Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) ²²⁶₈₈Ra and (b) ²²⁰₈₆R

Exercises · 13.12

\alpha

(a)

^{226}_{88}\text{Ra}

and

(b)

^{220}_{86}\text{Rn}

.
Given

m(^{226}_{88}\text{Ra}) = 226.02540

m(^{222}_{86}\text{Rn}) = 222.01750

m(^{220}_{86}\text{Rn}) = 220.01137

m(^{216}_{84}\text{Po}) = 216.00189

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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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$Q = [m_{\text{parent}} - m_{\text{daughter}} - m_{\alpha}]c^2$ , and the alpha particle's kinetic energy is $Q$ scaled by $\frac{A_{\text{daughter}}}{A_{\text{daughter}}+4}$ from momentum conservation. Both decays are exothermic with $Q\approx4.9$ and $6.4$ MeV respectively.

(a) $^{226}_{88}\text{Ra} \rightarrow {}^{222}_{86}\text{Rn} + {}^{4}_{2}\text{He}$

Q = \left[m(^{226}\text{Ra}) - m(^{222}\text{Rn}) - m(^{4}\text{He})\right] \times 931.5\ \text{MeV}

= [226.02540 - 222.01750 - 4.002603] \times 931.5

= 0.005297\ \text{u} \times 931.5 = 4.934\ \text{MeV}

Since the recoiling Rn-222 nucleus must carry away momentum equal and opposite to the alpha's, the kinetic energy splits in inverse proportion to mass. Treating mass numbers as proportional to mass:

KE_\alpha = Q \times \frac{A_{\text{daughter}}}{A_{\text{daughter}} + 4} = 4.934 \times \frac{222}{226} = 4.847\ \text{MeV}

(b) $^{220}_{86}\text{Rn} \rightarrow {}^{216}_{84}\text{Po} + {}^{4}_{2}\text{He}$

Q = \left[m(^{220}\text{Rn}) - m(^{216}\text{Po}) - m(^{4}\text{He})\right] \times 931.5

= [220.01137 - 216.00189 - 4.002603] \times 931.5

$= 0.006877\ \text{u} \times 931.5 = 6.406\ \text{MeV}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
$\alpha$
›Reveal solutionSolution
A positron is the antiparticle of the electron, and it is emitted specifically in $\beta^+$ (positron) decay. Answer: $\beta^+$ decay.

Concept and Intuition

Radioactive decay comes in several types distinguished by what's emitted: $\alpha$ decay emits a helium nucleus, $\beta^-$ decay emits an electron (from neutron $\to$ proton conversion), $\beta^+$ decay emits a positron (from proton $\to$ neutron conversion, only possible inside a nucleus using its binding energy), and $\gamma$ decay emits a high-energy photon with no change in atomic/mass number. The positron is the electron's antiparticle — same mass, opposite (positive) charge — and is uniquely associated with $\beta^+$ decay.

Step-by-Step Solution

$\alpha$ decay: emits a $^4_2He$ nucleus — no positron.

$\beta^+$ decay: $p \to n + e^+ + \nu_e$ — emits a positron ( $e^+$ ). This is the process asked about. …
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AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
$\beta$
›Reveal solutionSolution
Wolfgang Pauli postulated the neutrino to save energy conservation in $\beta$ -decay. Answer: (B).

Concept and Intuition

In $\beta$ -decay, a nucleus emits an electron, but the electron's energy was observed to vary continuously rather than take one fixed value — apparently violating energy conservation if only the nucleus and electron were involved. This puzzle needed a third, undetected particle to carry away the missing energy and momentum.

Step-by-Step Solution

Experiments showed $\beta$ -decay electrons emerge with a continuous range of energies, unlike $\alpha$ - or $\gamma$ -decay which are essentially mono-energetic.

In 1930, Wolfgang Pauli proposed (in a famous letter) that a light, neutral, weakly-interacting particle is also emitted, carrying away the 'missing' energy and momentum, thereby rescuing conservation laws.

Enrico Fermi later incorporated this particle into his theory of $\beta$ -decay and named it the 'neutrino' (little neutral one). …
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AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
$Q. Positron is the antiparticle of (A) proton (B) electron (C) neutron (D) photon$
›Reveal solutionSolution
The positron is the electron's antiparticle — identical mass and spin, opposite charge.

Concept and Intuition

Every fundamental particle has an antiparticle with the same mass but opposite charge (and opposite values of other additive quantum numbers). The positron ( $e^+$ ) is the antiparticle counterpart of the electron ( $e^-$ ): it carries the same rest mass ( $0.511\,\text{MeV}/c^2$ ) and spin- $\tfrac12$ , but a $+e$ charge instead of $-e$ . When an electron and a positron meet, they annihilate, converting their combined rest mass into gamma-ray photon energy.

Step-by-Step Solution

Recall the defining property of antiparticles: same mass, opposite charge (and opposite lepton number, etc.).

The positron's mass and spin match the electron exactly.

Its charge is $+e$ , exactly opposite the electron's $-e$ . …
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