Q.Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes:
[!FORMULA]
88223Ra→82209Pb+614C
[!FORMULA]
88223Ra→86219Rn+24He
Calculate the Q-values for these decays and determine that both are energetically allowed.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Nuclear Reaction Balancing
Nuclear Reaction Balancing: The Intuition
Think of a nuclear reaction like a game of atomic Lego. You start with a certain set of blocks (the reactants), and after the reaction, you end up with a different set of blocks (the products). The fundamental rule is: you cannot lose or gain any Lego pieces. You can rearrange them, break some apart, or fuse them together, but the total number of each type of piece must stay the same.
In the atomic world, the "pieces" are:
- Protons (positive charge, found in the nucleus)
- Neutrons (neutral charge, also in the nucleus)
- Energy (which can appear or disappear, but that's a separate story)
The nucleus of an atom is made of protons and neutrons. When a nuclear reaction happens, the nuclei change. But the total number of protons and the total number of neutrons must be conserved — they cannot be created or destroyed.
This is different from chemical reactions, where atoms themselves are conserved. In nuclear reactions, atoms can change into different elements, but the nucleons (protons + neutrons) are conserved.
The Precise Statement
A nuclear reaction is balanced when two quantities are equal on both sides of the reaction arrow:
- Mass number (A) — the total number of nucleons (protons + neutrons). This is the superscript number.
- Atomic number (Z) — the total number of protons. This is the subscript number.
For any nuclear reaction:
Reactant1+Reactant2→Product1+Product2+…
The balancing conditions are:
∑Areactants=∑Aproducts
∑Zreactants=∑Zproducts
Nuclear Reaction Balancing Rules
Total mass number (A) on left=Total mass number (A) on right
Total atomic number (Z) on left=Total atomic number (Z) on right
How to Write a Nuclear Equation
Every nuclear particle is written as:
ZAX
Where:
- X = chemical symbol of the element
- A = mass number (top left)
- Z = atomic number (bottom left)
Common particles you'll encounter:
| Particle | Symbol | A | Z |
|---|---|---|---|
| Alpha particle | α or 24He | 4 | 2 |
| Beta particle | β− or −10e | 0 | -1 |
| Gamma ray | γ or 00γ | 0 | 0 |
| Neutron | n or 01n | 1 | 0 |
| Proton | p or 11p | 1 | 1 |
| Positron | β+ or +10e | 0 | +1 |
A common mistake: forgetting that beta particles have Z=−1 (for β−) or Z=+1 (for β+). This is because a neutron turns into a proton (or vice versa), and the beta particle carries away the "missing" charge.
Worked Example
Problem: Balance the following alpha decay reaction:
92238U→90234Th+?
Step 1: Identify what's missing. We have an unknown particle on the right.
Step 2: Balance mass numbers (A).
Left: A=238
Right: A=234+Aunknown
So 238=234+Aunknown⟹Aunknown=4
Step 3: Balance atomic numbers (Z).
Left: Z=92
Right: Z=90+Zunknown …
Why this formula?
Why Nuclear Reaction Balancing Works
Nuclear reaction balancing rests on a single, non-negotiable principle: conservation laws are absolute. In every nuclear reaction — whether natural decay, artificial transmutation, or fission/fusion — two quantities never change:
- Total mass number (A) — the sum of protons + neutrons
- Total atomic number (Z) — the sum of protons
These aren't arbitrary rules. They follow from deeper physics: baryon number conservation (protons and neutrons are baryons, and their total count is fixed) and charge conservation (electric charge cannot be created or destroyed). A nuclear reaction is just a rearrangement of nucleons; the number of nucleons stays constant, and the total charge stays constant.
For a reaction Z1A1X+Z2A2Y→Z3A3W+Z4A4Z:
A1+A2=A3+A4
Z1+Z2=Z3+Z4
The Reasoning Behind Each Conservation Law
Mass number conservation (A conserved):
A nucleon (proton or neutron) can change identity — a neutron can beta-decay into a proton, or a proton can capture an electron and become a neutron — but it cannot vanish or appear from nothing. The total count of nucleons before the reaction equals the total count after. This is why, for example, in alpha decay:
92238U→90234Th+24He
The left side has A=238; the right side has 234+4=238. The alpha particle carries away exactly 4 nucleons.
