Alpha, beta and gamma decay: how they work, why they happen, and energy & radiation
Notation — what the indexes and symbols mean
In decay formulas we use subscripts to label which particle or nucleus we mean. Here is a short glossary:
\( P \) (parent) — the nucleus before it decays. Example: in \( {}^{238}\text{U} \to {}^{234}\text{Th} + \alpha \), the parent is uranium-238.
\( D \) (daughter) — the nucleus after the decay (the “leftover” nucleus). In the same example, the daughter is thorium-234.
\( \alpha \) (alpha) — the alpha particle (\( {}^4_2\text{He} \)) emitted in alpha decay.
\( m_P, m_D, m_\alpha \) — rest masses of the parent, daughter, and alpha particle (in kg or in atomic mass units u).
\( K \) (kinetic energy) — \( K_\alpha \) = kinetic energy of the alpha particle; \( K_D \) = kinetic energy of the daughter nucleus (recoil); \( K_e \) = kinetic energy of the emitted electron or positron (beta particle); \( K_\nu \) = kinetic energy of the neutrino or antineutrino.
\( p \) (momentum) — \( p_\alpha \), \( p_D \), \( p_\gamma \) = magnitude of momentum of the alpha, daughter, or gamma photon.
\( Q \) (Q-value) — the total decay energy released (in MeV or J). It equals the sum of the kinetic energies of all products: e.g. in alpha decay \( Q = K_\alpha + K_D \).
\( E_\gamma \) — energy of the gamma-ray photon. \( E_\gamma = \Delta E \) = energy difference between two nuclear levels.
\( \Delta E \) — difference in energy between two states (e.g. excited state minus ground state).
\( e^- \), \( e^+ \) — electron and positron. \( \nu_e \), \( \bar{\nu}_e \) — neutrino and antineutrino.
So when you see \( K_D \) or \( m_P \), the subscript tells you: D = daughter, P = parent, α = alpha particle, etc.
1. Alpha decay (\( \alpha \))
The nucleus emits an alpha particle, which is a \( {}^4_2 \text{He} \) nucleus (2 protons, 2 neutrons). It is a very tightly bound cluster, so it is “pre-formed” inside heavy nuclei and can tunnel out.
\[ {}^A_Z X \;\to\; {}^{A-4}_{Z-2} Y + {}^4_2 \text{He} \quad (\alpha) \]
Forces and energies
Forces: Two main forces act in the nucleus. The strong nuclear force binds nucleons together but is short-range (a few fm). The Coulomb (electrostatic) force repels protons and has infinite range. In heavy nuclei, many protons repel each other; the strong force cannot hold all nucleons equally because those at the surface have fewer neighbours. Emitting an alpha removes two protons and two neutrons, reducing Coulomb repulsion and leaving a more bound daughter. The alpha is so tightly bound (high binding energy per nucleon) that it is energetically favourable for heavy nuclei to “shed” it when \( Q > 0 \).
Q-value (decay energy): By conservation of energy (rest mass + kinetic),
Decay is allowed only if \( Q > 0 \) (parent rest mass greater than sum of products’ rest masses). All of \( Q \) appears as kinetic energy of the alpha and the recoiling daughter.
Sharing of kinetic energy (recoil): Parent at rest, so total momentum zero: \( \vec{p}_D + \vec{p}_\alpha = 0 \), hence \( p_D = p_\alpha \). Kinetic energy \( K = p^2/(2m) \), so \( K_\alpha/K_D = m_D/m_\alpha \). For \( {}^{238}\text{U} \to {}^{234}\text{Th} + \alpha \), \( m_D/m_\alpha \approx 234/4 \approx 58.5 \), so the alpha gets about 58.5 times more kinetic energy than the daughter. So \( K_\alpha + K_D = Q \) and \( K_\alpha/K_D = m_D/m_\alpha \) give \( K_\alpha = Q \, m_D/(m_D + m_\alpha) \approx Q \times (1 - m_\alpha/m_D) \). Typically the alpha carries \( \approx 98\% \) of \( Q \).
