Radioactivity

On a November morning in 1896, Henri Becquerel found that a photographic plate wrapped in black paper had fogged after lying next to a uranium salt he had left in a dark drawer — with no sunlight anywhere near it. The uranium was doing something to that plate all by itself, in the dark, with no outside trigger. That accidental discovery opened a door nobody knew existed: some nuclei are simply unstable, and they announce it by spontaneously firing off radiation and turning into a different nucleus. Marie and Pierre Curie chased the trail further, grinding through tonnes of pitchblende ore to isolate two new radioactive elements, polonium and radium — work that cost Marie Curie two Nobel Prizes and, tragically, her health. This article stays close to what the current CBSE Class 12 syllabus actually asks of you for this topic: a qualitative picture of why a nucleus becomes unstable, what the three classic radiations (alpha, beta, gamma) look like when you test them with a magnetic field or a sheet of paper, and why the whole process is fundamentally different from any chemical reaction you have met before. What we add here, staying strictly inside the syllabus's other prescribed tool — the mass–energy relation $E = mc^2$ — is the one quantitative question a curious student always asks next: where does the energy in radioactive decay actually come from? The answer turns out to be one of the most elegant applications of Einstein's equation you will meet at this level, and it is worked out in full below with real atomic mass data for radium-226.

Key takeaways

Radioactivity is the spontaneous emission of radiation by an unstable nucleus, turning it into a different, more stable nucleus — discovered by Becquerel (1896) and studied further by the Curies.
Nuclei become unstable when they carry an unfavourable neutron-to-proton ratio, excess internal energy, or are simply too large for the short-range nuclear force to hold together against proton–proton repulsion.
Three classically distinguished radiations exist: alpha (positive, heavy, least penetrating, most ionising), beta (negative, light, medium penetration), and gamma (uncharged electromagnetic radiation, most penetrating, least ionising).
A magnetic or electric field sorts the three by deflection direction and amount — alpha and beta deflect oppositely, gamma does not deflect at all.
Radioactivity is unaffected by ordinary conditions like temperature, pressure, or chemical bonding, because it is a nuclear process, not a chemical or atomic-shell one.
The energy released in a decay comes from a measurable mass defect: the total mass before decay is very slightly more than after, and that missing mass reappears as kinetic energy via $E = mc^2$ .
Key idea: an unstable nucleus spontaneously emits alpha, beta, or gamma radiation to move toward a more stable configuration, and the kinetic energy carried away by the products equals the mass defect of the decay converted to energy through $E=mc^2$ .

The core insight

Radioactivity is the spontaneous, nucleus-level emission of alpha, beta, or gamma radiation by an unstable nucleus, with the released kinetic energy fixed by the mass defect of the decay through $E = mc^2$ .

Where it fits in the syllabus

Radioactivity sits inside the broader "Nuclei" chapter alongside nuclear composition, the nuclear force, and binding energy. Under the CBSE 2025–26 rationalised syllabus, the quantitative decay machinery that older textbooks attach to this topic — the decay law $N = N_0 e^{-\lambda t}$ , half-life, mean life, and activity — is not listed as examinable theory this cycle, so this article does not build on it or expect you to use it. What is squarely in scope, and what this article leans on, is the mass–energy relation $E = mc^2$ and the idea of mass defect, both explicitly prescribed for this chapter. So you should expect exam questions that test the qualitative picture of alpha/beta/gamma radiation and questions that use $E=mc^2$ together with mass-defect data to find the energy released in a decay — but not a decay-law numerical.

The physical picture

A nucleus is a tightly packed cluster of protons and neutrons held together by the nuclear force, an attraction so strong at short range that it easily overcomes the electrostatic repulsion between the positively charged protons. But that force falls off sharply beyond about $2$ – $3$ femtometres, so as a nucleus gets larger or its proton-to-neutron balance drifts away from the stable range, the short-range attraction can no longer fully cancel the longer-range electrostatic repulsion. The nucleus is then unstable, and it relieves that instability by spontaneously emitting radiation. Three distinct kinds of radiation were identified early on, simply by watching how they behaved crossing a magnetic or electric field and how much material it took to stop them.

