Exercises — Fusion — solar fusion, tokamak (concept)

Rule kya kehta hai: do nuclei ko jodne se energy tabhi release hoti hai jab product Binding Energy per Nucleon Curve par reactants se upar ho — yaani sirf un nuclei ke liye jo iron peak se neeche hain.

(a) Deuterium aur tritium bahut halke hain (mass numbers 2 aur 3), iron se bahut neeche. Unhe fuse karna curve pe upar chadhna hai → energy release hoti hai. Haan.

(b) Uranium-235 iron se upar hai (mass number 235). Ek neutron add karke combine karna curve pe nahi chadhta; balki ${}^{235}\text{U}$ energy release karne ke liye split (fission) hota hai. Fusion yahan energy release nahi karta — Nahi.

Answer: (a) Haan, (b) Nahi.

(a) Coulomb (electrostatic) repulsion — dono protons positive hain aur same charges repel karte hain. (b) Strong nuclear force — attractive hai, lekin sirf bahut chhoti range par (~femtometres), isliye pehle protons bahut karib aane chahiye. (c) Quantum tunnelling — Quantum Tunnelling dekho; particles mein ek chhoti probability hoti hai Coulomb barrier ke andar se guzarne ki, bina itni classical energy ke ki wo uske upar se jaayein.

Step 1 — reactant mass. $2\times m_D=2\times 2.014102=4.028204$ u. Kyun sum? Rest mass dono inputs ki energy hold karta hai. Step 2 — product mass. $m_{{}^{3}He}+m_n=3.016029+1.008665=4.024694$ u. Products ko kyun sum karein? Do output particles mein locked rest mass woh hai jise hume inputs se compare karna hai — taaki dekh sakein kitna gayab hua. Step 3 — mass defect. $\Delta m=4.028204-4.024694=0.003510$ u. Products ko reactants se kyun subtract karein? Jo mass gayab hua woh energy ban gaya. Step 4 — convert. $Q=\Delta m\times 931.5=0.003510\times 931.5=3.27\text{MeV}$. $\times 931.5$ kyun? Yahi $1\text{u}\cdot c^2$ MeV mein hai, toh yeh mass-in-u ko energy-in-MeV mein convert karta hai.

Answer: $Q\approx 3.27\mathbf{MeV}$.

Step 1 — reactant mass. $2\times 2.014102=4.028204$ u. Kyun sum? Dono deuterons apna rest mass reaction mein laate hain, toh hum unhe add karte hain. Step 2 — product mass. $3.016049+1.007825=4.023874$ u. Products ko kyun sum karein? Tritium aur proton apna rest mass le jaate hain; unhe add karne se hum inputs aur outputs compare kar sakte hain. Step 3 — mass defect. $\Delta m=4.028204-4.023874=0.004330$ u. Subtract kyun? Rest mass mein jo kami hai wahi exactly woh mass hai jo energy mein convert hua. Step 4 — convert. $Q=0.004330\times 931.5=4.03\text{MeV}$. $\times 931.5$ kyun? Mass defect ko u mein $1\text{u}\cdot c^2=931.5$ MeV se multiply karne par woh gayab hua mass apne energy equivalent mein aa jaata hai.

Answer: $Q\approx 4.03\mathbf{MeV}$. Gaur karo ki dono D–D branches sirf ~3–4 MeV dete hain — D–T ke 17.6 MeV se bahut kam — aur isliye hi D–T preferred reactor fuel hai.

Step 1 — 1 g mein kitne D nuclei? Ek D atom ki mass $=2.014\times 1.6605\times 10^{-27}=3.344\times 10^{-27}$ kg. Kyun? u ko kg mein convert karo $1\text{u}=1.6605\times 10^{-27}$ kg se. Nuclei ki sankhya $N=\frac{10^{-3}\text{kg}}{3.344\times 10^{-27}\text{kg}}=2.99\times 10^{23}$. Step 2 — energy per reaction joules mein. $17.6\text{MeV}\times 1.602\times 10^{-13}\text{J/MeV}=2.82\times 10^{-12}$ J. Pehle joules mein kyun convert karein? Answer joules mein chahiye, aur MeV joules nahi hain; bina convert kiye MeV value ko $N$ se multiply karne par galat unit mein number milega, $1.602\times 10^{-13}$ ke factor se galat. Step 3 — total. $E=N\times (2.82\times 10^{-12})=2.99\times 10^{23}\times 2.82\times 10^{-12}=8.4\times 10^{11}$ J. $N$ se multiply kyun? Har $N$ reactions mein se ek hi energy ka packet release karta hai, toh total wahi per-reaction energy $N$ baar stacked hai.