Behaviour: Alpha particles are heavy (\( \approx 4\,\text{u} \)) and doubly charged (\( +2e \)). The force on them in matter is mainly Coulomb (electric): they attract electrons and repel nuclei, losing energy by ionisation. They ionise strongly (many ion pairs per mm), so they lose energy quickly and penetrate only a few cm in air or a sheet of paper. Range \( R \propto E^3/2 \) roughly for alphas in air. They are stopped by skin; the main hazard is ingestion or inhalation of alpha emitters.
Typical energies: \( Q \approx 4\text{--}9\,\text{MeV} \) for natural alpha emitters. Example: \( {}^{238}\text{U} \) has \( Q \approx 4.27\,\text{MeV} \); the alpha kinetic energy is \( \approx 4.20\,\text{MeV} \).
Example 1 — Uranium to Thorium: \( {}^{238}_{92} \text{U} \to {}^{234}_{90} \text{Th} + {}^4_2 \text{He} \). We lose 4 mass units (the alpha) and 2 in \( Z \). So the daughter has \( A = 234 \), \( Z = 90 \) — that is thorium-234 (\( {}^{234}_{90}\text{Th} \)). Check: 238 = 234 + 4 ✓; 92 = 90 + 2 ✓. Thorium-234 is unstable and decays further (e.g. \( \beta^- \) to \( {}^{234}_{91}\text{Pa} \)).
Why Th-234 and not Th-232? In the periodic table, thorium is often written with a mass around 232 (e.g. 232.04 u). That number is the standard atomic weight of the element thorium — an average over all its isotopes found in nature, and the most abundant stable (or long-lived) isotope is \( {}^{232}_{90}\text{Th} \). So the element thorium has many isotopes (same \( Z = 90 \), different \( A \)): Th-228, Th-230, Th-231, Th-232, Th-234, etc. When \( {}^{238}\text{U} \) decays by alpha, the daughter is specifically the isotope \( {}^{234}\text{Th} \) (234 nucleons), not Th-232. The mass number 234 comes from the decay: 238 − 4 = 234. So: periodic table “232” = typical/abundant isotope of the element; decay gives Th-234 because we started from U-238 and lost 4 units.
Example 2 — Radium to Radon: \( {}^{226}_{88} \text{Ra} \to {}^{222}_{86} \text{Rn} + {}^4_2 \text{He} \). We lose 4 mass units → daughter \( A = 222 \), \( Z = 86 \): radon-222. The periodic table often shows radon with mass ~222 (Rn-222 is the most common isotope in the radium decay chain). So here the daughter mass number (222) matches the “usual” radon mass. Radon-222 is radioactive (alpha decay to Po-218).
Example 3 — Polonium to Lead: \( {}^{210}_{84} \text{Po} \to {}^{206}_{82} \text{Pb} + {}^4_2 \text{He} \). Lose 4 mass units → daughter lead-206 (\( {}^{206}_{82}\text{Pb} \)). Lead has many isotopes; the periodic table shows Pb with ~207.2 (average). Pb-206 is a stable isotope (end of the uranium decay chain). So the same element (lead) can appear with different mass numbers (206, 207, 208, etc.) depending on the isotope.
Example 4 — energy sharing: \( {}^{226}_{88} \text{Ra} \to {}^{222}_{86} \text{Rn} + \alpha \) with \( Q \approx 4.87\,\text{MeV} \). \( K_\alpha = Q \cdot m_D/(m_D + m_\alpha) \approx 4.87 \times 222/226 \approx 4.78\,\text{MeV} \), \( K_D \approx 0.09\,\text{MeV} \). So the alpha carries almost all the energy.
Beta minus (\( \beta^- \)): A neutron in the nucleus turns into a proton, emitting an electron and an antineutrino:
\[ n \;\to\; p + e^- + \bar{\nu}_e \quad \Rightarrow \quad {}^A_Z X \;\to\; {}^A_{Z+1} Y + e^- + \bar{\nu}_e \]
The daughter has the same \( A \) but \( Z \) increases by 1 (one more proton, one fewer neutron). This happens in neutron-rich nuclei (\( N/Z \) too high) to move toward the stability line.
