Fusion — solar fusion, tokamak (concept): Worked examples

2.3.24 · D3 · Physics › Modern Physics › Fusion — solar fusion, tokamak (concept)

Yeh page parent Fusion note ke liye ek drill deck hai. Hum yahan theory re-derive nahi karenge — balki har tarah ke number dhundhenge jo ek fusion problem de sakta hai, aur unme se ek-ek solve karenge. Agar koi symbol unfamiliar lage, parent par wapas jaao, ya Mass-Energy Equivalence (E=mc^2) aur Binding Energy per Nucleon Curve dekho.

Scenario matrix

#	Cell (case class)	Kya cheez isse alag banati hai	Worked in
A	Standard exothermic fusion	$Δ m > 0$ , energy released	Ex. 1
B	Sign flip — endothermic	$Δ m < 0$ , energy supply karni padti hai	Ex. 2
C	Degenerate / zero case	$Δ m = 0$ , koi net energy nahi	Ex. 3
D	Binding energies use karna masses ke bajaye	$Q$ $B / A$ curve se, mass tables se nahi	Ex. 4
E	Per-nucleon / scaling	energy per kilogram fuel, "kitna coal?"	Ex. 5
F	Limiting behaviour	Coulomb barrier vs. temperature — order-of-magnitude	Ex. 6
G	Confinement (Lawson)	triple product $n T τ_{E}$ , kya plasma ignite karega?	Ex. 7
H	Real-world word problem	Sun ki lifetime mass budget se	Ex. 8
I	Exam-style twist	fraction of mass converted; unit trap	Ex. 9

Teen tools jinpar hum rely karte hain:

$Q$ ka sign hi poori kahani hai: $Q > 0$ ⇒ energy released (fusion kaam karta hai); $Q < 0$ ⇒ energy absorbed (spontaneously nahi hoga).

Alt-text: MeV mein Q-value ke liye ek horizontal number line. Origin par likha hai "Q = 0 (degenerate, Cell C)". Right half green shade mein hai aur label hai "energy RELEASED (fusion works)" ek green dot ke saath +17.6 MeV par jo Cells A/D/E se tagged hai. Left half red shade mein hai aur label hai "energy ABSORBED" ek red dot ke saath −0.09 MeV ke paas jo Cell B se tagged hai.

Example 1 — Standard exothermic fusion (Cell A)

Worked example D–T reaction ka

$Q$ -value $_{1}^{2} D +_{1}^{3} T \to_{2}^{4} He + n$ . Masses (u): $m_{D} = 2.014102$ , $m_{T} = 3.016049$ , $m_{H e} = 4.002603$ , $m_{n} = 1.008665$ . $Q$ nikalo.

Forecast: Products mein ek bada tightly-bound helium aur ek neutron hai — reactants mein do loose light nuclei hain. Guess: mass girti hai, toh $Q$ ek healthy positive number hona chahiye, tens of MeV.

Step 1 — Reactant masses add karo. $2.014102 + 3.016049 = 5.030151$ u. Yeh step kyun? Rest mass mein rest energy store hoti hai; inputs ka total "energy account" is sum mein hai.

Step 2 — Product masses add karo. $4.002603 + 1.008665 = 5.011268$ u. Kyun? Jo bahar aata hai uski bhi same accounting.

Step 3 — Mass defect. $Δ m = 5.030151 - 5.011268 = 0.018883$ u. Is tarah subtract kyun? Missing mass hi woh hai jo products ki kinetic energy mein convert hui.

Step 4 — Energy mein convert karo. $Q = 0.018883 \times 931.5 = 17.59 MeV$ . $\times 931.5$ kyun? Woh $1 u \cdot c^{2}$ MeV mein express hai — yeh hamare liye $\times c^{2}$ kar deta hai.

Verify: $Q > 0$ ✓ (forecast se match). Units: (u) $\times$ (MeV/u) = MeV ✓. Yeh ITER fuel ka famous $\approx 17.6$ MeV hai.