Answer: $E\approx 8.4\times 10^{11}\mathbf{J}$ — ek gram deuterium se 200 tonnes of TNT jitni energy. Point: yahi staggering energy density fusion power ka pura promise hai.

Step 1 — He per joules mein energy. $26.7\text{MeV}\times 1.602\times 10^{-13}=4.28\times 10^{-12}$ J. Convert kyun? $L$ watts mein hai (joules per second), toh $Q$ bhi joules mein hona chahiye taaki division ka matlab bane. Step 2 — rate. Per second sankhya $=\frac{L}{Q}=\frac{3.8\times 10^{26}}{4.28\times 10^{-12}}=8.9\times 10^{37}$ per second. Divide kyun? Har helium event ek fixed packet $Q$ deliver karta hai; total power $÷$ packet size $=$ events per second.

Answer: $\approx 8.9\times 10^{37}$ helium nuclei per second. Kyunki har ek mein 4 protons lagte hain, Sun har second $\sim 3.6\times 10^{38}$ protons jalaata hai — aur phir bhi billions of years ke liye kaafi hai.

Step 1 — teenon multiply karo. $nT{\tau }_E=(1.0\times 10^{20})(10)(3)=3.0\times 10^{21}{\text{keV⋅s⋅m}}^{-3}$. Teenon ko kyun multiply karein? Losses tab kam hote hain jab density zyada ho, temperature zyada ho, ya leakage slow ho; product measure karta hai ki unka combined effect losses ko beat karta hai ya nahi. Isliye yeh ek product hai, sum nahi — kisi bhi ek factor ko improve karne par pura product multiply hota hai. Step 2 — compare karo. $3.0\times 10^{21}\ge 3\times 10^{21}$ → bilkul threshold par hai. (b) Haan, marginally ignition par. (c) Kyunki yeh exactly threshold reach karta hai, koi change zaruri nahi; lekin agar robust margin chahiye (maan lo $2\times$ threshold se upar) toh tum, maslan, ${\tau }_E$ ko double karke 6 s kar sakte ho, ya $n$ double karo, ya $T$ double karo — koi bhi ek triple product ko 2 se multiply kar deta hai.

Answer: (a) $3.0\times 10^{21}$; (b) ignition reach karta hai (barely); (c) $n$, $T$, ya ${\tau }_E$ mein se kisi ko bhi double karne par product double ho jaata hai.

Step 1 — work ka matlab. Ek force dwara kaam karne ki rate (power) $P=\vec{F}\cdot \vec{v}$ hai — force motion ki direction mein. Dot product kyun? Yeh sirf velocity ke parallel force ka component pick karta hai; sidewise force koi energy transfer nahi karta. Step 2 — Lorentz plug in karo. $P=(q\vec{v}\times \vec{B})\cdot \vec{v}$. Step 3 — key geometric fact. Cross product $\vec{v}\times \vec{B}$ definition se $\vec{v}$ ke perpendicular hota hai. $\vec{v}$ ke perpendicular vector ko $\vec{v}$ se dot karne par $0$ milta hai. Toh $P=q(\vec{v}\times \vec{B})\cdot \vec{v}=0$ hamesha, kisi bhi $\vec{v}$, $\vec{B}$ ke liye. Step 4 — matlab. Zero work → field kabhi particle ki speed (kinetic energy) nahi badalta, sirf uski direction badalta hai. Toh magnetic cage confine karta hai (paths ko field lines ke around circles/helices mein mod deta hai) lekin heat nahi kar sakta. Heating doosre sources se aani chahiye: currents (ohmic), radio-frequency waves, neutral-beam injection.

Answer: $P=0$ identically; magnetic field confine karta hai lekin koi heating nahi karta.

Fusion, per kg. Step 1: energy per reaction $=17.6\text{MeV}\times 1.602\times 10^{-13}=2.82\times 10^{-12}$ J. Step 2: fuel mass per reaction $=5.03\text{u}\times 1.6605\times 10^{-27}=8.35\times 10^{-27}$ kg. Step 3: reactions per kg $=1/8.35\times 10^{-27}=1.198\times 10^{26}$. Step 4: energy per kg $=1.198\times 10^{26}\times 2.82\times 10^{-12}=3.38\times 10^{14}$ J/kg.