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

where: $X$ is the parent nucleus (mass number $A$ , atomic number $Z$ ), $Y$ is the daughter nucleus, and the emitted alpha particle is simply a helium-4 nucleus — two protons and two neutrons bound together, carrying a charge of $+2e$ . Because the alpha particle is comparatively heavy and slow, it interacts strongly with matter along its short path, ionising intensely but travelling only a few centimetres in air before it is stopped by something as thin as a sheet of paper. Beta radiation, by contrast, is a stream of much lighter, faster, negatively charged particles (electrons, physically indistinguishable from atomic electrons but emitted from the nucleus itself during the nuclear transformation, not from the electron cloud around it). Being light and fast, beta particles penetrate further — through paper easily, but stopped by a few millimetres of aluminium — while ionising less per unit path length than alpha radiation does. Gamma radiation carries no charge at all: it is high-energy electromagnetic radiation, essentially a very energetic photon emitted as the daughter nucleus settles from an excited state to a lower one. Carrying no charge, it is undeflected by any magnetic or electric field, and being pure electromagnetic radiation it is the most penetrating of the three — attenuated only by dense, thick shielding such as several centimetres of lead — while being the least ionising per unit path length.

Figure

Three particle tracks entering from the left labelled alpha beta and gamma passing through zones marked paper aluminium sheet and thick lead block, with alpha stopped at the paper zone, beta stopped at the aluminium zone, and gamma passing through all three zones

A useful summary is that penetrating power rises in the order alpha, then beta, then gamma, while ionising power per unit path length falls in exactly that same order. The unifying idea beneath all three is that radioactivity is a spontaneous, random process happening inside an individual nucleus — it does not depend on temperature, pressure, or the chemical compound the atom sits in, because it is governed by the nuclear force and nuclear energy states, not by the chemistry of the electron cloud. Whenever a nucleus decays, the products fly apart carrying real kinetic energy — alpha particles from natural sources typically emerge with a few million electron-volts of energy, far more than any chemical reaction could ever supply. Einstein's mass–energy relation explains exactly where that energy comes from.

E = mc^2

where: $E$ is energy, $m$ is mass, and $c = 3\times10^{8}\ \text{m/s}$ is the speed of light in vacuum. Read physically, this equation says mass and energy are two faces of the same quantity — a small amount of mass is equivalent to an enormous amount of energy, because $c^2$ is such a large conversion factor. In a nuclear reaction, if you carefully add up the rest masses of everything going in and everything coming out, you find the totals do not quite match: the products are very slightly lighter than the parent. That missing mass — the mass defect $\Delta m$ — has not vanished; it has been converted into the kinetic energy carried away by the decay products, exactly as this equation prescribes.

Derivation AThe energy released in a nuclear decay, from the mass defect

The goal here is to show, using nothing more than the mass–energy relation $E=mc^2$ and conservation of mass–energy, that the kinetic energy released when an unstable nucleus decays is fixed entirely by comparing the rest mass of the parent nucleus to the combined rest mass of everything it decays into — and that this comparison, called the mass defect, can be computed from tabulated atomic masses without needing to know anything about the decay law, half-life, or the timing of the decay itself. This is the one quantitative handle the syllabus gives you on radioactivity, and it is exactly the same logic used for binding energy elsewhere in this chapter.

Step 1. Write the decay as a bookkeeping equation for mass. Take a general alpha decay, ${}^{A}_{Z}X \rightarrow {}^{A-4}_{Z-2}Y + {}^{4}_{2}\text{He}$ , and treat every particle as carrying a definite rest mass: $M_X$ for the parent, $M_Y$ for the daughter, and $M_{\alpha}$ for the emitted alpha particle. Before the decay, the total rest mass present is simply $M_X$ .

Step 2. Compare that to the total rest mass after the decay, which is $M_Y + M_{\alpha}$ . Careful mass measurements (by mass spectrometry) show these two totals are not equal — the parent is always the heavier one — so define the mass defect of the decay as $\Delta m = M_X - (M_Y + M_{\alpha})$ , a small positive number typically a few thousandths of an atomic mass unit.

Step 3. Invoke $E = mc^2$ as a statement of energy conservation across the reaction. Total energy (rest-mass energy plus kinetic energy) must be the same before and after. Since the parent nucleus is normally taken to be at rest, all of its energy is rest-mass energy, $M_X c^2$ . After the decay, the products carry both their own rest-mass energy $(M_Y + M_{\alpha})c^2$ and some kinetic energy $Q$ shared between them. Equating the two totals, $M_X c^2 = (M_Y + M_{\alpha})c^2 + Q$ .

