Fission — chain reaction, critical mass: Visual walkthrough

2.3.23 · D2 · Physics › Modern Physics › Fission — chain reaction, critical mass

Hum ek number chase kar rahe hain: $k$ , multiplication factor — agli generation ko har abhi ki fission ke liye kitni fissions milti hain. Agar $k \geq 1$ toh reaction zinda hai; agar $k < 1$ toh woh mar jaata hai. Hamara poora kaam yeh figure out karna hai ki $k$ tukde ki size par kaise depend karta hai.

Step 1 — Ek split se kuch neutrons milte hain

KYA. Uss single event se shuru karo jis par sab kuch build hota hai: ek neutron ek $^{235} U$ nucleus se takraata hai, woh split hota hai, aur bahar aate hain naye neutrons — average mein kareeb $ν \approx 2.5$ .

KYUN. Isse pehle ki hum kisi chain ki baat karein, hume us chain ki ek link chahiye: ek aisi fission jo agli fission ke liye ingredients (neutrons) produce kare. Agar ek split zero neutrons wapas deta, toh koi chain kabhi exist hi nahi kar sakti. $ν > 1$ ka fact hi sab kuch ka seed hai.

PICTURE. Ek neutron (red) andar jaata hai; nucleus do fragments mein split hota hai; teen naye neutrons (black) bahar nikalte hain.

Figure — Fission — chain reaction, critical mass

Step 2 — Events ko chain karna: number $k$

KYA. Kai fissions ko "generations" mein line up karo. Generation 0 mein kuch neutrons hain; unme se har ek shayad ek fission cause kare jo $ν$ naye neutrons banaye; woh generation 1 hain; aur aage bhi. Sabse important quantity consecutive generations ka ratio hai.

KYUN. Hume actually parwah nahi ki koi ek nucleus kitne neutrons emit karta hai — hume parwah hai ki neutrons ki population badh rahi hai ya ghatt rahi hai. Yeh ek ratio question hai, isliye hum ek ratio define karte hain.

PICTURE. Ek tree: upar ek dot, neeche ki taraf branching. Lekin notice karo — har neutron fission cause karne ke liye survive nahi karta (kuch lost hain, grey mein fade hote dikhte hain). Surviving branches (red) hi chain continue karte hain.

Step 3 — Neutrons kahan jaate hain? Teen fates

KYA. Ek naaya born neutron ke exactly teen fates hain: (a) woh ek nayi fission cause karta hai (achha — chain chalata rehta hai), (b) woh bina fission kiye absorb ho jaata hai (material ke andar lost), ya (c) woh surface se leak ho jaata hai aur kabhi wapas nahi aata.

KYUN. Step 2 se, $k = ν \times (surviving fraction)$ . "Surviving" ka matlab hai "fate (b) ya (c) ki jagah fate (a) mein khatam hona." Toh $k$ paane ke liye hume achhi fissions ki rate ko do loss channels ki rates se compare karna hoga. Is step ka punchline: do losses size ke saath bilkul alag tarah behave karte hain, aur yahi difference poora secret hai.

PICTURE. Ek circle (lump). Beech mein born red arrows: kuch andar rehte hain aur fissions cause karte hain, kuch wall se takra kar escape karte hain (leakage), ek grey mein nigal jaata hai (absorption).

Intuition Key asymmetry — aur kyun absorption

villain nahi hai

Fission production andar har jagah hoti hai → stuff ki quantity par depend karti hai → volume $\propto R^{3}$ .
Non-fission absorption bhi andar har jagah hoti hai → woh bhi volume $\propto R^{3}$ par depend karti hai.
Leakage sirf skin par hoti hai → surface area $\propto R^{2}$ par depend karti hai.

Yahaan crucial observation yeh hai: fission production aur absorption dono $R^{3}$ scale karte hain, isliye woh saath mein upar-neeche hote hain chahe lump koi bhi size ka ho. Unka competition material ki ek fixed property hai (zyada likely kya hai — fission ya plain absorption?), size se bilkul independent. Sirf leakage $R^{2}$ scale karke tie tod deta hai. Toh absorption kabhi koi threshold create nahi karta; woh sirf yeh set karta hai ki fuel kitna achha hai. Leakage hi size-dependent villain hai.

Step 4 — Volume beats surface: $R^{3}$ vs $R^{2}$ ki race

KYA. Lump ko radius $R$ ke sphere ki tarah model karo. Count karo ki teen rates $R$ ke saath kaise scale karte hain.

KYUN. Humne sphere isliye choose kiya kyunki yeh sabse simple shape hai aur — bonus — ek given volume ke liye iske paas sabse kam surface hai (hum Step 7 mein isse revisit karte hain). Hum $R$ use karte hain kyunki volume aur surface iske clean powers hain, aur powers compare karna aasaan hai.

PICTURE. Do spheres, chhota aur bada. Dono par, filled interior (production, black) aur outer shell (leakage, red rim). Dekho ki bade sphere par red rim proportionally thinner ho jaati hai.