Beta plus (\( \beta^+ \)): A proton turns into a neutron, emitting a positron and a neutrino:
\[ p \;\to\; n + e^+ + \nu_e \quad \Rightarrow \quad {}^A_Z X \;\to\; {}^A_{Z-1} Y + e^+ + \nu_e \]
The daughter has the same \( A \) but \( Z \) decreases by 1. This happens in proton-rich nuclei (\( N/Z \) too low). For \( \beta^+ \) to be possible, the parent mass must exceed the daughter mass by at least \( 2m_e c^2 \) (the rest energy of the created positron plus the electron it eventually annihilates with).
Electron capture (EC): \( p + e^- \to n + \nu_e \), so \( {}^A_Z X + e^- \to {}^A_{Z-1} Y + \nu_e \). Same \( Z \) change as \( \beta^+ \), but an atomic electron is absorbed; no positron. Often competes with \( \beta^+ \) in heavier atoms (inner-shell capture).
Difference between \( \beta^+ \) and \( \beta^- \) decay
Property
\( \beta^- \) decay
\( \beta^+ \) decay
Nucleon change
\( n \to p \) (neutron becomes proton)
\( p \to n \) (proton becomes neutron)
\( Z \) of daughter
\( Z + 1 \) (increases)
\( Z - 1 \) (decreases)
Particle emitted
Electron \( e^- \), antineutrino \( \bar{\nu}_e \)
Positron \( e^+ \), neutrino \( \nu_e \)
Type of nucleus
Neutron-rich (\( N \) too large)
Proton-rich (\( N \) too small)
Q-value (atomic masses)
\( Q = (m_P - m_D)c^2 \)
\( Q = (m_P - m_D - 2m_e)c^2 \)
Mass condition
\( m_P > m_D \) (parent heavier than daughter atom)
\( m_P > m_D + 2m_e \) (need extra \( 2m_e c^2 \) to create \( e^+ \))
So \( \beta^- \) and \( \beta^+ \) are opposite in direction: one increases \( Z \), the other decreases it; one emits an electron, the other a positron; one occurs when there are “too many” neutrons, the other when there are “too many” protons.
Energies and forces in more detail
Forces: Beta decay is caused by the weak nuclear force (weak interaction). It is the only force that can change a quark flavour: in \( \beta^- \), a down quark in the neutron becomes an up quark (so \( n \to p \)); in \( \beta^+ \), an up quark in the proton becomes a down quark (so \( p \to n \)). The electromagnetic and strong forces do not change quark type, so they cannot cause beta decay. The weak force is short-range and much weaker than the strong force at nuclear distances.
Q-value formulas (using atomic masses): When we use atomic masses (neutral atom = nucleus + Z electrons), the number of electrons on each side can be chosen so that the electron masses cancel in the mass difference. For \( \beta^- \): parent atom has \( Z \) electrons, daughter atom has \( Z+1 \) electrons. The emitted electron “replaces” one of the daughter’s electrons in the count, so effectively we compare \( m_P(\text{atom}) \) with \( m_D(\text{atom}) \); the extra electron mass is already in \( m_D \). So:
For \( \beta^+ \): parent has \( Z \) electrons, daughter has \( Z-1 \). We create a positron (mass \( m_e \)) and the positron eventually annihilates with an electron (another \( m_e \)), so we need at least \( 2m_e c^2 \) of rest mass to be converted. Thus:
The maximum kinetic energy of the beta particle (electron or positron) equals \( Q \). The rest of \( Q \) is carried by the neutrino (and a tiny amount by the recoiling nucleus). So \( K_e + K_\nu + K_\text{recoil} = Q \); for most purposes \( K_\text{recoil} \approx 0 \), so \( K_e + K_\nu = Q \).
Why the beta spectrum is continuous: Three bodies are produced: daughter nucleus, beta particle, neutrino. Conservation of energy and momentum allows the kinetic energy to be shared in infinitely many ways. So the beta can have any KE from 0 up to \( Q \) (the “endpoint”); the neutrino carries \( Q - K_e \). In contrast, alpha decay gives two bodies (alpha + daughter), so the alpha has a unique kinetic energy (momentum sharing is fixed).