Example 2 — Sign flip: ek endothermic "fusion" (Cell B)

Worked example Jab join karna

energy kharach kare Socho $_{2}^{4} He +_{2}^{4} He \to_{4}^{8} Be$ . Masses (u): $m_{H e} = 4.002603$ , $m_{B e - 8} = 8.005305$ . $Q$ nikalo aur uske sign ki interpretation karo.

Forecast: Beryllium-8 famously unstable hai — do heliums se thoda upar energy mein. Guess: mass actually badh sakti hai, deta $Q < 0$ . Agar aisa hai, toh yeh reaction energy release nahi karta.

Step 1 — Reactant mass. $4.002603 \times 2 = 8.005206$ u. Kyun? Do identical helium nuclei.

Step 2 — Product mass. $8.005305$ u (given). Kyun? Single beryllium-8 nucleus.

Step 3 — Mass defect. $Δ m = 8.005206 - 8.005305 = - 0.000099$ u. Negative sign kyun rakho? Ek negative defect ka matlab hai product bhaari hai — mass daalna pada, toh energy absorb hui.

Step 4 — Convert. $Q = - 0.000099 \times 931.5 = - 0.092 MeV \approx - 92 keV$ .

Verify: $Q < 0$ ✓. Interpretation: $^{8} Be$ almost turant do heliums mein wapas toot jaata hai. Yahi reason hai ki Sun do-at-a-time helium fuse nahi kar sakta — use triple-alpha trick chahiye. Units: keV ✓, aur $∣ Q ∣$ chhota ✓ (barely bound).

Example 3 — Degenerate zero case (Cell C)

Worked example Agar

$Δ m = 0$ ho toh? Ek hypothetical rearrangement imagine karo jahan total product mass exactly total reactant mass ke barabar ho, e.g. $m_{in} = 6.000000$ u, $m_{out} = 6.000000$ u. $Q$ nikalo.

Forecast: Koi mass change nahi hoti ⇒ kuch convert nahi hota ⇒ $Q$ exactly zero hona chahiye. Yeh fusion aur non-reaction ke beech ki boundary hai.

Step 1 — Mass defect. $Δ m = 6.000000 - 6.000000 = 0$ u. Yeh note kyun karo? Yeh hamein upar figure ki number line ke origin par exactly bithata hai.

Step 2 — Convert. $Q = 0 \times 931.5 = 0 MeV$ . Kyun yeh matter karta hai: Na energy released hoti hai na absorbed. Binding Energy per Nucleon Curve par yeh iron ke paas flat top hai — wahan nuclei combine karne se kuch nahi milta. Yahi exact reason hai ki fusion iron par ruk jaata hai: uske baad, $Δ m$ negative ho jaata hai (fission territory).

Verify: $Q = 0$ ✓. Yeh degenerate point woh physical "peak" hai jo cell A ( $Q > 0$ , light nuclei) ko cell B ( $Q < 0$ , iron se upar) se alag karta hai.

Example 4 — Binding energies se $Q$ , masses se nahi (Cell D)

Worked example Same answer, alag toolbox

$^{2} D +^{3} T \to^{4} He + n$ ke liye, masses ki jagah binding energies use karo: $B (D) = 2.224$ MeV, $B (T) = 8.482$ MeV, $B (^{4} He) = 28.296$ MeV, $B (n) = 0$ . $Q$ nikalo.

Forecast: Binding energy hai "woh energy jo pehle hi release hui jab nucleus bana." Product kaafi zyada bound hai, toh $Q =$ (product binding) $-$ (reactant binding) Example 1 ki tarah $\approx 17.6$ MeV ke paas hona chahiye.

Step 1 — Products ki total binding. $28.296 + 0 = 28.296$ MeV. Kyun? Ek free neutron ke koi partners nahi, toh $B (n) = 0$ .

Step 2 — Reactants ki total binding. $2.224 + 8.482 = 10.706$ MeV. Kyun? D aur T mein pehle se stored binding sum karo.

Step 3 — Subtract. $Q = 28.296 - 10.706 = 17.59 MeV$ . Yahan products minus reactants kyun (mass formula se ulta order)? Zyada binding = kam mass. Product mein extra binding exactly woh energy hai jo free hoti hai.

Verify: $17.59$ MeV Example 1 se do decimals tak match karta hai ✓. Do alag tables, ek physics.