Atomic number conservation (Z conserved):
Charge is strictly conserved. The total positive charge (proton count) before equals the total after. In the same alpha decay, Z goes from 92 to 90+2=92. If charge weren't conserved, atoms would spontaneously change their chemical identity — which never happens in a closed system.
A common mistake is to think mass number conservation means mass is conserved. It does not. Mass-energy is conserved, but the rest mass can change (and usually does, releasing energy). The mass number A is a count of nucleons, not a measure of mass in kilograms.
How to Apply It: A Worked Example
Suppose you see: 92235U+01n→56141Ba+??Kr+301n
You know the total A on the left: 235+1=236.
On the right, you have 141+AKr+3(1)=144+AKr. …
Q for each decay is the mass difference between parent and products converted to energy; the real NCERT exercise does not restate these isotope masses in its own text (it expects the standard atomic-mass appendix table), so standard nuclear-data values are used here. …
Using standard atomic mass values for Ra-223, Pb-209, C-14 and Rn-219 (this exercise's real textbook printing does not restate these masses inline — it relies on the book's own Appendix mass table, unlike most other exercises in this chapter), Q≈31.8 MeV for carbon-14 emission and Q≈5.98 MeV for ordinary alpha emission — both positive, confirming both channels are allowed, though the far larger Q for alpha decay is why it dominates in practice.
Masses used (standard nuclear data, since not given in the exercise's own printed text):
m(223Ra)≈223.018502 u,m(209Pb)≈208.981091 u
m(14C)≈14.003242 u,m(219Rn)≈219.009480 u,m(4He)=4.002603 u (given)
Channel 1: 88223Ra→82209Pb+614C
Q1=[m(223Ra)−m(209Pb)−m(14C)]×931.5
=[223.018502−208.981091−14.003242]×931.5
=0.034169×931.5=31.83 MeV
Channel 2: 88223Ra→86219Rn+24He
Q2=[m(223Ra)−m(219Rn)−m(4He)]×931.5
=[223.018502−219.009480−4.002603]×931.5 …
- CBSE 2024Set ANNUAL1 markMCQQ.When ₃Li⁷ is bombarded by protons and the resultant nuclei are ₄Be⁸, the emitted particles are :(a) α-particle(b) β-particle(c) γ-photon(d) neutron
›Reveal solutionSolution
Mass number and atomic number already balance between reactants and product, so only energy (a gamma photon) is released — no material particle.
The reaction is:
37Li+ 11p→ 48Be+X
Check conservation of mass number (A) and atomic number (Z):
- Mass number: 7+1=8, and 48Be already has mass number 8 — balanced.
- Atomic number: 3+1=4, and 48Be already has atomic number 4 — balanced. …
- CBSE 2024Set ANNUAL1 markMCQQ.A nucleus of ₄Be⁹ absorbs an alpha particle and emits a neutron. The resulting nucleus will be -(a) ₆C¹²(b) ₄Be⁸(c) ₅C¹²(d) ₆C¹³
›Reveal solutionSolution
Conserve mass number and atomic number across the nuclear reaction 4Be9+2He4→X+0n1.
Mass number: 9+4=13=AX+1⇒AX=12
Atomic number: 4+2=6=ZX+0⇒ZX=6
…
- CBSE 2022Set I1 markMCQQ.Which of the following equations is correct? (A) 238/92 U → 234/90 U + 4/2 He (B) 238/92 U → 234/90 Th + 4/2 He (C) 238/92 U → 239/90 Th + 4/2 He (D) 238/92 U → 242/90 Th + 4/2 He
›Reveal solutionSolution
Correct α-decay: ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He.
In α-decay a nucleus emits a ⁴₂He nucleus, so the mass number decreases by 4 and the atomic number by 2. Applying conservation:
- Mass number: 238 = 234 + 4 ✓
- Atomic number: 92 = 90 + 2 ✓ …
- CBSE 2022Set ANNUAL1 markQ.Complete the following nuclear reactions:(a) 1532P→1632S+____+νˉ (½)(b) 1122Na→1022Ne+____+ν (½)
›Reveal solutionSolution
Both reactions are radioactive beta decays. Reaction (a) is β− decay (a neutron converts to a proton, emitting an electron and an antineutrino); reaction (b) is β+ decay (a proton converts to a neutron, emitting a positron and a neutrino).
