For the $\beta^{+}$ (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K-shell, is captured by the nucleus and a neutrino is emitted). $$^{A}_{Z}X + e^{-} \rightarrow {}^{A}_{Z-1}Y + \nu$$ Show that if $\beta^{+}$ emission is energetically allowed, electron capture is necessarily allowed but not vice-versa.

For the β⁺ (positron) emission from a nucleus, there is another competing process known as electron capture (e

Exercises · 13.22

\beta^{+}

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Concept understanding — Alpha Beta Gamma Decay

Alpha Beta Gamma Decay

Why Nuclei Decay at All

A nucleus is a tight ball of protons and neutrons held together by the short-range strong nuclear force, which must overcome the electric repulsion between protons. For very large nuclei, or ones with the wrong proton-to-neutron ratio, this balance is unstable, and the nucleus sheds energy or particles to reach a more stable state. This is radioactive decay, and it happens in three main forms: alpha, beta, and gamma.

Alpha Decay — The Nucleus Ejects a Helium Core

An alpha particle is a helium-4 nucleus (2 protons + 2 neutrons), a very stable package. A heavy nucleus (atomic number greater than about 82), where the strong force can no longer hold together against the mutual repulsion of so many protons, lowers its energy by ejecting one.

The nucleus $X$ loses 4 units of mass number and 2 units of atomic number, becoming a different element two places to the left on the periodic table:

_{Z}^{A}X \;\rightarrow\; _{Z-2}^{A-4}Y \;+\; _{2}^{4}\alpha

Note

The alpha particle is written as $_{2}^{4}\text{He}^{2+}$ — a helium ion. It is positively charged and relatively heavy, so it travels only a few centimetres in air and is stopped by a sheet of paper.

Example: Uranium-238 decays to thorium-234:

_{92}^{238}\text{U} \;\rightarrow\; _{90}^{234}\text{Th} \;+\; _{2}^{4}\alpha

Beta Decay — A Neutron Turns Into a Proton (or Vice Versa)

Beta decay is more subtle: the nucleus doesn't eject a pre-existing particle. A neutron inside it, in a nucleus with too many neutrons for stability, converts into a proton, an electron, and an antineutrino, lowering the nucleus's energy. The electron is ejected as a beta particle ( $\beta^-$ ); the new proton stays behind.

_{0}^{1}n \;\rightarrow\; _{1}^{1}p \;+\; _{-1}^{0}e \;+\; \bar{\nu}_e

Mass number stays the same (a neutron and proton weigh almost the same), but atomic number increases by 1 — the element moves one place to the right:

_{Z}^{A}X \;\rightarrow\; _{Z+1}^{A}Y \;+\; _{-1}^{0}\beta \;+\; \bar{\nu}_e

Watch out

The beta particle is not an orbital electron — it is created inside the nucleus at the moment of decay. The antineutrino carries away some energy and momentum; it is nearly massless and barely interacts with matter.

Example: Carbon-14 decays to nitrogen-14:

_{6}^{14}\text{C} \;\rightarrow\; _{7}^{14}\text{N} \;+\; _{-1}^{0}\beta \;+\; \bar{\nu}_e

A proton-rich nucleus does the reverse via beta-plus decay ( $\beta^+$ ): a proton converts into a neutron and emits a positron (the electron's antimatter twin) plus a neutrino.

_{1}^{1}p \;\rightarrow\; _{0}^{1}n \;+\; _{+1}^{0}e \;+\; \nu_e

Gamma Decay — The Nucleus Sheds Excess Energy

After alpha or beta decay, the daughter nucleus is often left in an excited state — extra energy, same number of protons and neutrons. It sheds this energy by emitting a high-energy photon, a gamma ray ( $\gamma$ ), dropping to a lower energy level much as an electron drops to a lower orbit — except here the energy is millions of times larger. Neither the mass number nor the atomic number changes:

_{Z}^{A}X^* \;\rightarrow\; _{Z}^{A}X \;+\; \gamma

Important

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Compare the two processes' Q-value conditions in terms of atomic mass difference $\Delta M = M(^A_ZX) - M(^A_{Z-1}Y)$ : $\beta^+$ needs $\Delta M > 2m_e$ , electron capture only needs $\Delta M > -m_e$ — a much weaker requirement — so $\beta^+$ being allowed guarantees electron capture is allowed, but the reverse is not guaranteed.

Setting up both Q-value expressions (using atomic masses)

For $\beta^{+}$ emission:

^{A}_{Z}X \rightarrow {}^{A}_{Z-1}Y + e^{+} + \nu

Since the parent atom's neutral mass includes $Z$ electrons and the daughter's includes $Z-1$ , and a positron is also created:

Q_{\beta^+} = \left[M(^A_ZX) - M(^A_{Z-1}Y) - 2m_e\right]c^2

For this decay to be energetically allowed, $Q_{\beta^+} > 0$ , i.e.:

M(^A_ZX) - M(^A_{Z-1}Y) > 2m_e \quad \text{...(1)}

For electron capture:

^{A}_{Z}X + e^{-} \rightarrow {}^{A}_{Z-1}Y + \nu

Here an extra electron is consumed (added to the left side), so:

Q_{EC} = \left[M(^A_ZX) + m_e - M(^A_{Z-1}Y)\right]c^2

For electron capture to be allowed, $Q_{EC} > 0$ , i.e.:

M(^A_ZX) - M(^A_{Z-1}Y) > -m_e \quad \text{...(2)}

Comparing the two conditions

Let $\Delta M = M(^A_ZX) - M(^A_{Z-1}Y)$ . Condition (1) requires $\Delta M > 2m_e$ ; condition (2) requires only $\Delta M > -m_e$ . Since $2m_e > -m_e$ (both being real, $m_e>0$ ), condition (1) is strictly the more demanding one:

\Delta M > 2m_e \implies \Delta M > -m_e

So whenever $\beta^{+}$ emission is energetically possible ( $\Delta M > 2m_e$ ), the mass difference automatically also satisfies $\Delta M > -m_e$ , meaning electron capture is necessarily allowed too. …

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