Fission — chain reaction, critical mass: Exercises

2.3.23 · D4 · Physics › Modern Physics › Fission — chain reaction, critical mass

Shuru karne se pehle, ek figure woh geometry fix karta hai jis par is set ke aadhe problems tike hain — neutron production (volume ke saath badhta hai) aur neutron leakage (surface ke saath badhta hai) ke beech ka tug-of-war.

Figure — Fission — chain reaction, critical mass

Do coloured curves dekho: burnt-orange wala (production, $\propto R^{3}$ ) chhoti ball ke liye teal wale (leakage $\propto R^{2}$ ) se neeche shuru hota hai, phir usse aage nikal jaata hai. Jahan ye cross karte hain, $k = 1$ — woh crossing radius hi critical radius hai.

Level 1 — Recognition

Problem 1.1

Ek-ek line mein batao ki fissile material ke ek lump ke saath kya hota hai jab (a) $k = 0.8$ , (b) $k = 1.0$ , (c) $k = 1.3$ .

Recall Solution

$k$ = (naye generation mein neutrons) ÷ (purane generation mein neutrons).

(a) $k = 0.8 < 1$ : har generation chhoti hoti hai → chain khatam ho jaati hai (subcritical).
(b) $k = 1.0$ : har generation same size ki → steady, self-sustaining (critical — reactor ki normal state).
(c) $k = 1.3 > 1$ : har generation badi hoti hai → exponential growth (supercritical — bomb / reactor start-up).

Problem 1.2

Ek fission event mein lagbhag $200$ MeV release hoti hai. Average mein ek fission se kitne neutrons nikalta hai, aur kyun woh number (energy nahi) decide karta hai ki chain reaction possible hai ya nahi?

Recall Solution

Average mein $ν \approx 2.5$ neutrons per fission (reaction $2$ – $3$ release karta hai). Energy cheezein garam karti hai, lekin emitted neutrons hi hain jo agla fission trigger karne jaate hain. Chain ke liye, average mein, kam se kam ek surviving neutron per fission chahiye jo kisi naye nucleus se takraye. Toh neutrons ka bachna — joules release hona nahi — chain ko sustain karta hai.

Level 2 — Application

Problem 2.1

Ek reactor $P = 300$ MW thermal power par chal raha hai. Har fission $200$ MeV release karta hai. Har second kitne fissions hote hain?

Recall Solution

Step 1 — ek fission joules mein. Kyun? Power joules per second hai, toh energy-per-event bhi joules mein honi chahiye. $E = 200 MeV = 200 \times 1.6 \times 1 0^{- 13} J = 3.2 \times 1 0^{- 11} J .$ Step 2 — divide karo. Kyun? (total energy rate) ÷ (energy per event) = events per second. $fissions/s = \frac{300 \times 1 0 ^{6}}{3.2 \times 1 0 ^{- 11}} = 9.375 \times 1 0^{18} s^{- 1} .$

Problem 2.2

$N_{0} = 1000$ neutrons aur $k = 1.02$ se shuru karo, $50$ generations ke baad kitne neutrons bachenge?

Recall Solution

Step 1 — growth law. Har generation $k$ se multiply hoti hai, toh $n$ generations ke baad $N_{n} = N_{0} k^{n}$ . Step 2 — plug in karo. Kyun power mein raise karte hain? Same factor se baar baar multiply karna hi exponentiation hai. $N_{50} = 1000 \times 1.0 2^{50} = 1000 \times 2.6916 \approx 2692 neutrons .$

Problem 2.3

$1$ kg $^{235} U$ completely fission ho jaata hai. Kitni energy (joules mein) release hoti hai? ( $200$ MeV per fission lo, molar mass $235$ g/mol.)

Recall Solution

Step 1 — kitne nuclei? Kyun? Energy = (fissions ki sankhya) × (har ek ki energy), aur har nucleus ek baar fission hota hai. $N = \frac{1000 g}{235 g/mol} \times 6.022 \times 1 0^{23} = 2.5626 \times 1 0^{24} nuclei .$ Step 2 — total energy. $E = N \times 3.2 \times 1 0^{- 11} J = 2.5626 \times 1 0^{24} \times 3.2 \times 1 0^{- 11} \approx 8.20 \times 1 0^{13} J .$ Yeh ek kilogram se lagbhag $20$ kilotonnes of TNT hai — fission weapon ka scale.

Level 3 — Analysis

Problem 3.1

Ek $^{235} U$ atom aur incoming neutron ka total mass, products ke total mass se $Δ m = 0.2154 u$ zyada hai. Energy $Q$ nikalo aur " $200$ MeV" rule of thumb se compare karo.

