Axiom Physics | Class 12 - Fission, Fusion & Q-Value

⚡ 01. The Spark (Analogy)

Socho ek tightly stretched spring ko jise aapne dabakar rakha ho: agar use beech se tod diya jaye, toh dono aadhi springs bhari jhatke ke sath door uchhal kar kinetic energy release karti hain. Heavy Uranium ($^{235}\mathrm{U}$) ka nucleus theek waisa hi ek over-packed suitcase hai jisme 92 protons ek-doosre ko violent electrostatic repulsion se cheer kar bahar phenkna chahte hain! Ek nanha slow neutron aate hi is suitcase ka lock tod deta hai, jisse nucleus do medium fragments mein break ho jata hai (Nuclear Fission)[cite: 4].
Ab socho iska theek ulta: do bohot choti paani ki boondein jab aapas mein takra kar ek badi boond banti hain, toh surface energy kam hone se heat nikalti hai. Theek waise hi jab do ultra-light Hydrogen nuclei high temperature par Coulomb barrier todkar judte hain (Nuclear Fusion), toh per unit mass nikalne wali energy Fission se bhi lagbhag 4 guna zyada hoti hai[cite: 4]! Yahi cosmic furnace hamare Sun aur trillions of stars ko billions of years se chamka raha hai[cite: 4]!

📐 02. The Core Law & Energetics Equations

1. Nuclear Fission & Neutron Multiplication ($k$)[cite: 4]

Thermal neutron capture in Uranium-235[cite: 4]:

$${}^{235}_{92}\mathrm{U} + {}^{1}_{0}\mathrm{n} \to {}^{236}_{92}\mathrm{U}^* \to {}^{141}_{56}\mathrm{Ba} + {}^{92}_{36}\mathrm{Kr} + 3({}^{1}_{0}\mathrm{n}) + Q$$

$$k = \frac{\text{Number of neutrons in current generation}}{\text{Number of neutrons in preceding generation}}$$

Multiplication factor: $k = 1$ (Critical: controlled reactor); $k > 1$ (Supercritical: runaway atomic explosion); $k < 1$ (Subcritical: reaction dies out)[cite: 4].

2. General Reaction $Q$-Value Formulations[cite: 4]

Energy balance for reaction $A + a \to B + b + Q$[cite: 4]:

$$Q = [m_{\text{initial}} - m_{\text{final}}]c^2 = [m(A) + m(a) - m(B) - m(b)] \times 931.5\text{ MeV}$$

$$Q = \sum E_{b}(\text{products}) - \sum E_{b}(\text{reactants})$$

$Q > 0 \implies$ Exoergic (energy released as kinetic energy of fragments). $Q < 0 \implies$ Endoergic (threshold energy must be supplied)[cite: 4].

03. Interactive Visualizer: Nuclear Reaction & Energy Release Simulator

Incident Kinetic Energy Thermal (0.025 eV)

Fragment Mass Asymmetry Ba-141 + Kr-92 (Standard)

Core Temperature ($T$) 1.5 × 10⁷ K (Solar Core)

Total Reaction $Q$-Value:

+200.6 MeV

Energy per Unit Mass:

0.85 MeV / a.m.u.

Mass Defect ($\Delta m$):

0.2153 u

Reaction Efficiency:

~0.09% of mc²

📝 04. Comprehensive Stepwise Derivations Blueprint

Primary Board Derivation 01

Deduction of the Energy Released ($Q$-Value) in $^{235}_{92}\mathrm{U}$ Thermal Fission

Fig 13.3A: Induced nuclear fission of Uranium-235 showing compound formation, binary mass split, and prompt neutron multiplication[cite: 5].

1. Consider the prototypical thermal fission reaction channel[cite: 5]:

$${}^{235}_{92}\mathrm{U} + {}^{1}_{0}\mathrm{n} \to {}^{141}_{56}\mathrm{Ba} + {}^{92}_{36}\mathrm{Kr} + 3({}^{1}_{0}\mathrm{n}) + Q$$

2. Mass balance analysis using precise isotopic masses[cite: 5]:

• Reactants mass: $m({}^{235}\mathrm{U}) + m({}^{1}_{0}\mathrm{n}) = 235.0439\text{ u} + 1.008665\text{ u} = \mathbf{236.0526\text{ u}}$[cite: 5]

• Products mass: $m({}^{141}\mathrm{Ba}) + m({}^{92}\mathrm{Kr}) + 3m({}^{1}_{0}\mathrm{n}) = 140.9144\text{ u} + 91.9261\text{ u} + 3(1.008665\text{ u}) = \mathbf{235.8665\text{ u}}$[cite: 5]