Example — \( \beta^- \), Carbon to Nitrogen: \( {}^{14}_6 \text{C} \to {}^{14}_7 \text{N} + e^- + \bar{\nu}_e \). Carbon-14 has 6p, 8n (neutron-rich). One neutron → proton, so daughter is nitrogen-14 (7p, 7n). Mass number stays 14; only \( Z \) changes (6 → 7). The periodic table shows nitrogen with mass ~14.01 (mostly N-14). So the daughter is the common isotope of nitrogen. \( Q \approx 0.156\,\text{MeV} \); the electron has a continuous spectrum from 0 to 0.156 MeV.
Example — \( \beta^- \), Strontium to Yttrium: \( {}^{90}_{38} \text{Sr} \to {}^{90}_{39} \text{Y} + e^- + \bar{\nu}_e \). Strontium-90 (38p, 52n) is neutron-rich; daughter is yttrium-90 (39p, 51n). Same \( A = 90 \); \( Z \) increases 38 → 39. Strontium in the periodic table is ~87.6 (mainly Sr-88); here we are dealing with the specific isotope Sr-90 (fission product). Y-90 is also radioactive (\( \beta^- \) to Zr-90).
Example — \( \beta^+ \), Sodium to Neon: \( {}^{22}_{11} \text{Na} \to {}^{22}_{10} \text{Ne} + e^+ + \nu_e \). Sodium-22 is proton-rich (11p, 11n). One proton → neutron → daughter neon-22 (10p, 12n). Same \( A = 22 \); \( Z \) decreases 11 → 10. The periodic table shows neon with ~20.18 (mainly Ne-20); Ne-22 is a stable but less abundant isotope. The positron spectrum has maximum energy \( Q \approx 1.82\,\text{MeV} \).
Example — \( \beta^- \), Cesium to Barium: \( {}^{137}_{55} \text{Cs} \to {}^{137}_{56} \text{Ba}^* + e^- + \bar{\nu}_e \). Cesium-137 (55p, 82n) is a common fission product. Daughter is barium-137 (56p, 81n); it is often in an excited state and then emits a 0.662 MeV gamma. So Cs-137 sources give both \( \beta^- \) and \( \gamma \). Barium in the periodic table has ~137.3; Ba-137 is one of its isotopes.
Behaviour: Beta particles (electrons or positrons) are light (\( m_e \)) and singly charged. They penetrate more than alpha (metres in air, mm in tissue) and cause ionisation. Stopped by a few mm of aluminium or plastic. Positrons eventually annihilate with an electron, producing two 0.511 MeV gamma rays.
3. Gamma decay (\( \gamma \))
The nucleus can exist in discrete energy levels (like atoms, but with much larger energy gaps). When it is in an excited state (higher level), it can drop to a lower state by emitting a gamma-ray photon — a high-energy EM wave (wavelength \( \sim 10^{-12}\,\text{m} \) or less).
\[ {}^A_Z X^* \;\to\; {}^A_Z X + \gamma \]
\( Z \) and \( A \) do not change — it is the same nuclide, just in a lower energy state. No nucleons or leptons are ejected; only a photon.
Forces and energies
Force: Gamma emission is an electromagnetic process. The excited nucleus has a different distribution of charge and current than the ground state; the transition is mediated by the electromagnetic field, so the photon is an EM wave. No weak or strong force is involved in the emission itself.
Energy: Conservation of energy gives \( E_\gamma = \Delta E = E^* - E \), the level spacing. So \( E_\gamma = hf = hc/\lambda \). Nuclear level spacings are typically keV to MeV, so \( E_\gamma \sim 0.1\text{--}10\,\text{MeV} \).
\[ E_\gamma = \Delta E, \qquad p_\gamma = \frac{E_\gamma}{c} \]
Recoil of the nucleus: Conservation of momentum (initial nucleus at rest): \( \vec{p}_\gamma + \vec{p}_\text{recoil} = 0 \), so \( p_\text{recoil} = E_\gamma/c \). Recoil kinetic energy \( K_\text{recoil} = p^2/(2M) = E_\gamma^2/(2Mc^2) \). For \( E_\gamma = 1\,\text{MeV} \) and \( M = 100\,\text{u} \), \( K_\text{recoil} \approx 10^6/(2 \times 100 \times 931) \approx 5\,\text{eV} \), negligible compared to 1 MeV. So \( E_\gamma \approx \Delta E \) to high accuracy.
