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 …
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.Complete the Nuclear fission equation given below. n(0,1) + U(92,235) -> Sb(51,133) + ___ + 4 n(0,1)(a) Ba(56,144)(b) Kr(36,89)(c) Nb(41,99)(d) Sr(38,94)
›Reveal solutionSolution
In any nuclear reaction, total mass number (A) and total atomic number (Z) are separately conserved; applying this to the given fission equation identifies the missing fragment.
n(0,1) + U(92,235) -> Sb(51,133) + X + 4 n(0,1)
Mass number balance: 1 + 235 = 133 + A_X + 4(1)
236 = 133 + A_X + 4
A_X = 236 - 137 = 99
…
- GUJCET 2025Set 031 markMCQQ.Choose correct option to complete the net effect of fusion reaction occurs in the Sun. 411H+2e−→+2v+6γ+. (A) 23He, 5.49 MeV (B) 24He, 26.7 MeV (C) 24He, 22.86 MeV (D) 23He, 0.42 MeV
›Reveal solutionSolution
[!TLDR]
The overall p–p chain yields 24He plus about 26.7 MeV of energy.
Concept
In the Sun, hydrogen fuses via the proton–proton chain whose net result converts four hydrogen nuclei into one helium-4 nucleus, with the released energy carried off by positrons/neutrinos and gamma rays.
Solution
The net reaction is
411H+2e−→24He+2ν+6γ+26.7 MeV. …
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.Complete the nuclear fission equation given below. ¹₀n + ²³⁵₉₂U → ²³⁶₉₂U → ___ + ⁹⁴₃₈Sr + ___.(a) ¹⁴⁴₅₆Ba, 3(¹₀n)(b) ¹⁴⁰₅₄Xe, 2(¹₀n)(c) ¹³³₅₁Sb, 4(¹₀n)(d) ¹⁴⁰₅₄Xe, 3(¹₀n)
›Reveal solutionSolution
A nuclear fission equation must conserve total mass number (A) and total atomic number (Z) on both sides.
¹₀n + ²³⁵₉₂U → ²³⁶₉₂U → X + ⁹⁴₃₈Sr + y(¹₀n)
Charge balance: 92 = Z(X) + 38 + 0 ⟹ Z(X) = 54 (Xenon).
Mass balance with y = 2 neutrons: 236 = A(X) + 94 + 2 ⟹ A(X) = 140.
…
- GUJCET 2024Set 131 markMCQQ.Find the value of x and y from below given nuclear reaction 92235U+01n→x133Sb+41yNb+401n (A) (133, 41) (B) (51, 95) (C) (92, 1) (D) (51, 99)
›Reveal solutionSolution
Conserving charge gives x=51; conserving mass gives y=99, so (x,y)=(51,99).
Concept. In a nuclear reaction both the atomic number (charge) and mass number are conserved.
Steps. For 92235U+01n→x133Sb+41yNb+401n: …
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.Choosing the correct option, complete the given nuclear fusion reaction that occurs in the sun. ³₁H + ³₁H → ⁴₂He + ¹₁H + ¹₁H + ___.(a) 0.42 MeV(b) 1.02 MeV(c) 12.86 MeV(d) 5.49 MeV
›Reveal solutionSolution
The energy released in a nuclear reaction equals the mass defect (Δm = mass of reactants − mass of products) converted using E = Δm c², i.e. 931.5 MeV per atomic mass unit.
Reaction: ³₁H + ³₁H → ⁴₂He + ¹₁H + ¹₁H + Q
Masses (in u): m(³H) = 3.016049, m(⁴He) = 4.002603, m(¹H) = 1.007825.
Mass of reactants = 2 × 3.016049 = 6.032098 u
Mass of products (excluding Q) = 4.002603 + 2(1.007825) = 6.018253 u
…
- GUJCET 2021Set 151 markMCQQ.How many neutrons will produced for a given following nuclear fission reaction? 01n+92235U→92236U→56144Ba+3689Kr+(?)01n (A) 1 (B) 3 (C) 2 (D) 4
›Reveal solutionSolution
Conserve mass number to count the released neutrons.
Concept. Mass number is conserved across the fission. …
- GUJCET 2020Set 071 markMCQQ.In which process neutron is converted into proton? (A) β− decay (B) β+ decay (C) α-decay (D) γ decay
›Reveal solutionSolution
β− decay converts a neutron into a proton, emitting an electron and antineutrino.
Concept: In β− decay:
n→p+e−+νˉ. …
- GUJCET 2019Set 131 markMCQQ.For the following nuclear disintegration process 92238U→82206Pb+x[24He]+6[−10e] the value of x is........... (A) 10 (B) 4 (C) 6 (D) 8
›Reveal solutionSolution
Conserve mass number to find the number of α particles.
Concept first: Each α decay removes 4 mass units; β decay does not change mass number. …
- GUJCET 2015Set C1 markMCQQ.A radioactive element X disintegrates successively as under: Xβ−X1αX2β−X3αX4. If atomic number and atomic mass number of X are respectively 72 and 180, what are the corresponding values for X4? (A) 69, 172 (B) 69, 176 (C) 71, 176 (D) 70, 172
›Reveal solutionSolution
[!TLDR]
Applying β−,α,β−,α to 72X180 gives 70X4172 — option (D).
Concept
In β− decay the atomic number increases by 1 while the mass number is unchanged; in α decay the atomic number decreases by 2 and the mass number decreases by 4.
Solution
Start: Z=72,A=180.
- β−: Z=73,A=180. …
- GUJCET 2015Set C1 markMCQQ.If by successive disintegration of 92U238, the final product obtained is 82Pb206, then how many number of α and β particles are emitted? (A) 6 and 8 (B) 8 and 6 (C) 12 and 6 (D) 8 and 12
›Reveal solutionSolution
[!TLDR] 8 α (from the mass loss of 32) and 6 β− (to fix the atomic number) are emitted.
Concept
Each α reduces A by 4 and Z by 2; each β− leaves A unchanged and raises Z by 1. Use the total change in A to find the number of α, then use Z to find the number of β.
Solution
Mass number change: 238−206=32. Number of α=32/4=8.
Effect of 8 α on Z: 92−2×8=92−16=76. …
- GUJCET 2014Set A1 markMCQQ.In the radio active transformation ZXA⟶ Z+1X1A⟶ Z−1X2A−4⟶ Z−3X3A−8 Which are successively emitted radioactive radiations? (A) α,β−,β− (B) β−,α,β− (C) β−,α,α (D) α,β−,α
›Reveal solutionSolution
[!TLDR] The changes (+1,0),(−2,−4),(−2,−4) in (Z,A) correspond to β−, then α, then α.
Concept
In radioactive decay: α-emission lowers Z by 2 and A by 4; β−-emission raises Z by 1 with A unchanged.
Solution
- Step 1: ZXA→ Z+1X1A: ΔZ=+1, ΔA=0 → β−.
- Step 2: Z+1X1A→ Z−1X2A−4: ΔZ=−2, ΔA=−4 → α. …
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