Alpha Decay & the Q-Value

Alpha decay left the laboratory and went to work. A tiny pellet of americium-241 sits in almost every household smoke detector: its alphas ionise the air in a small chamber and smoke particles interrupt the resulting current, tripping the alarm. Spacecraft too far from the Sun for solar panels — Voyager, Cassini, New Horizons past Pluto — run on radioisotope thermoelectric generators powered by the heat of plutonium-238 alpha decay, whose alphas conveniently deposit their energy locally as warmth. In medicine, targeted alpha therapy turns the very property that makes alphas so damaging — they dump all their energy within a few cell diameters — into a scalpel, killing cancer cells while sparing tissue a hair's breadth away (radium-223, actinium-225). And Gamow's tunnelling runs in reverse in the cores of stars, where nuclei tunnel through their mutual Coulomb barrier to fuse — the process that lets the Sun burn at "only" fifteen million degrees. The little wave that leaks out of a uranium nucleus and the sunlight on your face are the same quantum effect.

An alpha particle is nothing exotic: it is a helium-4 nucleus, two protons and two neutrons locked together, charge $+2e$, mass about $4\ \text{u}$. Alpha decay is the spontaneous ejection of one of these clusters from a nucleus, almost always a heavy one — uranium, thorium, radium, polonium, and their kin above lead. When it happens, the nucleus keeps every conservation law: charge and nucleon number are both conserved, so if the parent is $^{A}_{Z}X$ the daughter must be

The daughter has two fewer protons (an element two places back in the periodic table) and four fewer nucleons. Uranium-238 ($Z=92$) becomes thorium-234 ($Z=90$); radium-226 becomes radon-222; polonium-210 becomes stable lead-206. This element-changing step is what the word transmutation means — Rutherford and Soddy's great insight of 1902–03 that radioactivity is one element turning into another.

The alphas do not dribble out at random energies. Every alpha from a given isotope emerges with essentially the same, sharp kinetic energy — uranium-238 alphas all carry very close to $4.20\ \text{MeV}$, polonium-210 alphas all carry about $5.3\ \text{MeV}$. That sharpness is the signature of a two-body split from a nucleus with definite energy levels, and it is the crucial contrast with beta decay's smeared spectrum (the subject of the next article).

The engine of alpha decay is $E=mc^2$. Weigh the parent nucleus, then weigh the daughter and the alpha separately, and the parent is found to be very slightly heavier than its two products together. That missing mass, $\Delta m$, has not vanished — it has become kinetic energy, shared between the flying alpha and the recoiling daughter. The energy released is called the Q-value:

where: $M(X)$ is the parent's mass, $M(Y)$ the daughter's, $M(\alpha)$ the alpha's, and $c$ the speed of light. Physically, $Q>0$ means the decay is exothermic — it can happen spontaneously and releases energy; $Q<0$ would mean the parent is stable against alpha emission because the products weigh more than it does. The convenient unit is the atomic mass unit and its energy equivalent, fixed by the constant every board quotes,

There is a bookkeeping subtlety that trips up half the exam hall, and dealing with it cleanly is worth a mark every time. Equation (1) is written for bare nuclei, but tables list atomic masses — the nucleus plus its full complement of electrons. Do we have to correct for the electrons? For alpha decay, wonderfully, no. The parent atom carries $Z$ electrons; the daughter carries $Z-2$ and the neutral helium atom carries $2$, for $Z-2+2 = Z$ on the product side — exactly matching. The electron masses cancel identically, so you may drop the neutral atomic masses straight into equation (2) and never think about electrons again. (This tidy cancellation is special; beta-plus decay, for instance, does not enjoy it.)

The Q-value is the total released energy, but it is not the energy the alpha carries away — a distinction examiners love. When the parent sits at rest and splits into two pieces, momentum conservation ties their kinetic energies together, and because the alpha is so much lighter than the daughter, it runs off with almost — but not quite — all of $Q$.

Was the non-relativistic $KE = p^2/2m$ legitimate? Check the alpha: its $4.2\ \text{MeV}$ of kinetic energy is only about $0.1\%$ of its rest energy $4\ \text{u}\times931.5 = 3726\ \text{MeV}$, giving $v/c \approx \sqrt{2\times4.2/3726} \approx 0.047$ — under five per cent of light speed. Alpha decay is comfortably non-relativistic, and the school-algebra split above is exact to a fraction of a per cent.