Example 5 — Fuel ke per kilogram energy (Cell E)

Worked example 1 gram D–T mein kitni energy?

Har D–T fusion $17.6$ MeV deta hai. Ek fusion mein $2.014 + 3.016 = 5.030$ u fuel consume hota hai. 1 gram D–T mixture se kitni energy? ( $1 u = 1.6605 \times 1 0^{- 27}$ kg, $1 MeV = 1.602 \times 1 0^{- 13}$ J.)

Forecast: Ek gram mein atoms ki bahut badi sankhya hai, aur har ek pure MeV-scale burst deta hai. Guess: saikdon gigajoules — kaafi saalon ghar chalane ke liye.

Step 1 — Mass per reaction in kg. $5.030 \times 1.6605 \times 1 0^{- 27} = 8.352 \times 1 0^{- 27}$ kg. Kyun? Hume "1 g mein kitne reactions fit hote hain" chahiye, toh pehle ek reaction ki mass nikalo.

Step 2 — 1 g mein reactions ki sankhya. $N = \frac{1 0 ^{- 3}}{8.352 \times 1 0 ^{- 27}} = 1.197 \times 1 0^{23}$ . Divide kyun? Total mass ÷ mass-per-reaction = reactions ki sankhya.

Step 3 — Energy per reaction in joules. $17.6 \times 1.602 \times 1 0^{- 13} = 2.820 \times 1 0^{- 12}$ J. Convert kyun? Hume joules mein real-world energy chahiye, MeV nahi.

Step 4 — Total energy. $E = N \times (2.820 \times 1 0^{- 12}) = 1.197 \times 1 0^{23} \times 2.820 \times 1 0^{- 12} \approx 3.4 \times 1 0^{11}$ J.

Verify: $\approx 3.4 \times 1 0^{11}$ J = 340 GJ sirf ek gram se ✓ (forecast: saikdon GJ). Scale ke liye, 1 g coal jalane se sirf around $3 \times 1 0^{4}$ J (≈30 kJ) milta hai — toh fusion energy per gram mein roughly $1 0^{7}$ × denser hai. Units: (count) $\times$ (J) = J ✓.

Example 6 — Limiting behaviour: barrier vs. temperature (Cell F)

Worked example Kyun

$1 0^{7}$ K "cold" hai — ek order-of-magnitude check Do protons ko strong force ke liye $r \approx 3 \times 1 0^{- 15}$ m tak approach karna hoga. Us separation par Coulomb (electric) potential energy hai $U = \frac{k e ^{2}}{r}$ jahan $k = 8.99 \times 1 0^{9}$ , $e = 1.602 \times 1 0^{- 19}$ C. $U$ ko thermal energy $\sim k_{B} T$ se compare karo $T = 1.5 \times 1 0^{7}$ K par ( $k_{B} = 1.381 \times 1 0^{- 23}$ J/K).

Forecast: Agar $U ≫ k_{B} T$ , toh protons classically reach nahi kar sakte — Sun sirf Quantum Tunnelling se jalta hai. Guess: $U$ thermal energy se saikdon ke factor se zyada hai.

Step 1 — Coulomb barrier height. $U = \frac{( 8.99 \times 1 0 ^{9} ) ( 1.602 \times 1 0 ^{- 19} ) ^{2}}{3 \times 1 0 ^{- 15}}$ . Numerator: $8.99 \times 1 0^{9} \times 2.566 \times 1 0^{- 38} = 2.307 \times 1 0^{- 28}$ . $3 \times 1 0^{- 15}$ se divide karo: $U = 7.69 \times 1 0^{- 14}$ J. Yeh formula kyun? Do like charges repel karte hain; unki stored electric energy woh "wall" hai jo climb karni hai.

Step 2 — Thermal energy. $k_{B} T = 1.381 \times 1 0^{- 23} \times 1.5 \times 1 0^{7} = 2.07 \times 1 0^{- 16}$ J. $k_{B} T$ kyun? Yeh temperature $T$ par ek particle ki typical kinetic energy set karta hai.