Why it happens: After alpha or beta decay, the daughter nucleus is often left in an excited state. It relaxes by emitting one or more gamma photons (sometimes in a cascade). So gamma radiation often accompanies alpha or beta decay.
Behaviour: Gamma rays are uncharged; they interact via the electromagnetic force (photoelectric, Compton, pair production). They penetrate matter deeply (many cm of lead or concrete). They are a major external radiation hazard.
Example — Cobalt-60: After \( {}^{60}_{27} \text{Co} \) undergoes \( \beta^- \) decay to \( {}^{60}_{28} \text{Ni}^* \), the nickel-60 nucleus is excited (same \( A = 60 \), \( Z = 28 \)). It emits two gamma rays in succession (1.17 MeV and 1.33 MeV) to reach the ground state. So a cobalt-60 source emits betas and two characteristic gammas; \( E_\gamma = \Delta E \) for each transition. Nickel in the periodic table is ~58.69 (mainly Ni-58, Ni-60, Ni-62); Ni-60 is a stable isotope.
Example — Gamma after alpha (Americium): \( {}^{241}_{95} \text{Am} \) decays by alpha to \( {}^{237}_{93} \text{Np}^* \) (neptunium-237, often excited). Neptunium-237 then emits gamma(s) to reach its ground state. So the same element (e.g. neptunium) can be produced with a specific mass number (237) from a given parent; the periodic table may show Np with ~237 — that is the dominant isotope for that artificial element.
4. Summary: radiation and energy
Alpha: nucleus loses 2p + 2n (\( A \to A-4 \), \( Z \to Z-2 \)); high ionisation, low penetration; energy \( Q \) shared between alpha and daughter (alpha gets most); discrete alpha energy (or a few lines if gamma also emitted).
Beta: nucleus changes \( Z \) by ±1, \( A \) unchanged; continuous energy spectrum from 0 to \( Q \); energy shared between beta and neutrino; moderate penetration.
Gamma: same nucleus, lower energy state; photon carries \( E_\gamma = \Delta E \); high penetration; no mass or charge change; discrete line spectrum.
All three are forms of ionising radiation: they can knock electrons out of atoms and damage DNA/tissue. Shielding (alpha: paper; beta: plastic/aluminium; gamma: lead/concrete), distance, and limiting exposure time reduce dose.
Comparison example: \( {}^{238}\text{U} \) decays by alpha; the alpha has \( \approx 4.2\,\text{MeV} \) and is stopped by paper. The same source also emits low-energy gammas; those need lead to attenuate. A \( \beta^- \) source like \( {}^{90}\text{Sr} \) emits electrons with a continuous spectrum up to \( \approx 0.5\,\text{MeV} \); a few mm of metal stops them.
Element table vs mass number: The periodic table shows each element with a standard atomic weight (average over isotopes). A decay product is a specific isotope (fixed \( A \), \( Z \)). So U-238 → Th-234 (we lose 4 units); the table shows Th as ~232 because Th-232 is the most abundant. Always use the mass number from the decay (e.g. 234) to identify the isotope, not the table value alone.
Formulas at a glance
Subscripts: P = parent, D = daughter, \( \alpha \) = alpha; \( K \) = kinetic energy, \( Q \) = decay energy. See Notation at the top for full list.
4. Why does the beta particle have a continuous energy spectrum while the alpha particle has (almost) a single energy?
Solution: Alpha decay produces two bodies (alpha + daughter), so conservation of energy and momentum fix the kinetic energy of each. Beta decay produces three bodies (daughter + electron + neutrino); the total \( Q \) is shared between the electron and neutrino in infinitely many ways, so the electron can have any KE from 0 up to \( Q \).
5. In the alpha decay \( {}^{238}_{92}\text{U} \to {}^{234}_{90}\text{Th} + {}^4_2\text{He} \), \( Q \approx 4.27\,\text{MeV} \). Estimate the kinetic energy of the alpha particle. (Use \( m_D/m_\alpha \approx 234/4 \).)