Here is the puzzle a sharp student always asks. If a heavy nucleus is unstable, why does it fling out a whole four-nucleon package rather than shedding a single proton or neutron, which sounds easier? The answer is energy, and it turns on how tightly the helium-4 nucleus is bound. Recall that a bound nucleus weighs less than its separate nucleons, and the mass deficit times $c^2$ is its binding energy — the energy that would be needed to pull it apart. Helium-4 is a standout: its binding energy is $28.3\ \text{MeV}$, an enormous $7.07\ \text{MeV}$ per nucleon, a local peak far above its neighbours. When two protons and two neutrons leave a heavy nucleus already assembled as an alpha, that $28.3\ \text{MeV}$ of alpha binding energy is "cashed in" and helps make $Q$ positive. Ejecting a single proton or neutron cashes in no such cluster binding; instead you must pay the nucleon's separation energy (typically $6$–$8\ \text{MeV}$ for a heavy nucleus), so the Q-value for single-nucleon emission comes out negative — the process simply cannot happen spontaneously. The alpha wins because it is the most tightly bound light cluster nature offers: emitting it is essentially the one nucleon-shedding channel that releases energy for most heavy nuclei. (Splitting into two large fragments — fission — is the other, taken up in later chapters.)

The Q-value tells us alpha decay is allowed, but not how fast it goes — and the rates are staggeringly varied. Polonium-212 alpha-decays with a half-life of under a microsecond; uranium-238 takes four and a half billion years. What sets this astronomical range? Picture the alpha already formed inside the parent, sitting in the deep attractive well of the strong nuclear force. To escape it must get past the electrostatic repulsion between its charge $+2e$ and the daughter's charge $+(Z-2)e$. That repulsion builds a Coulomb barrier; its height (higher-tier: JEE-Adv/A-Level/IB-HL, not core NEET/IGCSE), right at the nuclear surface $R \approx R_0\big(A_Y^{1/3} + 4^{1/3}\big)$ with $R_0 = 1.2\ \text{fm}$, is

where: $k(2e)(Z{-}2)e$ is the Coulomb potential energy of the two charges at separation $R$, and the handy constant $ke^2 = 1.44\ \text{MeV}\,\cdot\,\text{fm}$ packages $1/4\pi\varepsilon_0$ into nuclear units. For the thorium daughter of uranium ($Z-2 = 90$, $R \approx 9.3\ \text{fm}$) this gives $V_B \approx 28\ \text{MeV}$. Read that again against the alpha's energy: the wall is about $28\ \text{MeV}$ high, and the alpha arrives with only $4.27\ \text{MeV}$. Classically it is trapped for eternity — a ball rolling at a hill many times too tall never gets over.

Classically, but not quantum-mechanically. In 1928 George Gamow (and, independently, Ronald Gurney and Edward Condon) realised the alpha behaves as a wave, and a wave does not stop dead at a barrier it cannot surmount — a little of it leaks through and emerges on the far side. The alpha has a small but non-zero probability of appearing outside the barrier on each of the trillions of times per second it batters against the wall from inside; sooner or later it tunnels out. This quantum tunnelling is the whole of alpha decay's dynamics, and it explains the enormous spread of half-lives at a stroke: the leakage probability is fiercely sensitive to the barrier's height and width, so a modest rise in $Q$ — which both lowers the effective barrier and thins it — shrinks the half-life by many powers of ten. That is the quantitative content of the old empirical Geiger–Nuttall law (1911), which linked a higher alpha energy to a shorter half-life long before anyone understood why. You can see it directly in the numbers: as $Q$ climbs from $4.27\ \text{MeV}$ (uranium-238) to $4.87\ \text{MeV}$ (radium-226) to $5.41\ \text{MeV}$ (polonium-210), the half-life collapses from $4.5$ billion years, to $1600$ years, to $138$ days. The exact tunnelling probability requires solving the Schrödinger equation for the wave through the barrier — Gamow's 1928 calculation, a piece of university quantum mechanics well beyond this article's algebra. We quote the picture and its consequence, not the integral. For a high-school or first-year exam you are expected to describe tunnelling and use the barrier idea qualitatively; deriving the Gamow factor is JEE-Advanced, A-Level or IB-HL enrichment, not core.

Board level — the alpha energy from radium-226 (CBSE and ICSE pattern)

Radium-226 alpha-decays to radon-222. Given atomic masses $M(^{226}\text{Ra}) = 226.025410\ \text{u}$, $M(^{222}\text{Rn}) = 222.017578\ \text{u}$, $M(^{4}\text{He}) = 4.002603\ \text{u}$ and $1\ \text{u} = 931.5\ \text{MeV}$, find (a) the daughter nuclide, (b) the Q-value, and (c) the kinetic energy of the emitted alpha.

Step 1. Balance the equation. Charge and nucleon number conserve: $^{226}_{\,88}\text{Ra} \to {}^{222}_{\,86}\text{Rn} + {}^{4}_{2}\text{He}$. The daughter is radon-222 ($Z=86$, $A=222$), two places back in the periodic table.

Step 2. Mass defect, using atomic masses (electrons cancel, $88 = 86+2$): $\Delta m = 226.025410 - 222.017578 - 4.002603 = 0.005229\ \text{u}$.