Chemical (carbon), per kg. Step 1: energy per atom $=4\text{eV}=4\times 1.602\times 10^{-19}=6.41\times 10^{-19}$ J. Step 2: mass per atom $=12\times 1.6605\times 10^{-27}=1.993\times 10^{-26}$ kg. Step 3: atoms per kg $=1/1.993\times 10^{-26}=5.02\times 10^{25}$. Step 4: energy per kg $=5.02\times 10^{25}\times 6.41\times 10^{-19}=3.22\times 10^7$ J/kg.

Ratio. $\frac{3.38\times 10^{14}}{3.22\times 10^7}\approx 1.0\times 10^7$.

Answer: fusion chemical burning se roughly das million guna zyada energy per kilogram deta hai. Kyun: nuclear binding energies (MeV) chemical bond energies (eV) se roughly ek million ka factor se badi hoti hain per particle, aur halke fuel mass se aur amplify hota hai.

(a) Mass-to-energy rate. Mass-Energy Equivalence (E=mc^2) se, $E=m{c}^2\Rightarrow \dot{m}=L/c^2$. $\dot{m}=\frac{3.8\times 10^{26}}{(3.0\times 10^8{)}^2}=\frac{3.8\times 10^{26}}{9.0\times 10^{16}}=4.2\times 10^9\text{kg/s}$. $c^2$ se kyun divide karein? $c^2$ mass se energy ka "exchange rate" hai; energy-per-second ko isse divide karne par mass-per-second milta hai. (b) Poori life mein total. Seconds in $10^{10}$ years $=10^{10}\times 3.156\times 10^7=3.156\times 10^{17}$ s. Total mass converted $=4.2\times 10^9\times 3.156\times 10^{17}=1.33\times 10^{27}$ kg. Fraction $=\frac{1.33\times 10^{27}}{2.0\times 10^{30}}=6.6\times 10^{-4}\approx 0.07\%$.

Answer: (a) $\approx 4.2\times 10^9$ kg/s ($\sim$40 lakh tonne/second); (b) $\approx 1.3\times 10^{27}$ kg total, sirf Sun ke mass ka 0.07%. Woh tiny fraction 10 billion years ki sunlight power karta hai — yahi $c^2$ mein chhupi energy ki pehchaan hai.

Reasoning chain (har link agla earn karta hai):

1. Classically impossible. $1.3\text{keV}\ll$ kaafi sau keV barrier, toh almost koi bhi proton repulsive Coulomb barrier ke upar chadhne ke liye kaafi energy nahi rakhta. (Cross-check: thermal distribution ka fast tail bhi rarely barrier-top tak pahuncha hai.)
2. Tunnelling ek darwaza kholti hai. Quantum Tunnelling se, ek proton mein barrier ke andar se guzarne ki chhoti lekin nonzero probability hoti hai. Yeh probability energy ke saath steeply badhti hai, toh fusion rare, above-average-energy protons (the "Gamow peak") se dominate hoti hai.
3. Weak-force bottleneck. Tunnelling ke baad bhi, pehla step $p+p\to {}^2\text{D}+e^{+}+{\nu }_e$ require karta hai ki ek proton weak interaction ke zariye neutron mein convert ho — yeh intrinsically slow process hai.
4. Result: har individual proton average par billions of years tak wait karta hai fuse hone ke liye. Sun steadily isliye chamakta hai kyunki usme astronomical number ($\sim 10^{57}$) protons hain, toh total rate enormous hai bawajood iske ki per-proton rate tiny ho.

Answer: fusion quantum tunnelling ke zariye plasma ke hot ionized core mein hoti hai, slow weak interaction se gated hoti hai; per proton slowness + huge proton count = steady multi-billion-year burn.

Fusion Q aata hai ==mass defect $\times c^2$ se; D–T deta hai 17.6 MeV; magnetic field zero work karta hai (confine karta hai, kabhi heat nahi karta); ignition ke liye triple product $nT{\tau }_E$ threshold se upar chahiye; Sun fuse karta hai sirf quantum tunnelling== ke zariye.