the\ angle\ between\ \vec{p_{e^-}},\vec{p_{B}})\big)$ **but four coupled equations** *. so we can't solve them **uniquely**. That's also what happens physically. You will get **different values** of $p_{e^-},p_{\nu},p_{B},\theta,\phi$ satisfying the **four coupled equations**. Hence different $\beta$ particles will have different energy($\sqrt{p_{e^-}^2+m_{e^-}^2}$) maintaining the statistics of decay process. Hence the continuous spectra.

*The **four coupled equations** are equ.2 and three equations which we can get by taking **Dot** products of $\vec{p_{e^-}},\vec{p_{\nu}}\rm\ and\ \vec{p_{B}}$ with the momentum conservation($\vec{p_B}+\vec{p_{e^-}}+\vec{\nu_e}=0$
) and remember they lie in a plane so two angles ($\theta\rm\ and\ \phi$) are sufficient.

## Best Answer

The decay of potassium-40 to argon-40 is either a $\beta^+$ decay in which what is emitted is not an electron but a positron $$ {}^{40}{\rm K} \to {}^{40}{\rm Ar} + e^+ + \nu_e $$ or, more frequently (if we have whole atoms), an electron capture that you mentioned in which no charged leptons are emitted at the end! About 11% of the potassium-10 decays proceed in this way. $$ {}^{40}{\rm K} + e^- \to {}^{40}{\rm Ar} + \nu_e $$ The remaining 89% of the decays of potassium-40 go to calcium-40 (the beta-plus decay is a small fraction of a percent). Note that the two reactions above differ by moving the positron from the right hand side to the left hand side so that the sign has to be changed.

Potassium-40 has 19 protons and 21 neutrons. Argon-40 has 18 protons and 22 neutrons. So if we focus on the "minimum part" of the nuclei, the reactions above may be reduced either to $$ p \to n + e^+ + \nu_e$$ or $$ p+e^- \to n + \nu_e $$ which are the standard reactions switching protons and neutrons. In particular, the second reaction displayed right above this line is the more "microscopic" description of the electron capture you're primarily interested it.

These reactions preserve the electric charge, baryon number, and lepton number. They also have to preserve energy. A free proton couldn't decay to the neutron and other two particles because it's lighter. Even a proton and a low-velocity electron wouldn't have enough mass/energy to produce the neutron (plus the neutrino) as in the second reaction.

But when the protons and neutrons are parts of whole nuclei, the energies of the initial and final nuclei are affected by the nuclear interactions. In particular, the argon-40 nucleus (and especially atom) is highly bound which means lighter and the reactions where argon-40 appears as a product are therefore "more possible".

To summarize, the electron capture (=falling of the electron) simply means that the proton has a nonzero probability to meet with one of the electrons – probably in the inner shells – and merge into a new particle, a neutron, plus a neutrino. This process can't occur in the vacuum due to the energy conservation but in the context of the nucleus, the interactions with other neutrons and protons make the final state with the new neutron favorable.

On the contrary, alpha decays are more rare. Among 24 isotopes of potassium, only potassium-36 may alpha-decay. Carbon-14 doesn't alpha-decay, either. Among the isotopes of carbon, only carbon-9 alpha-decays. Both of these alpha-decays must be preceded by a beta-decay. Usually just heavy enough nuclei (with too small an excess of neutrons) alpha-decay.