Recall Solution

Step 1 — mass-energy equivalence. Kyun $E = m c^{2}$ ? Fission thodi si lost mass ("mass defect") ko energy mein convert karta hai; dekho Mass-Energy Equivalence E=mc^2. $Q = Δ m \cdot c^{2} = 0.2154 \times 931.5 MeV \approx 200.6 MeV .$ Step 2 — compare karo. Yeh $\approx 200$ MeV estimate se match karta hai jo humne Binding Energy per Nucleon Curve se nikali thi ( $0.9$ MeV/nucleon $\times 235$ ). Do independent routes — mass defect aur binding-energy difference — agree karte hain, number confirm hota hai.

Problem 3.2

Ek reactor exactly $k = 1$ par hai. Ek operator control rod withdraw karta hai, $k$ jump karke $1.01$ ho jaata hai. Effective generation time (delayed neutrons se dominate hoti hai) $τ = 0.1$ s hai. Neutron population double hone mein kitna time lagega?

Recall Solution

Step 1 — double hone ke liye generations. $N_{n} = N_{0} k^{n}$ se, ratio $2$ set karo: $k^{n} = 2 \Rightarrow n = \frac{l n 2}{l n k} = \frac{0.6931}{l n 1.01} = \frac{0.6931}{0.009950} \approx 69.7 generations .$ Kyun log lete hain? Unknown $n$ exponent mein hai; logarithm woh tool hai jo exponent ko neeche kheench ke solve karta hai. Step 2 — time mein convert karo. Kyun $τ$ se multiply? Har generation $τ$ seconds chalti hai. $t = n τ \approx 69.7 \times 0.1 \approx 7.0 s .$ Saat second aasaani se control ho sakta hai — yeh delayed neutrons ka gift hai. Sirf prompt neutrons se ( $τ \sim 1 0^{- 8}$ s) same $k$ par ek microsecond se pehle double ho jaata. Dekho Nuclear Reactor.

Problem 3.3

Toy model $k (R) = \frac{a}{b} R$ use karo (production $\propto R^{3}$ , surface leakage $\propto R^{2}$ ), radius $R = 8$ cm wale sphere ka $k = 0.8$ hai. Critical radius $R_{c}$ nikalo.

Recall Solution

Step 1 — constant pin karo. Kyun? Is toy model mein $k$ $R$ mein linear hai, toh ek data point slope $a / b$ fix kar deta hai. $\frac{a}{b} = \frac{k}{R} = \frac{0.8}{8 cm} = 0.1 cm^{- 1} .$ Step 2 — $k = 1$ set karo. Kyun? Critical matlab production exactly leakage ko balance kare. $1 = \frac{a}{b} R_{c} = 0.1 R_{c} \Rightarrow R_{c} = 10 cm .$ Hamara $8$ cm ball $R_{c}$ se neeche hai, jo uske subcritical $k = 0.8$ ke consistent hai.

Level 4 — Synthesis

Problem 4.1

Critical mass $M_{c} \propto 1/ ρ^{2}$ density $ρ$ ke saath scale karta hai. Ek implosion device plutonium core ko apni normal density ke $2.5 \times$ tak compress karta hai. Critical mass kitne factor se girti hai, aur kyun yeh ek subcritical lump ko supercritical bana deta hai?

Recall Solution

Step 1 — scaling apply karo. Kyun $1/ ρ^{2}$ ? Zyada density nuclei ko paas pack karti hai, toh neutron kisi se takraane se pehle shorter distance chalta hai (chhota mean free path); leakage set karne wali geometry dono chhoti radius aur tighter packing mein sikodte hain — $ρ$ ke do factors. $\frac{M _{c} ( compressed )}{M _{c} ( normal )} = \frac{1}{2. 5 ^{2}} = \frac{1}{6.25} = 0.16.$ Toh critical mass apni original value ki $16%$ tak gir jaati hai. Step 2 — supercritical kyun hota hai. Core ka actual mass unchanged hai. Agar woh fixed mass purani critical mass se thodi neeche thi, toh critical mass ko $16%$ tak slice karne par fixed mass ab naye threshold se kaafi upar hai → $k > 1$ → supercritical. Yahi ek implosion bomb ki puri trick hai.

Problem 4.2

$^{235} U$ ka ek sample ordinary radioactive decay bhi karta hai half-life $7.0 \times 1 0^{8}$ years ke saath. Maano ki decay karne ki bajay, har nucleus ek supercritical assembly mein effective doubling time $1 0^{- 8}$ s ke saath chain fission karta. Dono timescales contrast karo — ek microsecond mein kitne doublings hote hain, aur usi microsecond mein ordinary radioactivity se kitna fraction nuclei decay karta hai.