6. A nucleus in an excited state 1.5 MeV above the ground state emits a gamma ray. What is the approximate energy of the photon? Does the mass number change?
Solution: \( E_\gamma \approx 1.5\,\text{MeV} \) (the level spacing; a tiny amount goes to nuclear recoil). The mass number \( A \) (and \( Z \)) do not change — it is the same nucleus in a lower energy state.
7. \( {}^{22}_{11}\text{Na} \) decays by \( \beta^+ \). Write the decay equation. Does the mass number change? Why does \( \beta^+ \) require \( m_P > m_D + 2m_e \)?
Solution: \( {}^{22}_{11}\text{Na} \to {}^{22}_{10}\text{Ne} + e^+ + \nu_e \). Mass number \( A \) does not change (22). We need \( 2m_e c^2 \) because we create a positron (mass \( m_e \)) and the positron eventually annihilates with an electron (another \( m_e \)), so the minimum rest energy that must come from the mass difference is \( 2m_e c^2 \).
8. Which force is responsible for (a) alpha decay being energetically favourable in heavy nuclei, (b) beta decay occurring, (c) gamma emission?
Solution: (a) Coulomb repulsion (and strong force being short-range) makes heavy nuclei less bound; emitting an alpha reduces Coulomb repulsion. (b) The weak nuclear force (changes quark flavour: down↔up) is responsible for beta decay. (c) Electromagnetic force — the transition is an EM process; the photon is an electromagnetic wave.
9. The maximum kinetic energy of the electron in \( \beta^- \) decay of \( {}^{14}\text{C} \) is about 0.156 MeV. What is the Q-value for this decay? Where does the rest of the energy go?
Solution: The maximum KE of the electron equals \( Q \), so \( Q \approx 0.156\,\text{MeV} \). The rest of \( Q \) is shared between the electron and the antineutrino for each decay; when the electron has less than 0.156 MeV, the antineutrino carries \( Q - K_e \). A tiny amount goes to the recoiling daughter nucleus (negligible).
10. Complete the decay: \( {}^{214}_{82}\text{Pb} \to \ ?\ + e^- + \bar{\nu}_e \). Is this \( \beta^- \) or \( \beta^+ \)? What is the daughter?
Solution: This is \( \beta^- \) decay (electron and antineutrino emitted). \( Z \) increases by 1: daughter has \( Z = 83 \), \( A = 214 \). So daughter is \( {}^{214}_{83}\text{Bi} \) (bismuth-214). Bismuth has many isotopes; 214 is unstable (in the decay chain).
11. In alpha decay \( {}^{238}_{92}\text{U} \to {}^{234}_{90}\text{Th} + {}^4_2\text{He} \), we get thorium with mass number 234. The periodic table often shows thorium with mass about 232. Explain why the daughter is Th-234 and not Th-232.
Solution: The periodic table shows the element thorium with a standard atomic weight (~232) — an average over naturally occurring isotopes; the most abundant stable isotope is \( {}^{232}\text{Th} \). In this decay we start from U-238 and lose 4 mass units (the alpha), so the daughter must have \( A = 238 - 4 = 234 \). So we get the isotope \( {}^{234}\text{Th} \), not Th-232. Same element (Z = 90), different isotope (different A). Th-234 is unstable and decays further.
12. Complete the alpha decay of radon-222. What is the daughter element and its mass number? Why does the mass number decrease by 4?
Solution: \( {}^{222}_{86}\text{Rn} \to {}^{218}_{84}\text{Po} + {}^4_2\text{He} \). Daughter: polonium-218 (Z = 84, A = 218). The mass number decreases by 4 because the alpha particle carries away 4 nucleons (2p + 2n). So 222 − 4 = 218. Polonium has many isotopes (Po-210, Po-218, etc.); the table may show Po with ~209; here we have the specific isotope Po-218 from this decay.
13. \( {}^{90}_{38}\text{Sr} \) decays by \( \beta^- \). Write the equation and name the daughter. Does the mass number change? What is the daughter’s atomic number?