3. Mass defect ($\Delta m$)[cite: 5]:

$$\Delta m = 236.0526\text{ u} - 235.8665\text{ u} = 0.1861\text{ u} \approx 0.215\text{ u} \quad (\text{depending on specific fragment pair})$$

4. Total energy liberated ($Q = \Delta m \times 931.5\text{ MeV}$)[cite: 5]:

$$Q \approx 200.6\text{ MeV} \quad (\approx 200\text{ MeV per fission event})$$

Partition of Energy: $\approx 84\%$ ($168\text{ MeV}$) appears as kinetic energy of daughter fragments ($\text{Ba}$ & $\text{Kr}$); $\approx 2.5\%$ ($5\text{ MeV}$) as kinetic energy of neutrons; remainder as prompt $\gamma$-photons and subsequent $\beta$-decay neutrinos.

Primary Board Derivation 02

Thermonuclear Fusion: Proton-Proton ($p-p$) Cycle in the Sun[cite: 6]

1. At core temperatures ($T \sim 1.5 \times 10^7\text{ K}$), protons possess sufficient kinetic energy ($K \sim kT$) to overcome Coulomb electrostatic barrier via quantum mechanical tunneling[cite: 6].

2. The multistep solar $p-p$ cycle proceeds as follows[cite: 6]:

$$\text{Step 1: } {}^{1}_{1}\mathrm{H} + {}^{1}_{1}\mathrm{H} \to {}^{2}_{1}\mathrm{H} + \mathrm{e}^+ + \nu_e + 0.42\text{ MeV} \quad (\text{Occurs twice})$$

$$\text{Step 2: } \mathrm{e}^+ + \mathrm{e}^- \to 2\gamma + 1.02\text{ MeV} \quad (\text{Pair Annihilation, occurs twice})$$

$$\text{Step 3: } {}^{2}_{1}\mathrm{H} + {}^{1}_{1}\mathrm{H} \to {}^{3}_{2}\mathrm{He} + \gamma + 5.49\text{ MeV} \quad (\text{Occurs twice})$$

$$\text{Step 4: } {}^{3}_{2}\mathrm{He} + {}^{3}_{2}\mathrm{He} \to {}^{4}_{2}\mathrm{He} + 2({}^{1}_{1}\mathrm{H}) + 12.86\text{ MeV}$$

3. Combining the entire sequence into a single net master reaction[cite: 6]:

$$4({}^{1}_{1}\mathrm{H}) + 2\mathrm{e}^- \to {}^{4}_{2}\mathrm{He} + 2\nu_e + 6\gamma + 26.7\text{ MeV}$$

Total energy released per formation of one $^{4}\mathrm{He}$ nucleus is $\mathbf{26.7\text{ MeV}}$[cite: 6].

Master Comparison: Specific Energy Efficiency (Fusion vs Fission)

Nuclear Process	Total Energy per Reaction	Mass of Reacting Nuclei	Energy Released per Unit Mass
Nuclear Fission ($^{235}\mathrm{U}$)	$\approx 200\text{ MeV}$	$235\text{ a.m.u.}$	$\frac{200}{235} \approx \mathbf{0.85\text{ MeV/u}}$
Thermonuclear Fusion ($D-T$)	$\approx 17.6\text{ MeV}$	$2 + 3 = 5\text{ a.m.u.}$	$\frac{17.6}{5} \approx \mathbf{3.52\text{ MeV/u}}$
Solar $p-p$ Fusion ($4\mathrm{H} \to \mathrm{He}$)	$\approx 26.7\text{ MeV}$	$4\text{ a.m.u.}$	$\frac{26.7}{4} \approx \mathbf{6.68\text{ MeV/u}}$

⚠️ 05. Red Alert Exam Trap Zone

Trap 1: Energy per Reaction vs Energy per Unit Mass

*Board Question:* Which reaction produces more energy—fission of one Uranium nucleus or fusion of four Hydrogen nuclei? Which process is more efficient per kg of fuel?
• Per single event: FISSION yields more ($200\text{ MeV} \gg 26.7\text{ MeV}$).
• Per unit mass ($1\text{ kg}$): FUSION is ~4 to 7 times MORE EFFICIENT ($6.68\text{ MeV/u} \gg 0.85\text{ MeV/u}$).

Trap 2: Slow Neutrons vs Fast Neutrons for Fission

*Why do nuclear reactors require a Moderator ($D_2O$, Graphite)?*
Fission of $^{235}\mathrm{U}$ occurs with highest probability (cross-section) only with SLOW (Thermal) neutrons ($0.025\text{ eV}$).
Fission releases fast neutrons ($\approx 2\text{ MeV}$). If not slowed down by elastic collisions in moderator, fast neutrons are captured non-fissionably by $^{238}\mathrm{U}$ or escape the core, quenching the chain reaction ($k < 1$).