Recall Solution

Step 1 — $1 μ$ s mein chain doublings. Kyun divide? Doublings ki sankhya = (elapsed time) ÷ (doubling time). $\frac{1 0 ^{- 6} s}{1 0 ^{- 8} s} = 100 doublings \Rightarrow growth factor 2^{100} \approx 1.27 \times 1 0^{30} .$ Step 2 — $1 μ$ s mein ordinary decay. Kyun exponential-decay law? Radioactivity $N = N_{0} e^{- λ t}$ follow karta hai $λ = ln 2/ T_{1/2}$ ke saath; dekho Radioactive Decay and Half-life. $λ = \frac{0.6931}{7.0 \times 1 0 ^{8} \times 3.156 \times 1 0 ^{7} s} \approx 3.14 \times 1 0^{- 17} s^{- 1} .$ Fraction decayed $\approx λ t = 3.14 \times 1 0^{- 17} \times 1 0^{- 6} \approx 3.1 \times 1 0^{- 23}$ — bilkul negligible. Conclusion: Usi microsecond mein, chain reaction $\sim 1 0^{30}$ se multiply hoti hai jabki natural decay $1 0^{23}$ mein ek hisse ko bhi mushkil se touch karta hai. Chain fission bahut hi zyada fast hai — isliye assembled critical mass essentially instantaneously apni energy release kar deti hai.

Level 5 — Mastery

Problem 5.1

Ek "gun-type" weapon do subcritical $^{235} U$ pieces ko ek supercritical mass banane ke liye saath crash karta hai jiska $k = 1.05$ aur prompt generation time $τ = 1 0^{- 8}$ s hai. Ek single stray neutron se shuru karke, estimate karo ki population $1 0^{20}$ neutrons tak pahunchne mein kitna time lagega (fuel ka significant fraction fission karne ke liye enough).

Recall Solution

Step 1 — growth law se real time. Real time mein, $N (t) = N_{0} k^{t / τ}$ . $N_{0} = 1$ , $N (t) = 1 0^{20}$ set karo: $1 0^{20} = 1.0 5^{t / τ} .$ Step 2 — exponent free karne ke liye log lo. Kyun phir logs? Same wajah jaise L3: unknown exponent mein trapped hai. $\frac{t}{τ} = \frac{l n 1 0 ^{20}}{l n 1.05} = \frac{20 \times 2.3026}{0.04879} = \frac{46.05}{0.04879} \approx 943.8 generations .$ Step 3 — convert karo. $t = 943.8 \times 1 0^{- 8} s \approx 9.44 \times 1 0^{- 6} s \approx 9.4 μ s .$ Das microseconds se kam mein ek neutron se full-scale energy release tak — dharti ka koi control system intervene nahi kar sakta. Isliye hi ek bomb deliberately fast (unmoderated) neutrons aur $k > 1$ use karta hai, jabki ek reactor delayed neutrons par rely karta hai aur $k = 1$ rakhta hai.

Problem 5.2

Design-reasoning question. Tumhare paas $60$ kg $90%$ -enriched $^{235} U$ hai (bare-sphere critical mass $\approx 52$ kg). Surface-to-volume aur cross-section reasoning use karke, teen independent tarike explain karo is assembly ko fuel add kiye bina zyada reactive banane ke, aur ek tarika safely subcritical banane ka.

Recall Solution

Zyada reactive (raise $k$ ):

Compress karo (density badhao). $M_{c} \propto 1/ ρ^{2}$ : squeezing critical mass ko giraata hai, toh hamara fixed $60$ kg threshold se aur upar aa jaata hai. Physically, chhota mean free path matlab neutrons nucleus milne se pehle leak hone se pehle milte hain (Problem 4.1).
Neutron reflector / tamper se wrap karo. Ek tamper bhaagne-wale neutrons ko wapas core mein bounce karta hai, leakage term ( $\propto R^{2}$ ) ko cut karta hai. Surface par kam neutrons lost → zyada $k$ , effective critical mass kam.
Sphere ki shape do. Sphere ka apne volume ke liye sabse chhota surface area hota hai, leakage minimize karta hai; koi bhi non-spherical shape (slab, rod) volume per unit zyada surface rakhta hai aur zyada leak karta hai. Upar figure mein crossing point dekho.

(Cross-section angle:) moderator se neutrons slow karna $^{235} U$ ka fission Neutron Cross-section badhata hai, zyada fissions per neutron ka ek aur raasta — halanki ek fast weapon deliberately prompt rehne ke liye isse avoid karta hai.

Safely subcritical:

Mass ko separated sub-pieces mein tod do (ya thin slab mein flatten karo). Sphere ko do door lumps mein todna, har ek $52$ kg se kam, har piece ka surface-to-volume ratio bahut zyada badha deta hai → leakage dominate karti hai → har ek ka $k < 1$ . Exactly aise hi fissile material store ki jaati hai.

Kyun sab ek idea mein reduce hota hai: yahan har lever ya toh production badhata hai (density, cross-section) ya leakage kaata hai (reflector, spherical shape) — ya, safety ke liye, deliberately leakage maximize karta hai (splitting, flattening). Reactivity production versus loss ka ledger hai.