Write nuclear reaction equations for (i) α-decay of ²²⁶₈₈Ra (ii) α-decay of ²⁴²₉₄Pu (iii) β⁻-decay of ³²₁₅P (i

Exercises · 13.16

Q. Write nuclear reaction equations for

(i)

\alpha

-decay of

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

(ii)

\alpha

-decay of

^{242}_{94}\text{Pu}

(iii)

\beta^{-}

-decay of

^{32}_{15}\text{P}

(iv)

\beta^{-}

-decay of

^{210}_{83}\text{Bi}

(v)

\beta^{+}

-decay of

^{11}_{6}\text{C}

(vi)

\beta^{+}

-decay of

^{97}_{43}\text{Tc}

(vii) Electron capture of

^{120}_{54}\text{Xe}

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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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Balance mass number $A$ and atomic number $Z$ on both sides for each decay type; the seven equations are listed below with the physics of each mode explained.

(i) $\alpha$ -decay of $^{226}_{88}\text{Ra}$

An alpha particle ( $^4_2\text{He}$ nucleus) is emitted, so $A$ drops by 4 and $Z$ by 2:

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

(ii) $\alpha$ -decay of $^{242}_{94}\text{Pu}$

^{242}_{94}\text{Pu} \rightarrow {}^{238}_{92}\text{U} + {}^{4}_{2}\text{He}

(iii) $\beta^{-}$ -decay of $^{32}_{15}\text{P}$

A neutron inside the nucleus converts to a proton, emitting an electron and an antineutrino; $A$ is unchanged, $Z$ increases by 1:

^{32}_{15}\text{P} \rightarrow {}^{32}_{16}\text{S} + e^{-} + \bar{\nu}

(iv) $\beta^{-}$ -decay of $^{210}_{83}\text{Bi}$

^{210}_{83}\text{Bi} \rightarrow {}^{210}_{84}\text{Po} + e^{-} + \bar{\nu}

(v) $\beta^{+}$ -decay of $^{11}_{6}\text{C}$

Here a proton converts to a neutron, emitting a positron and a neutrino; $Z$ decreases by 1:

^{11}_{6}\text{C} \rightarrow {}^{11}_{5}\text{B} + e^{+} + \nu

(vi) $\beta^{+}$ -decay of $^{97}_{43}\text{Tc}$

^{97}_{43}\text{Tc} \rightarrow {}^{97}_{42}\text{Mo} + e^{+} + \nu

(vii) Electron capture of $^{120}_{54}\text{Xe}$ …

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

KEAM 2026Set eng-2026-04214 marksMCQ
$^{236}_{94}X$
›Reveal solutionSolution

Two $\alpha$ and two $\beta$ decays: $a=228$ , $b=92$ .

Each $\alpha$ removes 4 mass units and 2 protons; each $\beta^-$ raises $Z$ by 1 with no mass change.

Mass: $236 - 2(4) = 228$ . …
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KEAM 2025Set eng-2025-04274 marksMCQ
$^{197}_{80}X \rightarrow\ ^{197}_{79}Y + Z + \nu$
›Reveal solutionSolution
$^{197}_{80}X \to\ ^{197}_{79}Y + Z + \nu$ : mass number unchanged, atomic number falls by 1, and a neutrino (not antineutrino) accompanies it — the signature of $\beta^{+}$ decay.

Balancing the reaction:
$^{197}_{80}X \rightarrow\ ^{197}_{79}Y + Z + \nu.$

Mass number: $197 = 197 + A_Z \implies A_Z = 0$ .

Atomic number: $80 = 79 + Z_Z \implies Z_Z = +1$ . …
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KEAM 2025Set pha-2025-0424F4 marksMCQ
$\alpha$
›Reveal solutionSolution

Alpha emission reduces the atomic number by 2 ( $Z\to Z-2$ ), so the daughter lies two places lower in the periodic table.

An $\alpha$ -particle is a helium nucleus $\left(^{4}_{2}\text{He}\right)$ . Its emission changes the parent nucleus as

$^{A}_{Z}X\rightarrow\ ^{A-4}_{Z-2}Y+\ ^{4}_{2}\text{He}.$ …
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KEAM 2025Set pha-2025-0429F4 marksMCQ
$^{22}_{11}Na \rightarrow\, ^{22}_{10}Ne + e^+ + X$
›Reveal solutionSolution

The process is positron ( $\beta^+$ ) decay: $^{22}_{11}\mathrm{Na}\to{}^{22}_{10}\mathrm{Ne}+e^+ + \nu$ . Conservation of lepton number requires the emitted particle $X$ to be a neutrino.

In $\beta^+$ decay a proton converts to a neutron, emitting a positron and a neutrino:

$p \to n + e^+ + \nu.$ …
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KEAM 2024Set eng-2024-06064 marksMCQ
$Q. In gamma emission, the nucleus emits (A) a photon (B) a neutron (C) a neutrino (D) an electron (E) a positron$
›Reveal solutionSolution

A $\gamma$ -transition emits a photon; the nucleon numbers $Z$ and $A$ are unchanged.

In gamma emission an excited nucleus drops to a lower energy state and releases the surplus energy as an electromagnetic quantum, i.e. a photon (gamma ray). No nucleon is added or removed, so $A$ and $Z$ stay the same. …
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KEAM 2024Set eng-2024-06094 marksMCQ
$_{Z}^{A}X \rightarrow _{Z+1}^{A}Y + e^{-} + q$
›Reveal solutionSolution

In $\beta^-$ (electron) emission a neutron converts to a proton, emitting an electron and an electron antineutrino.

The process shown, $^{A}_{Z}X \to\,^{A}_{Z+1}Y + e^- + q$ , is $\beta^-$ decay, at the nucleon level:

$n \to p + e^- + \bar{\nu}_e.$ …
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KEAM 2024Set pha-2024-06104 marksMCQ
$Q.In a nuclear decay, after the emission of one $\alpha$-particle and one $\beta$-particle (A) atomic number remains unchanged (B) mass number is reduced by 4 units (C) mass number is reduced by 8 units (D) mass number increases by 4 units (E) atomic number is increased by 2 units$
›Reveal solutionSolution
Track mass number A and atomic number Z through both emissions.

$\alpha$ -emission: $A \to A-4$ , $Z \to Z-2$ .

$\beta$ -emission: $A \to A$ (unchanged), $Z \to Z+1$ .

Net effect after one $\alpha$ and one $\beta$ :

Mass number: $A - 4$ (reduced by 4 units). …
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KEAM 2021Set eng-2021-P1-A14 marksMCQ
$\beta^-$
›Reveal solutionSolution
$\beta^-$ decay increases the proton number.

Concept and Intuition

In $\beta^-$ decay, $n\to p+e^-+\bar\nu_e$ . A neutron becomes a proton, so $Z$ rises by 1 and the neutron number falls by 1; the mass number $A$ stays the same.

Step-by-Step Solution

$^{A}_{Z}X\to{}^{A}_{Z+1}Y+e^-+\bar\nu_e$ .

Mass number $A$ unchanged, so options (A)/(E) are out.

Neutron number decreases, so (B) is out. …
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Q.Write nuclear reaction equations for

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