In alpha decay, why does the unstable nucleus emit a $^{4}_{2}\text{He}$ nucleus? Why does it not emit four separate nucleons?

In alpha decay, why does the unstable nucleus emit a ⁴₂He nucleus? Why does it not emit four separate nucleons

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^{4}_{2}\text{He}

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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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Step 1. The disintegration energy released in any decay is $Q=[m_{\text{parent}}-m_{\text{products}}]c^2$ ; for spontaneous decay, $Q$ must be positive.

Step 2. Because the $^{4}_{2}He$ nucleus is itself strongly bound (has a large binding energy), its mass is significantly less than the combined mass of two free protons and two free neutrons - so emitting it as one bound unit gives a positive $Q$ .

Step 3. If the same parent nucleus instead tried to emit four completely separate, unbound nucleons, the total mass of the products would work out to be greater than the parent's mass, giving a negative $Q$ . …

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

Showing the 12 most recent of 26 on this concept.

CBSE 2026Set A1 markMCQ
$Q. Radioactivity was discovered by (A) Marie Curie (B) Pierre Curie (C) Ernest Rutherford (D) Henri Becquerel$
›Reveal solutionSolution

Radioactivity was discovered by Henri Becquerel in 1896.

In 1896 Henri Becquerel found that uranium salts spontaneously emitted a penetrating radiation that fogged photographic plates even in the dark, without any external excitation. This spontaneous emission was named radioactivity. Marie and Pierre Curie later extended the work by isolating polonium and radium, and Rut …

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CBSE 2026Set ANNUAL1 mark
$Q. Write True or False: In γ-decay low energy photons are emitted.$
›Reveal solutionSolution

False — γ-decay emits HIGH energy photons.

After an alpha or beta decay, a daughter nucleus is often left in an excited (higher energy) state. It reaches the ground state by emitting the excess energy as an electromagnetic photon — a gamma ray. Because the energy differences between nuclear levels are very large (of the order of MeV), these gamma photons are of very high ener …

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CBSE 2025Set D1 markMCQ
$Q. Which one of the following is not a type of radioactive decay? (A) Alpha decay (B) Beta decay (C) Gamma decay (D) Muon decay$
›Reveal solutionSolution
The three radioactive decays of a nucleus are alpha, beta and gamma; muon decay is not one of them.

Radioactivity is the spontaneous disintegration of an unstable nucleus. It occurs by three modes:

Alpha (α) decay — emission of a helium nucleus (⁴₂He).

Beta (β) decay — emission of an electron (β⁻) or positron (β⁺) with a neutrino/antineutrino. …
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CBSE 2025Set D1 markMCQ
$Q. Which of the following is chargeless particle? (A) α-particle (B) β-particle (C) Proton (D) Photon$
›Reveal solutionSolution
Among the listed particles only the photon is chargeless.

Checking each:

α-particle — a helium nucleus, charge +2e.

β-particle — an electron, charge −e.

Proton — charge +e. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following particle has mass equal to that of electron ?$
(a) neutron
(b) proton
(c) positron
(d) neutrino
›Reveal solutionSolution

The positron is the antiparticle of the electron: identical mass, opposite (positive) charge.

A positron ( $e^+$ , emitted for instance in $\beta^+$ decay of a proton-rich nucleus) has exactly the same rest mass as an electron ( $m_e \approx 9.11\times10^{-31}$ kg), but carries a charge of $+e$ instead of $-e$ . Neutrons and protons are both fa …

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CBSE 2024Set A11 mark
$Q.Alpha particle is a ———————— nucleus. Fill in the blank choosing the appropriate answer from the bracket: (decreasing, interference, helium, greater, diffraction, increasing)$
›Reveal solutionSolution

helium nucleus. …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. If a radioactive atom emits a beta-particle then its atomic mass$
(a) increases by one unit
(b) remains unchanged
(c) decreases by two units
(d) increases by two units
›Reveal solutionSolution

Beta-minus decay converts a neutron into a proton inside the nucleus, so the total nucleon count (mass number) is unchanged even though the atomic number increases by one.