(a) 1532P→1632S+____+νˉ
Mass number stays 32→32 (unchanged), but atomic number increases 15→16. An increase in Z by 1 with no change in A, accompanied by an antineutrino, is the signature of β− (electron) decay:
n→p+e−+νˉ
So:
1532P→1632S+−10e+νˉ
(b) 1122Na→1022Ne+____+ν
…
- CBSE 2022Set ANNUAL1 markMCQQ.Number and type of nucleons in the nucleus of Helium (2He4) will be:(a) 2 protons(b) 2 protons and 2 neutrons(c) 2 protons and 2 electrons(d) 2 neutrons
›Reveal solutionSolution
Z=2 gives 2 protons; A−Z=4−2=2 gives 2 neutrons: option (B).
Nucleons are the particles inside a nucleus — protons and neutrons (electrons are NOT nucleons).
For helium 2He4:
- Atomic number Z=2 → number of protons =2. …
- CBSE 2021Set ANNUAL1 markMCQQ.What is missing in the following nuclear reaction? ₁H² + ₁H² → ₂He³ + ______ .(a) meson(b) electron(c) positron(d) neutron
›Reveal solutionSolution
This is a deuterium-deuterium fusion reaction; balancing mass and atomic numbers shows a free neutron is released.
Check conservation of mass number (A) and atomic number (Z):
Left side: 1H2+1H2 → A=2+2=4, Z=1+1=2
Right side: 2He3 → A=3, Z=2 …
- CBSE 2020Set ANNUAL1 markMCQQ.When 7-N-14 nuclei are bombarded by neutrons and the resultant nuclei are 6-N-14, the emitted particles will be(a) Proton(b) Neutrino(c) Deuteron(d) Electron
›Reveal solutionSolution
Balancing mass number and atomic number across the nuclear reaction shows the emitted particle must be a proton.
The reaction is:
7N14 + 0n1 -> 6C14 + X
Conserve mass number (A): 14 + 1 = 14 + A_X, so A_X = 1.
Conserve atomic number (Z): 7 + 0 = 6 + Z_X, so Z_X = 1.
…
- CBSE 2019Set ANNUAL1 markMCQQ.A nuclear reaction is given as 2He^4 + ZX^A → (Z+2)Y^(A+3) + W. The particle W is(a) electron(b) proton(c) neutron(d) positron
›Reveal solutionSolution
Conserving mass number and charge, W has A = 1 and Z = 0 → a neutron: option (c).
In a nuclear reaction, both the mass number (superscript) and the atomic number/charge (subscript) must be conserved.
Reaction: ₂He⁴ + ₓX^A → ₍Z₊₂₎Y^(A+3) + W.
Mass number balance:
4 + A = (A + 3) + (mass number of W)
⇒ mass number of W = 4 + A − A − 3 = 1.
Charge (atomic number) balance:
2 + Z = (Z + 2) + (charge of W) …
- CBSE 2018Set ANNUAL1 markMCQQ.The atomic number and mass number for a specimen are Z and A respectively. The number of neutrons in the atom will be -(a) A(b) Z(c) A+Z(d) A-Z
›Reveal solutionSolution
Neutrons = mass number − atomic number = A − Z.
The atomic number Z is the number of protons. The mass number A counts protons plus neutrons (the nucleons):
…
- CBSE 2018Set ANNUAL1 markMCQQ.The quantities, which remain conserved in a nuclear reaction -(a) Total Charge(b) Angular momentum(c) Linear momentum(d) All the above
›Reveal solutionSolution
Nuclear reactions conserve charge, linear momentum and angular momentum (and energy, nucleon number).
The fundamental conservation laws hold in every nuclear reaction:
- Total electric charge is conserved.
- Total linear momentum is conserved.
- Total angular momentum is conserved. …
- CBSE 2018Set ANNUAL1 markQ.What will be the value of neutron multiplication factor for controlled chain reactions?
›Reveal solutionSolution
A controlled chain reaction requires the multiplication factor K = 1.
The neutron multiplication factor K is the ratio of the number of neutrons (or fissions) in one generation to that in the preceding generation:
K=neutrons in the preceding generationneutrons produced in a generation
- If K>1, the number of fissions grows each generation → the reaction is uncontrolled (as in a bomb).
- If K<1, the reaction dies out. …
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