Fusion — D-T reaction, solar fusion (p-p chain), tokamak - ICF

5.2.7 · Chemistry › Nuclear & Radiochemistry

Fusion energy release KYUN karta hai?

Curve KYUN hota hai? $B / A$ (binding energy per nucleon) ko mass number $A$ ke against plot karo. Yeh halke nuclei ke liye teji se baḍhta hai, $A \approx 56$ (iron/nickel, $B / A \approx 8.8$ MeV) ke paas peak karta hai, phir dheere dheere girta hai.

Peak ki taraf left se move karna (halke nuclei fuse karna) → energy bahar → fusion.
Peak ki taraf right se move karna (bhaari nuclei toḍna) → energy bahar → fission.

Figure — Fusion — D-T reaction, solar fusion (p-p chain), tokamak - ICF

Released energy ko scratch se derive karna (Q-value)

Worked Example 1 — D–T reaction

$_{1}^{2} H +_{1}^{3} H \to_{2}^{4} He +_{0}^{1} n$

Masses (atomic mass units, $1 u = 931.5$ MeV/ $c^{2}$ ):

$m (^{2} H) = 2.014102 u$
$m (^{3} H) = 3.016049 u$
$m (^{4} He) = 4.002602 u$
$m (n) = 1.008665 u$

Step 1 — reactants ka sum. $2.014102 + 3.016049 = 5.030151 u$ . Yeh step kyun? Q sirf initial aur final total mass par depend karta hai.

Step 2 — products ka sum. $4.002602 + 1.008665 = 5.011267 u$ .

Step 3 — mass defect. $Δ m = 5.030151 - 5.011267 = 0.018884 u$ . Yeh step kyun? Jo mass "gayab" hua woh energy ban jaata hai.

Step 4 — convert karo. $Q = 0.018884 \times 931.5 \approx 17.6 MeV$ .

Step 5 — energy split karo. Momentum conservation: halka neutron zyada KE le jaata hai. $E_{n} : E_{α} = m_{α} : m_{n} \approx 4 : 1 \Rightarrow E_{n} \approx 14.1 MeV, E_{α} \approx 3.5 MeV$ Yeh step kyun? Rest ke paas shuru hone wali reaction mein, $p_{n} = p_{α}$ , isliye $E = p^{2} /2 m$ se KE mass ke inversely proportional hoti hai. 14 MeV neutron woh hai jo tritium breed karta hai aur blanket heat karta hai; 3.5 MeV $α$ plasma ko hot rakhne ke liye ruk jaata hai.

Solar fusion — proton–proton (p–p) chain

Sun D–T reaction ke liye bahut thanda hai; use plain protons fuse karne paḍte hain. Net result: $4_{1}^{1} H \to_{2}^{4} He + 2 e^{+} + 2 ν_{e} + energy (\approx 26.7 MeV)$

Worked Example 2 — Net solar Q-value

$Q = [4 m (^{1} H) - m (^{4} He)] c^{2} - (positron annihilation accounting)$ $4 (1.007825) - 4.002602 = 0.028698 u$ use karte hue, $Q \approx 0.028698 \times 931.5 \approx 26.7$ MeV. Yeh step kyun? Neutrinos escape karke ~2% le jaate hain; baaki (~26.7 MeV har helium per) woh hai jo hum sunlight ke roop mein feel karte hain.

Earth par fusion KAISE karte hain? The Coulomb barrier

Do engineering routes

Common mistake Common errors ko steel-man karna

"Fusion hoti hai kyunki nuclei ek doosre ko attract karte hain, isliye yeh aasaan hai." Kyun sahi lagta hai: strong force short range par attractive HAI, aur fusion energy release karti hai. Fix: strong force ke grab karne se pehle, positive charges repel karte hain (Coulomb barrier). Paas aane ke liye hi bahut bada KE/temperature chahiye — yahi mushkil hissa hai. Energy-favourable ≠ shuru karna aasaan.

"Iron bhi fusion se energy release kar sakta hai." Kyun sahi lagta hai: fusion lagta hai ki hamesha energy deta hai. Fix: $Fe$ $B / A$ ke peak par baitha hai. Iron se aage fuse karne mein energy lagti hai. Fusion sirf tab energy release karta hai jab $A$ peak se neeche ho.