Step 3 — Ratio lo. $\frac{U}{k _{B} T} = \frac{7.69 \times 1 0 ^{- 14}}{2.07 \times 1 0 ^{- 16}} \approx 371$ . Ratio kyun? Yeh batata hai ek typical proton barrier se kitna kam padata hai.

Verify: ratio $\approx 370 ≫ 1$ ✓ (forecast: saikdon). Toh ek typical solar proton ke paas ~1/370 energy hai jo chahiye — classically forbidden, sirf tunnelling se allowed. Yeh woh limiting case hai jo Sun ki slowness explain karta hai.

Example 7 — Confinement: kya plasma ignite karega? (Cell G)

Worked example Lawson triple product

Ek tokamak run mein density $n = 1.0 \times 1 0^{20} m^{- 3}$ , temperature $T = 15$ keV, confinement time $τ_{E} = 2.0$ s milta hai. D–T ignition threshold hai $n T τ_{E} ≳ 3 \times 1 0^{21} keV\cdots\cdotm^{- 3}$ . Kya yeh ignite karega?

Forecast: Teeno multiply karo; agar product $3 \times 1 0^{21}$ clear kar le, toh ignite hoga. Yeh near-ITER numbers hain, toh guess: barely threshold reach kar raha hai.

Step 1 — Triple product multiply karo. $n T τ_{E} = (1.0 \times 1 0^{20}) (15) (2.0)$ . Multiply kyun, add nahi? Losses kam hoti hain agar tum denser ya hotter ya longer confinement wale ho — requirements compound hoti hain, toh criterion ek product hai (parent note dekho).

Step 2 — Value compute karo. $1.0 \times 1 0^{20} \times 15 \times 2.0 = 1.0 \times 1 0^{20} \times 30 = 3.0 \times 1 0^{21} keV\cdots\cdotm^{- 3}$ . Yeh arithmetic kyun? Hum teeno factors ko ek number mein collapse karte hain taaki directly single threshold value se compare kar sakein — pehle do chhote factors multiply karo ( $15 \times 2.0 = 30$ ), phir power of ten lagao.

Step 3 — Compare. $3.0 \times 1 0^{21} \geq 3 \times 1 0^{21}$ ⇒ just ignites (marginally). "Marginal" kyun matter karta hai: $τ_{E}$ mein koi bhi giraahat (worse confinement) aur yeh fizzle ho jaayega — yahi reason hai ki bade machines banana (longer $τ_{E}$ ) pura game hai.

Verify: product $= 3.0 \times 1 0^{21}$ , exactly threshold par ✓. Units: $m^{- 3} \cdot keV \cdot s = keV\cdots\cdotm^{- 3}$ ✓ — criterion ke units se match karta hai.

Example 8 — Real-world word problem: Sun ki lifetime (Cell H)

Worked example Sun kitne time tak shine kar sakta hai?

Sun ( $M = 2.0 \times 1 0^{30}$ kg) apne core mein hydrogen fuse karta hai. Uski sirf about 10% mass core hydrogen hai jo fusion ke liye available hai, aur fusion us hydrogen ki mass ka about $0.7%$ energy mein convert karta hai. Woh $L = 3.8 \times 1 0^{26}$ W par radiate karta hai. Uski main-sequence lifetime years mein estimate karo.

Forecast: Mass ka sirf ek fraction ka fraction energy banta hai, lekin $c^{2}$ enormous hai. Guess: around ten billion years (textbook figure).

Step 1 — Fusible hydrogen mass. $M_{H} = 0.10 \times 2.0 \times 1 0^{30} = 2.0 \times 1 0^{29}$ kg. 10% kyun? Sirf hot dense core fuse karta hai; outer layers bahut cool hain.

Step 2 — Mass jo actually energy mein convert hoti hai. $Δ m = 0.007 \times 2.0 \times 1 0^{29} = 1.4 \times 1 0^{27}$ kg. $0.7%$ kyun? Yeh p–p chain ka mass-defect fraction hai (4 protons → helium mein ~0.7% mass kho jaata hai).

Step 3 — Total energy available. $E = Δ m c^{2} = 1.4 \times 1 0^{27} \times (3.0 \times 1 0^{8})^{2} = 1.26 \times 1 0^{44}$ J. $E = m c^{2}$ kyun? Vanished mass ko radiated energy mein convert karta hai — Mass-Energy Equivalence (E=mc^2) dekho.