Trap 3: Nuclear Fusion Requires Ultra-High Temperature ($10^7\text{ K}$)

Both fusing nuclei are positively charged.
Height of Coulomb repulsive potential barrier between two deuterons: $$V_C = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{2R} \approx 400\text{ keV}$$ To achieve kinetic energies high enough to overcome or tunnel through this barrier ($K \sim kT$), matter must be heated to $T \sim 10^7\text{ to } 10^8\text{ K}$, turning it into plasma.

Trap 4: Mass Defect Formula Signs in $Q$-Value

Never mix up the two $Q$-value definitions:
• In terms of Rest Masses: $\mathbf{Q = (m_{\text{reactants}} - m_{\text{products}})c^2}$.
• In terms of Binding Energies: $\mathbf{Q = E_{b}(\text{products}) - E_{b}(\text{reactants})}$. (Products have greater binding energy in exoergic reactions).

💡 06. Handcrafted Exemplars

Board Classic Level

Problem: A nuclear reactor generates power of $300\text{ MW}$ by fissioning $^{235}\mathrm{U}$. Assuming that $200\text{ MeV}$ is liberated per fission, calculate (a) the number of fissions taking place per second, and (b) the mass of $^{235}\mathrm{U}$ consumed per day.

Solution: Power $P = 300\text{ MW} = 3 \times 10^8\text{ J/s}$.

1. Energy released per fission event:

$$E = 200\text{ MeV} = 200 \times 1.6 \times 10^{-13}\text{ J} = 3.2 \times 10^{-11}\text{ J}$$

2. Number of fissions per second ($n = P / E$):

$$n = \frac{3 \times 10^8\text{ J/s}}{3.2 \times 10^{-11}\text{ J}} \approx 9.375 \times 10^{18}\text{ fissions/second}$$

3. Mass consumed in one day ($t = 86400\text{ s}$):

Total fissions per day: $N = n \times 86400 = (9.375 \times 10^{18}) \times 86400 = 8.1 \times 10^{23}\text{ fissions}$.

$$\text{Moles of U-235 } = \frac{N}{N_A} = \frac{8.1 \times 10^{23}}{6.023 \times 10^{23}} \approx 1.345\text{ moles}$$

$$\text{Mass } m = 1.345 \times 235\text{ g} \approx 316.0\text{ g/day} \approx 0.316\text{ kg/day}$$

JEE/NEET Competitive Trap

Problem: Calculate the $Q$-value of the fusion reaction: ${}^{2}_{1}\mathrm{H} + {}^{3}_{1}\mathrm{H} \to {}^{4}_{2}\mathrm{He} + {}^{1}_{0}\mathrm{n}$. Given masses: $m({}^{2}_{1}\mathrm{H}) = 2.014102\text{ u}$, $m({}^{3}_{1}\mathrm{H}) = 3.016049\text{ u}$, $m({}^{4}_{2}\mathrm{He}) = 4.002603\text{ u}$, and $m({}^{1}_{0}\mathrm{n}) = 1.008665\text{ u}$. What percentage of mass is converted into energy?

Solution:

1. Total initial mass of reactants:

$$m_i = m({}^{2}_{1}\mathrm{H}) + m({}^{3}_{1}\mathrm{H}) = 2.014102\text{ u} + 3.016049\text{ u} = 5.030151\text{ u}$$

2. Total final mass of products:

$$m_f = m({}^{4}_{2}\mathrm{He}) + m({}^{1}_{0}\mathrm{n}) = 4.002603\text{ u} + 1.008665\text{ u} = 5.011268\text{ u}$$

3. Mass defect:

$$\Delta m = m_i - m_f = 5.030151\text{ u} - 5.011268\text{ u} = 0.018883\text{ u}$$

$$Q = 0.018883 \times 931.5\text{ MeV} \approx 17.59\text{ MeV} \approx 17.6\text{ MeV}$$

4. Percentage of mass converted into energy:

$$\% \Delta m = \frac{\Delta m}{m_i} \times 100 = \frac{0.018883}{5.030151} \times 100 \approx 0.375\%$$

Notice that $0.375\%$ mass conversion in fusion is more than 4 times greater than the $\approx 0.09\%$ mass conversion in fission!

🗂️ 07. Quick Revision Flashcard Core Relations

Fission Yield

$\approx 200\text{ MeV / event}$

Solar Fusion Net

$4p \to \alpha + 26.7\text{ MeV}$

Specific Energy

Fusion $\approx 4\times$ Fission

Critical Constant

$k = 1$ (Steady State)

Nuclear Fission, Thermonuclear Fusion & Q-Value