In beta-minus ( $\beta^-$ ) decay, a neutron in the nucleus transforms into a proton, emitting an electron (the beta particle) and an antineutrino:

$n \rightarrow p + e^- + \bar{\nu}$

…

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CBSE 2023Set F1 markMCQ
$Q. Which of the following has the highest penetrating power? (A) alpha-rays (B) beta-rays (C) gamma-rays (D) Cathode rays$
›Reveal solutionSolution
γ-rays penetrate most; α least.

Penetrating power increases as ionising power decreases:

α-rays (helium nuclei): heavily ionising, stopped by paper/skin — least penetrating.

β-rays (electrons): moderate, stopped by a few mm of aluminium. …
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CBSE 2023Set TERM21 markMCQ
$Q. Study the following paragraph and answer questions number 8 and 9 based on it: The Radioactive Nuclei are unstable and emit alpha (2He^4), beta (-1e^0) and gamma (0γ^0) radiations to achieve states of greater stability. Ruther-ford and Soddy suggested the following rules governing the Radioactive decay.$
(i) The algebraic sum of atomic numbers before and after disintegration must be same.
(ii) The sum of mass numbers before and after disintegration must also be same. Choose the correct product of nuclear reaction: 90Th234 → 91Pa234 + ............
(a) α (Alpha)
(b) β (Beta)
(c) γ (Gamma)
(d) None of these.
›Reveal solutionSolution

Balancing mass number and atomic number in ${}_{90}\text{Th}^{234} \to {}_{91}\text{Pa}^{234} + X$ shows $X$ must be a $\beta$ -particle (electron).

By the Soddy–Fajans conservation rules given: mass number (A) is conserved and atomic number (Z) is conserved across the reaction.

Mass number: $234 = 234 + A_X \Rightarrow A_X = 0$ .

Atomic number: $90 = 91 + Z_X \Rightarrow Z_X = -1$ .

…

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CBSE 2023Set TERM21 markMCQ
$Q. Based on the radioactive-decay passage above (Q.8): In alpha (α) decay, A is mass number and Z is atomic number.$
(a) A decreases by 4 and Z increases by 2
(b) A decreases by 4 and Z decreases by 2
(c) A increases by 4 and Z decreases by 2
(d) A increases by 4 and Z increases by 2.
›Reveal solutionSolution

An alpha particle is a helium nucleus, ${}_2\text{He}^4$ , so emitting one removes 4 from the mass number and 2 from the atomic number.

Alpha decay is the emission of an alpha particle, ${}_2\text{He}^4$ , which itself has mass number 4 and atomic number (charge number) 2. By conservation of A and Z (as stated in the passage):

${}_Z X^A \to {}_{Z-2}Y^{A-4} + {}_2\text{He}^4$ …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. In a given reaction : zX^A \to (z+1)Y^A \to (z-1)K^(A-4) \to (z-1)K^(A-4), the radioactive radiations are emitted in the sequence of -$
(a) α, β, γ
(b) γ, α, β
(c) β, α, γ
(d) γ, β, α
›Reveal solutionSolution
Track how Z and A change at each step of the decay chain: an increase in Z with same A is β⁻ decay; a drop of 2 in Z with a drop of 4 in A is α decay; no change in Z or A but a distinct final state is γ emission.

Why / steps:

$_{Z}X^{A}\to {}_{Z+1}Y^{A}$ : mass number unchanged, atomic number increases by 1 → this is β⁻ decay (a neutron converts to a proton, emitting an electron and an antineutrino).

$_{Z+1}Y^{A}\to {}_{Z-1}K^{A-4}$ : atomic number drops by 2, mass number drops by 4 → this is α decay (loss of a helium-4 nucleus). …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. During alpha-decay, mass number ________ by ________.$
(a) decreases, 2
(b) increases, 2
(c) decreases, 4
(d) increases, 4
›Reveal solutionSolution

In alpha decay the mass number decreases by 4.

An alpha particle is a helium nucleus, 2 protons and 2 neutrons (mass number 4, charge +2). When a nucleus emits it, the parent loses 4 nucleons and 2 protons: mass number A decreases by 4 and atomic number Z decreases by 2. For examp …

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Q.In alpha decay, why does the unstable nucleus emit a 24He^{4}_{2}\text{He}24​He nucleus? Why does it not emit four separate nucleons?

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Q.In alpha decay, why does the unstable nucleus emit a $^{4}_{2}\text{He}$ nucleus? Why does it not emit four separate nucleons?

Unlock everything free for 14 days