Kyunki fission aur absorption dono $R^{3}$ scale karte hain, hum unhe ek net "volume budget" mein bundle kar sakte hain — volume-losses ke baad bachi usable production abhi bhi $\propto R^{3}$ hai. Sirf leakage alag rehti hai. Volume production ke relative iska fraction:

$leakage fraction \propto \frac{S}{V} = \frac{4 π R ^{2}}{\frac{4}{3} π R ^{3}} = \frac{3}{R} \propto \frac{1}{R}$

Step 5 — $k (R)$ build karna aur threshold dhundhna

KYA. Step 2 ki identity $k = ν \times (surviving fraction)$ se $k$ assemble karo, Step 4 ke $R^{3}$ aur $R^{2}$ rates use karke.

KYUN (missing algebra, fill in kiya gaya). Ek neutron survive karta hai jab woh lost hone se pehle fission cause kare. Ek generation mein:

neutrons produced per unit time $\propto ν \cdot p_{f} n_{density} σ_{f} \cdot V$ — saare size-independent constants ke clump ko $a$ bolte hain; toh production $= a R^{3}$ .
neutrons lost to leakage per unit time $\propto (leak speed) \cdot S$ — iske size-independent clump ko $b$ bolte hain; toh leakage $= b R^{2}$ .
neutrons lost to absorption $\propto V \propto R^{3}$ — pehle se $a$ mein fold kar liye gaye hain (yeh sirf $a$ ko thoda chhota karta hai, kyunki yeh same volume neutrons ke liye compete karta hai; Step 3 dekho).

Next-generation neutrons ka this-generation neutrons se ratio — jo exactly Step 2 ke hisaab se $k$ hai — hai (woh neutrons jo fission karne jaate hain) divided by (woh neutrons jo humne shuru kiye, yaani fission + leakage se consume hue). Parent note jo toy model use karta hai, usmein yeh volume production over size-breaking loss tak reduce ho jaata hai:

$k (R) = \frac{volume production}{surface loss} = \frac{a R ^{3}}{b R ^{2}} = \frac{a}{b} R$

toh har symbol ab earned hai, guess nahi kiya gaya:

$a$ = fission production strength = $ν \times (fission probability per neutron) \times (nuclei per unit volume n) \times (fission cross-section σ_{f})$ , pehle se un absorption se reduce kiya gaya jo same neutrons ke liye compete karte hain. Bada $ν$ , denser fuel, bada $σ_{f}$ → bada $a$ . ( $σ_{f}$ ke liye Neutron Cross-section dekho.)
$b$ = leakage strength = kitni tezi se neutrons unit surface se bahar stream karte hain. Leakier geometry → bada $b$ .
$R$ = radius — ek size knob.

PICTURE. Ek straight line $k = (a / b) R$ origin se upar jaati hai, lekin ceiling $k = ν$ par cap hoti hai Step 1 se (tum jitne born hue the usase zyada survive nahi kar sakte). $k = 1$ par ek horizontal dashed line break-even mark karti hai; crossing $R_{c}$ hai (red dot).

$k = 1$ set karo (production exactly leakage loss ko balance kare):

$\frac{a}{b} R_{c} = 1 ⟹ R_{c} = \frac{b}{a}$

Step 6 — Har case walk karna (poori line, uski ceiling samet)

KYA. Teeno regimes check karo taaki koi bhi reader koi unseen scenario na dekhe — aur top par saturation bhi.

KYUN. Contract: har case cover karo. Ek line ke $k = 1$ ke relative teen regions hain (neeche, par, upar), har ek real physical situation hai — plus physical ceiling $k \leq ν$ jo linear toy formula otherwise bade $R$ ke liye violate kar deta.

PICTURE. Wahi line, teen zones mein shaded (red-under-1 = marta hai, dot at 1 = steady, black-over-1 = badhta hai), aur ab $k = ν$ par flatten hone ke liye bend over karti hai — kyunki jab leakage negligible ho jaati hai, har surviving neutron ek aisi fission se aaya jo $ν$ banaye, aur tum uss se exceed simply nahi kar sakte.

Region	$k$	Generations par neutron population	Naam
$R < R_{c}$	$k < 1$	shrinks $N_{n} = N_{0} k^{n} \to 0$	subcritical (ek dud)
$R = R_{c}$	$k = 1$	constant	critical (reactor)
$R > R_{c}$	$1 < k \leq ν$	badhta hai $N_{n} = N_{0} k^{n}$	supercritical (bomb / startup)

Yahaan $N_{n} = N_{0} k^{n}$ , $n$ generations ke baad neutron count hai, $N_{0}$ se shuru karke. Power kyun? Har generation $k$ se multiply hoti hai (Step 2), aur same factor se $n$ baar multiply karna hi $n$ -th power tak raise karna hai.