"Sun D–T reaction chalaata hai." Kyun sahi lagta hai: D–T famous "fusion reaction" hai. Fix: Sun ordinary protons se shuru karta hai (p–p chain). D–T hamari Earth-lab choice hai kyunki achievable temperatures par iska sabse bada cross-section hota hai.

"D–T se saara 17.6 MeV useful charged-particle energy ke roop mein capture ho sakta hai." Kyun sahi lagta hai: total Q 17.6 MeV hai. Fix: ~14.1 MeV ek neutral neutron ke roop mein nikalta hai (magnetic confinement nahi hota) — use blanket mein slow karna padta hai; sirf 3.5 MeV $α$ plasma ko directly heat karta hai.

Recall Feynman: 12-saal ke bacche ko samjhao

Socho chhoti balls jo ek doosre se chipakna chahti hain aur jab chipakti hain, ek energy ka dhakka deti hain — jaise do strong magnets snap karke bund ho jaayein. Dikkat yeh hai: balls mein ek force-field bhi hai jo unhe alag dhakelta hai, isliye unhe ek doosre par pagalon ki tarah fast fenkna padta hai (super hot) "no entry" zone ko cross karne ke liye. Sun yeh hydrogen ke saath karta hai kyunki woh bahut bada aur patient hai. Earth par hum ya to ek magnetic doughnut banate hain hot gas pakadne ke liye (tokamak) ya ek fuel pellet ko lasers se crush karte hain (ICF) balls ko ek instant ke liye saath force karne ke liye.

Flashcards

Binding-energy-per-nucleon curve ki kaunsi property fusion ko exothermic banati hai?

B / A

A \approx 56

par peak ki taraf baḍhta hai; halke nuclei fuse hoke binding mein "upar" jaate hain, fark energy ke roop mein release hota hai.

D–T fusion reaction likho.

_{1}^{2} H +_{1}^{3} H \to_{2}^{4} He +_{0}^{1} n

, Q ≈ 17.6 MeV.

D–T ka 17.6 MeV products mein kaise split hota hai?

~14.1 MeV neutron ko, 3.5 MeV $α$ ko (KE mass ke inversely proportional kyunki momenta equal hain). Neutron D–T energy ka zyada hissa kyun le jaata hai? ::: Momentum conservation: $p_{n} = p_{α}$ aur $E = p^{2} /2 m$ , isliye halka neutron KE $m_{α} : m_{n} \approx 4 : 1$ ratio mein paata hai. Solar fusion ki net reaction? ::: $4^{1} H \to^{4} He + 2 e^{+} + 2 ν_{e} + \approx 26.7 MeV$ MeV. p–p chain mein bottleneck kaunsa step hai aur kyun? ::: $p + p \to^{2} H + e^{+} + ν$ — isme weak-interaction β-conversion chahiye, extremely slow, Sun ko lambi zindagi deta hai. Coulomb barrier kya hai aur kyun matter karta hai? ::: $U_{C} = \frac{Z _{1} Z _{2} e ^{2}}{4 π ε _{0} r _{0}}$ ; nuclei ko strong force ke bind karne se pehle yeh electrostatic repulsion overcome karni padti hai (high T + tunnelling ke zariye). Lawson/triple-product criterion batao. ::: $n T τ_{E}$ ek threshold (

3 \times 1 0^{21}

keV·s·m⁻³ for D–T) se zyada hona chahiye net energy gain ke liye.

Tokamak vs ICF strategy ek phrase mein?

Tokamak = low density, long confinement (magnetic doughnut); ICF = ultra-high density, ultra-short confinement (laser-imploded pellet).

Earth par p–p ki jagah D–T kyun prefer kiya jaata hai?

D–T ka achievable temperatures par sabse bada cross-section hai; p–p ka weak step reactor ke liye bahut slow hai.

Mass defect define karo.

Δ m = (Z m_{p} + N m_{n}) - m_{nucleus}

; jo mass gum hua woh

B = Δ m c^{2}

ke zariye binding energy ke barabar hai.

Connections

Binding Energy per Nucleon Curve
Nuclear Fission (usi $B / A$ peak ka doosra pehlu)
Mass-Energy Equivalence E=mc^2
Beta Decay & the Weak Interaction (p–p step 1 drive karta hai)
Maxwell–Boltzmann Distribution (high-energy tail fusion enable karta hai)
Quantum Tunnelling (sub-barrier fusion)
Plasma Physics & Magnetic Confinement
Stellar Nucleosynthesis