Step 3. Q-value: $Q = 0.005229 \times 931.5 = \boxed{4.87\ \text{MeV}}$.

Step 4. Alpha kinetic energy from equation (6): $KE_\alpha = Q(A-4)/A = 4.87 \times 222/226 = \boxed{4.78\ \text{MeV}}$. The radon recoil carries the small remainder, $4.87 - 4.78 = 0.09\ \text{MeV}$.

Competitive level — polonium-210, the recoil, and the alpha's speed (JEE and NEET pattern)

Polonium-210 (the isotope Marie Curie named after Poland) alpha-decays to stable lead-206. Atomic masses: $M(^{210}\text{Po}) = 209.982874\ \text{u}$, $M(^{206}\text{Pb}) = 205.974465\ \text{u}$, $M(^{4}\text{He}) = 4.002603\ \text{u}$. Find (a) the Q-value, (b) the alpha's kinetic energy and the lead recoil energy, and (c) the alpha's speed. Take $1\ \text{u} = 931.5\ \text{MeV} = 1.66\times10^{-27}\ \text{kg}$ and $1\ \text{MeV} = 1.602\times10^{-13}\ \text{J}$.

Step 1. Mass defect and Q-value: $\Delta m = 209.982874 - 205.974465 - 4.002603 = 0.005806\ \text{u}$, so $Q = 0.005806 \times 931.5 = \boxed{5.41\ \text{MeV}}$.

Step 2. Energy split, equation (6) with $A = 210$: $KE_\alpha = 5.41 \times 206/210 = \boxed{5.3\ \text{MeV}}$ and the lead recoil $KE_Y = 5.41 \times 4/210 = \boxed{0.10\ \text{MeV}}$. Note the recoil is $103\ \text{keV}$ — small, but enough to knock the lead atom clean out of its lattice site, which is how alpha emitters slowly damage nearby material.

Step 3. Alpha speed from $KE_\alpha = \tfrac12 m_\alpha v^2$. Convert: $KE_\alpha = 5.3\ \text{MeV} = 5.3\times1.602\times10^{-13} \approx 8.5\times10^{-13}\ \text{J}$; alpha mass $m_\alpha = 4\times1.66\times10^{-27} = 6.64\times10^{-27}\ \text{kg}$. Then $v = \sqrt{2\,KE_\alpha/m_\alpha} = \sqrt{2\times8.5\times10^{-13}/6.64\times10^{-27}} = \boxed{1.6\times10^{7}\ \mathrm{m\,s^{-1}}}$ — about $0.05c$, safely non-relativistic, confirming the algebra we used. Timing note: about two minutes. The single mass subtraction gives $Q$; after that everything is the two ratios $(A-4)/A$ and $4/A$ and one square root. The trap the clock rewards you for dodging is reaching for relativistic formulas — check $KE/mc^2 \approx 5.3/3726 \approx 0.14\%$ first and use $\tfrac12 mv^2$ with a clear conscience.

Competitive level — reading the Geiger–Nuttall trend (JEE Advanced and A-Level pattern)

Three alpha emitters have Q-values and half-lives: uranium-238 ($Q = 4.27\ \text{MeV}$, $T_{1/2} = 4.5\times10^{9}\ \text{yr}$), radium-226 ($Q = 4.87\ \text{MeV}$, $T_{1/2} = 1600\ \text{yr}$), polonium-210 ($Q = 5.41\ \text{MeV}$, $T_{1/2} = 138\ \text{days}$). Without any formula for the tunnelling rate, argue the direction of the trend and explain it physically.

Step 1. List the pattern. As $Q$ rises $4.27 \to 4.87 \to 5.41\ \text{MeV}$ — a change of barely $1\ \text{MeV}$ — the half-life falls $4.5\ \text{billion years} \to 1600\ \text{years} \to 138\ \text{days}$, a plunge of roughly ten powers of ten.

Step 2. Explain it by the barrier. A higher-energy alpha meets the Coulomb barrier at a lower point and has a thinner slice of barrier to cross (the classical exit radius, where the $1/r$ tail falls to $Q$, moves inward as $Q$ grows). Tunnelling probability depends exponentially on both the height and the width of the barrier, so a small rise in $Q$ multiplies the escape probability enormously — and the half-life, being inversely proportional to it, collapses.

Step 3. State the law. This monotonic anti-correlation between alpha energy and half-life is the Geiger–Nuttall law, an empirical fit of 1911 that Gamow's 1928 tunnelling theory finally explained. The physics lesson: nuclear energy differences of a fraction of a per cent can drive lifetime differences of billions — the exponential is unforgiving. Timing note: under a minute — this is a "state the trend and give the mechanism" question with no arithmetic. The examiner wants the words "higher Q, shorter half-life" and "tunnelling probability is exponentially sensitive to the barrier"; the extreme numerical spread is the punchline, not a calculation.