Step 4 — Lifetime = energy ÷ power. $t = \frac{1.26 \times 1 0 ^{44}}{3.8 \times 1 0 ^{26}} = 3.3 \times 1 0^{17}$ s. Divide kyun? Power energy per second hai; total energy ÷ rate = time.

Step 5 — Years mein convert karo. $t = \frac{3.3 \times 1 0 ^{17}}{3.156 \times 1 0 ^{7}} \approx 1.05 \times 1 0^{10}$ yr. $3.156 \times 1 0^{7}$ se divide kyun? Ek saal mein approximately $3.156 \times 1 0^{7}$ seconds hote hain ( $365.25 \times 24 \times 3600$ ), toh total seconds ko "seconds per year" se divide karne par years ki count milti hai.

Verify: $\approx 1.0 \times 1 0^{10}$ yr = 10 billion years ✓ (forecast mila — standard solar lifetime). Units: J ÷ W = s ✓, s ÷ (s/yr) = yr ✓.

Example 9 — Exam-style twist: fraction of mass converted (Cell I)

Worked example Unit-trap question

D–T reaction mein, initial rest mass ka kitna fraction energy mein convert hota hai? (Example 1 se $Δ m = 0.018883$ u aur reactant mass $5.030151$ u use karo.) Phir confirm karo ki yeh fraction times $c^{2}$ energy per unit mass reproduce karta hai.

Forecast: Fusion fission (~0.1%) se zyada efficient hai lekin absolute terms mein abhi bhi tiny hai. Guess: kuch tenths of a percent.

Step 1 — Fraction converted. $f = \frac{Δ m}{m _{reactants}} = \frac{0.018883}{5.030151} = 3.754 \times 1 0^{- 3}$ . Reactant mass se divide kyun? "Fraction converted" = mass lost ÷ woh mass jisse tumne start kiya.

Step 2 — Percentage mein. $f = 3.754 \times 1 0^{- 3} \times 100% = 0.375%$ . Yeh note kyun karo? Yeh nuclear fission mein fraction (~0.09%) se ~4× zyada hai — ek key reason kyun fusion prized hai (compare Nuclear Fission).

Step 3 — Energy per kg se cross-check. D–T fuel ke per kg energy released honi chahiye $f c^{2} = 3.754 \times 1 0^{- 3} \times (3.0 \times 1 0^{8})^{2} = 3.38 \times 1 0^{14}$ J/kg. Kyun? Kyunki $E = (f m) c^{2}$ hai, $m$ se divide karne par $E / m = f c^{2}$ milta hai — fraction tabhi energy density banta hai jab $c^{2}$ se multiply karo.

Step 4 — Example 5 se compare karo. Example 5 ne $3.4 \times 1 0^{11}$ J per gram diya, jo $3.4 \times 1 0^{14}$ J per kg hai. Yeh $f c^{2} = 3.38 \times 1 0^{14}$ J/kg se match karta hai. ✓

Verify: $f = 0.375%$ ✓; aur $f c^{2} = 3.38 \times 1 0^{14}$ J/kg Example 5 ki per-kg value se agree karta hai ✓ — twist aur direct method milte hain. Trap dodge kiya: "fraction" $f$ dimensionless hai; sirf $\times c^{2}$ ke baad iska energy density ke units (J/kg) aate hain.

Recall Matrix ko self-test karo

Kaun sa cell hai jahan $Q < 0$ hai, aur iska physically kya matlab hai? ::: Cell B — energy absorbed hoti hai; reaction spontaneously nahi chalega (e.g. $^{4} He +^{4} He \to^{8} Be$ ). $Q$ compute karne ke do tarike? ::: Masses se (reactants − products) $c^{2}$ , ya binding energies se (products − reactants). Lawson condition teen quantities ka product kyun hai? ::: Kyunki losses kam hoti hain agar tum density, temperature, ya confinement time badhaao — requirements compound hoti hain, add nahi. D–T mass ka roughly kitna fraction energy banta hai? ::: About $0.375%$ — fission se kuch times zyada.