Step 7 — Do real-world dials jo picture predict karta hai

KYA. Wahi $S / V$ picture instantly explain karta hai ki bomb-makers spheres aur compression kyun use karte hain, aur $M_{c} \propto 1/ ρ^{2}$ kyun hai.

KYUN. Ek achhi derivation ko payoff dena chahiye. Dono engineering tricks sirf "leakage fraction chhoti karo" hain — Step 4 se directly padhi gayi.

PICTURE. Left: equal volume ka ek cube aur ek sphere — cube extra red surface dikhata hai (zyada leakage → zyada mass chahiye). Right: ek sphere same mass par chhote $R$ mein squeeze kiya gaya (density up), red rim thinner.

Sphere kyun? Ek given volume wali sabhi shapes mein, sphere ki sabse chhoti surface hoti hai. Chhoti surface → chhoti leakage fraction → chhota lump $k = 1$ reach karta hai → chhota critical mass.

Mean-free-path idea ke liye Neutron Cross-section dekho, aur control rods exactly $k = 1$ par kaise baithe hain iske liye Nuclear Reactor dekho.

Ek-picture summary

Ek canvas par sab kuch: volume mein born neutrons ( $\propto R^{3}$ , black), surface se leak karte hue ( $\propto R^{2}$ , red), resulting rising line $k = (a / b) R$ jo apni ceiling $ν$ par flatten ho jaati hai, aur crossing point $R_{c}$ jo marta hai ko badhta hai se alag karta hai.

Recall Feynman retelling (kisi dost ko yeh sunao)

Ek neutron uranium nucleus ko split karta hai aur bahar aate hain kareeb $ν \approx 2.5$ naye neutrons. Woh aur nuclei split kar sakte hain — ek chain. Chain jiye ya mare yeh ek number hai, $k$ : kitne agli fission cause karne tak survive karte hain. Aur $k$ sirf losses se trim kiya hua $ν$ hai: $k = ν \times (surviving fraction)$ . Ab, neutrons lump ke andar har jagah bante hain — woh volume hai, $\propto R^{3}$ — aur woh har jagah nigale bhi ja sakte hain (absorption, woh bhi $\propto R^{3}$ ); kyunki woh dono same tarah scale karte hain, unka muqabla sirf material par depend karta hai, size par nahi. Woh ek loss jo size ki parwah karta hai woh hai skin se escape karna, $\propto R^{2}$ . Toh jaise tum lump bada karte ho, escaping fraction $1/ R$ ki tarah simar jaata hai: tiny lump, almost sab escape ho jaate hain, chain marta hai ( $k < 1$ ); itna bada lump, production jeet jaata hai ( $k > 1$ ). Bilkul beech mein ek exact size hai jahaan woh balance hote hain ( $k = 1$ ): critical radius, aur iska mass hai critical mass. Khas baat, $k$ forever grow nahi kar sakta — woh $ν$ par top out hota hai, kyunki tum jitne born hue usase zyada neutrons kabhi nahi rakh sakte. Spheres aur squeezing help karte hain kyunki dono leaky skin ko chhota karte hain: sphere ka volume ke liye sabse kam skin hota hai, aur squeezing neutron ke hops chhote karta hai isliye $R_{c} \propto 1/ ρ$ , aur $M_{c} \propto 1/ ρ^{2}$ banta hai.

Recall Quick self-test

Leakage fraction $1/ R$ kyun scale karta hai? ::: Leakage $\propto$ surface $\propto R^{2}$ ; production $\propto$ volume $\propto R^{3}$ ; ratio $= R^{2} / R^{3} = 1/ R$ . Absorption ko constant $a$ mein kyun fold kiya ja sakta hai? ::: Absorption volume $\propto R^{3}$ scale karta hai, exactly production ki tarah, isliye yeh kabhi koi size threshold create nahi karta — yeh sirf effective production strength $a$ ko reduce karta hai. Toy model mein $k (R) = \frac{a}{b} R$ , critical radius kya hai? ::: $R_{c} = b / a$ , $k = 1$ set karke nikala gaya. $k$ ki physical ceiling kya hai, aur kyun? ::: $k \leq ν$ : tum born hue se zyada surviving neutrons le hi nahi sakte, aur har fission sirf $ν \approx 2.5$ hi banati hai. Fuel compress karna critical mass ko $1/ ρ^{2}$ ki tarah kyun lower karta hai? ::: $R_{c} \propto λ \propto 1/ ρ$ , toh volume $\propto 1/ ρ^{3}$ aur $M_{c} = ρ \cdot$ volume $\propto 1/ ρ^{2}$ . Kya $k$ aur $ν$ same hain? ::: Nahi — $ν$ neutrons hai jo har fission mein born hote hain; $k = ν \times (surviving fraction)$ woh hain jo agli fission cause karne tak survive karte hain.

Related building blocks: Binding Energy per Nucleon Curve (energy release kyu hoti hai), E = mc² (mass defect ko energy mein convert karna), Radioactive Decay and Half-life (delayed-neutron timing), aur doosri taraf Nuclear Fusion.