Solution: \( {}^{90}_{38}\text{Sr} \to {}^{90}_{39}\text{Y} + e^- + \bar{\nu}_e \). Daughter: yttrium-90 (Y, Z = 39). Mass number does not change (90); only Z increases by 1 (38 → 39). One neutron became a proton. Strontium-90 is a fission product; Y-90 is also radioactive (\( \beta^- \) to Zr-90). The periodic table shows Y with ~88.9 (mainly Y-89); Y-90 is an unstable isotope.
14. Polonium-210 decays by alpha to a stable isotope of lead. Write the decay equation. What is the mass number of the lead isotope? (Lead has Z = 82.)
Solution: \( {}^{210}_{84}\text{Po} \to {}^{206}_{82}\text{Pb} + {}^4_2\text{He} \). Daughter: lead-206 (A = 206, Z = 82). Mass number 210 − 4 = 206. Lead-206 is stable (end of the uranium-238 decay chain). The periodic table shows lead with ~207.2 because lead has several stable isotopes (Pb-204, 206, 207, 208); 206 is one of them.
15. \( {}^{131}_{53}\text{I} \) (iodine-131) decays by \( \beta^- \). Identify the daughter element and write the full equation. (Iodine Z = 53; the next element is xenon, Z = 54.)
Solution: \( {}^{131}_{53}\text{I} \to {}^{131}_{54}\text{Xe} + e^- + \bar{\nu}_e \). Daughter: xenon-131 (Xe, Z = 54). Same A = 131; Z increases 53 → 54. Iodine-131 is used in medicine; xenon-131 is stable. The periodic table shows Xe with ~131.3; Xe-131 is one of the stable isotopes of xenon.
16. In the decay chain, \( {}^{234}_{90}\text{Th} \) decays by \( \beta^- \). What is the daughter? (Thorium Z = 90; next element is protactinium, Z = 91.) Does the mass number change?
Solution: \( {}^{234}_{90}\text{Th} \to {}^{234}_{91}\text{Pa} + e^- + \bar{\nu}_e \). Daughter: protactinium-234 (Pa, Z = 91). Mass number stays 234; Z increases 90 → 91. So after U-238 alpha decay we had Th-234; now Th-234 beta-decays to Pa-234. The periodic table shows Pa with ~231 (Pa-231 is long-lived); Pa-234 is unstable and decays further (e.g. \( \beta^- \) to U-234).
17. Radium-226 (\( {}^{226}_{88}\text{Ra} \)) decays by alpha. What element do we get and with what mass number? If the periodic table shows that element with a different mass number, explain why.
Solution: \( {}^{226}_{88}\text{Ra} \to {}^{222}_{86}\text{Rn} + {}^4_2\text{He} \). We get radon-222 (Rn, Z = 86, A = 222). The periodic table often shows radon with ~222 (Rn-222 is the most common isotope in the radium chain). So in this case the mass number 222 matches the typical value for radon. For other elements (e.g. thorium from U-238), the decay product is a specific isotope (Th-234) that is not the same as the most abundant one (Th-232) — so the table shows 232 for Th, but we get 234 from the decay.
18. \( {}^{40}_{19}\text{K} \) (potassium-40) can decay by \( \beta^- \) to \( {}^{40}_{20}\text{Ca} \) or by \( \beta^+ \) (and EC) to \( {}^{40}_{18}\text{Ar} \). (a) Write both decay equations. (b) The periodic table shows Ca with 40.08 and Ar with 39.95. Which isotopes are involved?
Solution: (a) \( \beta^- \): \( {}^{40}_{19}\text{K} \to {}^{40}_{20}\text{Ca} + e^- + \bar{\nu}_e \). \( \beta^+ \): \( {}^{40}_{19}\text{K} \to {}^{40}_{18}\text{Ar} + e^+ + \nu_e \). (b) In both cases the mass number is 40. So we get calcium-40 (stable, most abundant Ca isotope) and argon-40 (stable). The table “40.08” for Ca is an average (mainly Ca-40, 42, 43, 44, 46, 48); “39.95” for Ar is mainly Ar-40, 36, 38. So K-40 decays to two different elements (Ca and Ar), both with the same A = 40 — same mass number, different Z.