Reporting $Q$ as the alpha's energy. The detector reads $KE_\alpha = Q(A-4)/A$, not $Q$. For uranium-238 that is $4.20\ \text{MeV}$ versus $4.27\ \text{MeV}$. Losing the recoil term is the classic slip on CBSE, ICSE, JEE and NEET numericals — always split the energy.

Double-counting electrons. For alpha decay the neutral atomic masses already make the electrons cancel ($Z = (Z-2)+2$). Do not also subtract $Z\,m_e$; that is a second correction for a problem you have already solved, and it produces a wrong $Q$.

The daughter's numbers. Atomic number drops by 2 and mass number by 4: $^{A}_{Z}X \to {}^{A-4}_{Z-2}Y$. Writing $Z-4$, or forgetting to move the element two places back in the table, is a guaranteed mark lost on the "write the decay equation" question.

"The alpha climbs over the barrier." It does not — its energy is far below the barrier top. Alpha escape is tunnelling, a quantum effect. Saying the alpha "has enough energy to overcome the Coulomb barrier" is simply wrong and is penalised in any question that mentions the barrier.

Over-claiming the Gamow factor. Describing tunnelling qualitatively is expected; deriving the tunnelling probability needs university quantum mechanics and is out of scope for CBSE, NEET, IGCSE and IB-SL. Reserve any exponential-barrier formula for JEE-Advanced, A-Level or IB-HL enrichment, and never invent one under pressure.

Board scope. The transmutation rule ($Z-2$, $A-4$), the identity of the alpha as a helium nucleus, and the ideas of penetration and half-life are on every board including IGCSE 0625. But the quantitative Q-value — the $931.5\ \text{MeV}$ mass–energy conversion, the mass-defect arithmetic and the momentum-sharing split — is not on IGCSE (which keeps only the qualitative half) and is core on CBSE, ICSE, JEE, NEET, AP-2, IB (Theme E, SL and HL) and A-Level 9702. The nuclear-radius barrier estimate and Gamow tunnelling are higher-tier (JEE-Advanced, A-Level, IB-HL); do not present them as required for a NEET or IGCSE candidate.

Texts. NCERT Physics Class 12, Chapter 13 ("Nuclei"), the radioactivity and alpha-decay sections · H. C. Verma, Concepts of Physics Vol. 2, "The Nucleus" · Krane, Introductory Nuclear Physics, the alpha-decay chapter (for the Gamow theory at university level) · Halliday, Resnick and Walker, "Nuclear Physics".

Provenance. Atomic masses from standard tables; $1\ \text{u} = 931.5\ \text{MeV}$, $ke^2 = 1.44\ \text{MeV}\,\cdot\,\text{fm}$, $R_0 = 1.2\ \text{fm}$. Recompute: $^{238}\text{U}$ $\Delta m = 0.004584\ \text{u}$, $Q = 4.27\ \text{MeV}$, $KE_\alpha = Q\cdot234/238 = 4.20\ \text{MeV}$, recoil $= 0.072\ \text{MeV}$; alpha $v/c = 0.047$; $^{226}\text{Ra}$ $\Delta m = 0.005229\ \text{u}$, $Q = 4.87\ \text{MeV}$, $KE_\alpha = 4.78\ \text{MeV}$; $^{210}\text{Po}$ $\Delta m = 0.005806\ \text{u}$, $Q = 5.41\ \text{MeV}$, $KE_\alpha = 5.3\ \text{MeV}$, recoil $= 0.10\ \text{MeV}$, $v = 1.6\times10^{7}\ \mathrm{m\,s^{-1}}$; $^{4}\text{He}$ binding energy $= 28.3\ \text{MeV}$ ($7.07$ per nucleon); Coulomb barrier for the thorium daughter $R = 1.2(234^{1/3}+4^{1/3}) = 9.3\ \text{fm}$, $V_B = 28\ \text{MeV}$; half-life trend U-238 $4.5\times10^{9}\,\text{yr} >$ Ra-226 $1600\,\text{yr} >$ Po-210 $138\,\text{days}$ as $Q$ rises. All recomputed in verify_hs2-alpha-decay.py.

Prerequisites. Radioactivity: Alpha, Beta & Gamma — where the three radiations are introduced and compared · and Mass Defect & Nuclear Binding Energy, which supplies the $E=mc^2$ mass-ledger and the binding-energy idea this article turns into a Q-value. Leads to. Beta Decay & the Neutrino, whose continuous spectrum contrasts sharply with the alpha's single line, and the university seam Particle Decay Kinematics (go-deeper only), which treats two- and three-body decays relativistically and in full generality.