Step 4. Solve for the released energy. Rearranging gives $Q = \left[M_X - (M_Y + M_{\alpha})\right]c^2 = \Delta m\, c^2$ . This $Q$ -value is the total kinetic energy shared between the daughter nucleus and the alpha particle, and it is the direct nuclear analogue of binding energy per nucleon: exactly as a nucleus releases energy while it is being assembled from separate nucleons, an unstable nucleus releases energy while it falls apart into a lighter, more tightly bound configuration. A positive $\Delta m$ is precisely the condition for a decay to be able to happen spontaneously at all — if the products were heavier than the parent, energy conservation would forbid the decay outright. Every quantity here — $M_X$ , $M_Y$ , $M_{\alpha}$ — is an ordinary tabulated atomic mass in atomic mass units $u$ , with $1\ u = 931.5\ \text{MeV}/c^2$ , so $Q$ in MeV follows directly from $Q = \Delta m\ (\text{in } u) \times 931.5\ \text{MeV}$ .

Worked examples

Board level - identifying radiation by deflection and stopping power

Question: An unknown radioactive source emits radiation that deflects strongly towards the negative plate of a parallel-plate electric field arrangement, and is completely stopped by a single sheet of paper. Identify the type of radiation and justify the reasoning with both clues.

Step 1. Interpret the deflection. Deflection towards the negative plate means the radiation is positively charged, since opposite charges attract. This immediately rules out gamma radiation (uncharged, so undeflected by any field) and beta radiation (negatively charged, which would deflect towards the positive plate instead, not the negative one).

Step 2. Check the remaining candidate. Alpha radiation is indeed positively charged (charge $+2e$ ), which is fully consistent with the observed deflection towards the negative plate.

Step 3. Confirm with the second, independent clue. Alpha radiation is the least penetrating of the three classic radiations and is characteristically stopped by something as thin as a single sheet of paper, exactly matching the second observation.

Answer: Both independent clues — positive deflection and being stopped by paper — point consistently to alpha radiation.

Competitive level - the energy released in the alpha decay of radium-226

Question: Radium-226 decays by alpha emission to radon-222: ${}^{226}_{88}\text{Ra} \rightarrow {}^{222}_{86}\text{Rn} + {}^{4}_{2}\text{He}$ . Given the atomic masses $M(\text{Ra-226}) = 226.025403\ u$ , $M(\text{Rn-222}) = 222.017570\ u$ , and $M(\text{He-4}) = 4.002603\ u$ , find the total kinetic energy released in this decay.

Step 1. Compute the mass defect using the derivation above: $\Delta m = M_X - (M_Y + M_{\alpha}) = 226.025403\ u - (222.017570\ u + 4.002603\ u)$ .

Step 2. Add the product masses first: $222.017570\ u + 4.002603\ u = 226.020173\ u$ .

Step 3. Subtract to find the mass defect: $\Delta m = 226.025403\ u - 226.020173\ u = 0.005230\ u$ . This is small compared to the total mass, exactly as expected — the vast majority of the parent's mass survives in the products, and only a sliver converts to kinetic energy.

Step 4. Convert the mass defect to energy using $1\ u = 931.5\ \text{MeV}/c^2$ : $Q = \Delta m\,c^2 = 0.005230 \times 931.5\ \text{MeV} \approx 4.87\ \text{MeV}$ .

Answer: The decay releases about $4.87\ \text{MeV}$ of kinetic energy, shared between the recoiling radon-222 nucleus and the alpha particle — consistent with the multi-MeV alpha energies measured experimentally for radium sources, and far beyond anything a chemical reaction (typically a few eV per reaction) could ever supply. Timing note: In an exam, do not waste time re-deriving $Q = \Delta m\,c^2$ from scratch — quote it directly, then spend your effort on careful subtraction of the given masses to enough decimal places (mass defects here are typically only in the fourth or fifth decimal place of $u$ , so truncating early gives a badly wrong answer) and on the standard conversion $1\ u = 931.5\ \text{MeV}$ , which is worth memorising cold.

Common traps

Assuming beta particles are electrons pulled from the atom's outer electron shell — they are created and emitted from inside the nucleus during the nuclear transformation itself, entirely separate from the atom's electron cloud.
Ranking penetrating power and ionising power in the same order — they run opposite to each other: alpha is least penetrating but most ionising per unit path length, gamma is most penetrating but least ionising.
Thinking radioactivity can be sped up, slowed down, or triggered by heating, cooling, or pressurising a sample, or by putting the atom into a different chemical compound — being a nuclear process, it is essentially unaffected by any of these ordinary external conditions.
Rounding tabulated atomic masses too early when computing a mass defect — because $\Delta m$ is the small difference of two large, nearly equal numbers, truncating to three or four decimal places instead of six can change the computed $Q$ -value by a large fraction of its true size.
Expecting an exam question on decay law, half-life, mean life, or activity for this topic — under the current CBSE syllabus, that quantitative machinery is not the examined content here; the exam-safe quantitative tool for radioactivity is $Q = \Delta m\,c^2$ using $E=mc^2$ , not the decay law.

Common trapMass defect and energy release apply to individual decay events, not to a decay rate — do not reach for a half-life or decay-constant formula on this topic

Every quantity used above ( $\Delta m$ , $Q$ , $E=mc^2$ ) describes a single decay event: how much energy one parent nucleus releases when it becomes one daughter nucleus plus its emitted particle. None of it says anything about how fast a sample of many such nuclei decays, or what fraction remain after a given time — that is the separate question of decay rate, and the formulas for it (decay law, half-life, activity) are explicitly not part of the current CBSE theory syllabus for this chapter. If a question gives you a mass defect or asks for energy released, reach for $Q=\Delta m\,c^2$ ; if a question tries to ask about "how much sample remains after so many years," recognise that this sits outside the syllabus as currently rationalised and should not be answered with a memorised decay-law formula you were not asked to learn.

You should be able to…

Explain qualitatively why some nuclei are stable and others are unstable, in terms of the nuclear force, proton–neutron balance, and nuclear size.
Identify alpha, beta, and gamma radiation from their charge, typical penetrating power, and typical ionising power, including how each behaves in a magnetic or electric field.
Explain why radioactivity is a nuclear phenomenon, unaffected by temperature, pressure, or the chemical environment of the atom.
State and use the mass-defect relation $Q = \Delta m\,c^2$ to find the energy released in a given nuclear decay from tabulated atomic masses.
Convert between atomic mass units and MeV using $1\ u = 931.5\ \text{MeV}/c^2$ with the required precision to avoid losing the small mass defect to rounding error.
Recognise that decay-law, half-life, and activity numericals are outside the current syllabus's examined scope for this topic, and not reach for those formulas on an exam.

Ask the tutor: want a plain-language walkthrough of how Becquerel's accidental discovery and the Curies' painstaking chemistry actually pieced together the existence of radioactivity, and how that story connects to the mass-defect calculation above? Ask the AI tutor to tell it.

If beta particles are electrons, and every atom already has electrons around it, how are these different?

The electrons in the atom's electron cloud stay bound to the nucleus by electrostatic attraction and are not created or destroyed during radioactivity. A beta particle is a completely separate electron, created inside the nucleus itself at the instant of decay (through a neutron transforming, in effect, into a proton plus this ejected electron) and immediately flung outward at high speed. So the atom's own electron cloud is untouched by beta decay happening inside its nucleus — the emitted electron is a brand-new particle from a nuclear process, not one borrowed from the existing shell.

Why does mass just disappear and turn into energy? Doesn't mass always have to be conserved?

In ordinary chemistry and everyday mechanics, mass conservation is such a good approximation that it feels like an absolute law, because the energies involved are tiny compared to $mc^2$ for any object. In nuclear reactions the energies involved are millions of times larger, so the tiny fractional mass change becomes measurable and important. Nothing actually disappears — total energy (rest-mass energy plus kinetic energy) is still exactly conserved, as the derivation above shows. It is only rest mass by itself, not total energy, that changes, because some of it has simply converted into kinetic energy of the products.

Why isn't half-life or decay rate part of this chapter if it feels like the most natural next question?

It is a completely reasonable question to ask, and older editions of this same NCERT chapter did cover it in detail. Under the current CBSE rationalisation for 2025–26, that quantitative decay-rate content (the exponential decay law, half-life, mean life, activity) has been removed from the examined theory for this chapter, likely to trim syllabus load. The qualitative picture of alpha, beta, and gamma radiation, plus the energetics via $E=mc^2$ , remain in scope and are what this article covers. If a future syllabus revision restores the decay law, that content would need to be added back at that point.

Is the 4.87 MeV in the worked example the energy of just the alpha particle, or something else?

It is the total kinetic energy shared between both products, the recoiling radon-222 nucleus and the emitted alpha particle, because momentum conservation (with the parent initially at rest) requires the two fragments to fly apart in opposite directions and share the released energy between them. In practice, because the alpha particle is so much lighter than the radon nucleus, it carries away the large majority of that total kinetic energy, while the heavy radon nucleus recoils slowly with only a small share — a detail beyond what this syllabus asks you to calculate, but useful for a physical sense of the numbers.