Worked examples — Fission — chain reaction, critical mass

2.3.23 · D3 · Physics › Modern Physics › Fission — chain reaction, critical mass

Shuru karne se pehle, teen symbols jo tumhare paas hone chahiye (yeh sab sirf plain counting hai):

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The scenario matrix

Har fission problem in cells mein se ek hai. Neeche ke examples us cell ke saath tagged hain jo woh hit karte hain.

Cell	Kya vary karta hai	Example
A — energy per event	ek fission, $Q=\Delta m{c}^2$	Ex 1
B — many events / power	$N$ fissions/s from power $P$	Ex 2
C — subcritical $k<1$	reaction marti hai, count decay karta hai	Ex 3
D — critical $k=1$	steady state, count constant	Ex 3 (limit)
E — supercritical $k>1$	count badhta hai, doubling time	Ex 4
F — degenerate $k=0$	har neutron lost, one-shot	Ex 5
G — geometry & threshold	critical radius $R_c$, sign of $k-1$	Ex 6
H — limiting behaviour	$R\to \infty$ aur $R\to 0$	Ex 6 (limits)
I — real-world word problem	fuel burn-up over a day	Ex 7
J — exam twist	density/compression, $M_c\propto 1/{\rho }^2$	Ex 8

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Example 1 — Cell A: ek single fission ki energy

Step 1. Conversion $1u\cdot c^2=931.5$ MeV use karo. $Q=\Delta m{c}^2=0.2146\times 931.5\text{MeV}\approx 200\text{MeV}.$

Yeh step kyun? Mass aur energy ek hi currency hain (Mass-Energy Equivalence E=mc^2); ek atomic mass unit ke liye "exchange rate" exactly $931.5$ MeV hai, to hum bas multiply karte hain.

Step 2. MeV ko joules mein convert karo $1\text{MeV}=1.6\times 10^{-13}$ J use karke. $Q=200\times 1.6\times 10^{-13}\text{J}=3.2\times 10^{-11}\text{J}.$

Yeh step kyun? Joules tab chahiye hote hain jab hum power (watts $=$ joules/second) ko Ex 2 mein mix karte hain.

Verify: $0.2146\times 931.5=199.9\approx 200$ MeV. Sanity: parent note ne Binding Energy per Nucleon Curve ($235\times 0.9$ MeV) se $\approx 200$ MeV derive ki thi. Independent route, same answer. ✓

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Example 2 — Cell B: power se fissions per second tak

Step 1. Fissions per second $=\frac{P}{Q}$. $\frac{P}{Q}=\frac{3000\times 10^6\text{J/s}}{3.2\times 10^{-11}\text{J}}=9.375\times 10^{19}\text{fissions/s}.$

Yeh step kyun? Total energy rate ÷ energy per event $=$ events per rate. Units cancel ho ke $1/\text{s}$ aata hai.

Step 2. Fissions per day $=9.375\times 10^{19}\times 86400\text{s}=8.1\times 10^{24}$.

Yeh step kyun? Ek din $=86400$ s; ek per-second rate ko seconds se multiply karne par raw count milta hai.

Step 3. Har fissioned nucleus ka weight $235u=235\times 1.66\times 10^{-27}$ kg $=3.90\times 10^{-25}$ kg hai. Mass burned: $m=8.1\times 10^{24}\times 3.90\times 10^{-25}\text{kg}\approx 3.16\text{kg}.$

Yeh step kyun? Count × mass-per-nucleus $=$ total mass. Ek bade reactor ke liye roz lagbhag 3 kg uranium — yeh ek genuinely useful sanity anchor hai.

Verify: $9.375\times 10^{19}$ $6\times 10^{23}$ se kaafi kam hai, to per second ek mole se bhi kam nuclei split hote hain. Mass $\approx 3.2$ kg/day $\sim 3$ GW plant ke liye textbook figure hai. ✓

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Example 3 — Cells C & D: subcritical aur critical bookkeeping

Step 1 (a). $N_n=N_0{k}^n$ use karo (har generation $k$ se multiply hoti hai). $N_{200}=10^6\times 0.98^{200}.$

Yeh step kyun? Yeh parent ka exponential-growth law hai; $k<1$ ke saath yeh exponential decay ban jaata hai.

Step 2 (a). $0.98^{200}=e^{200\ln 0.98}=e^{200\times (-0.02020)}=e^{-4.04}\approx 0.0176$ compute karo. $N_{200}\approx 10^6\times 0.0176\approx 1.76\times 10^4.$

Yeh step kyun? Logs ek bade power ko ek easy exponential mein convert kar dete hain — Ex 4 mein same trick forward direction mein use hogi.

Step 3 (b). $k=1$ ke saath: $N_n=10^6\times 1^n=10^6$ har $n$ ke liye.

Yeh step kyun? Yeh exactly critical state hai: production aur loss balance karte hain, isliye population freeze ho jaati hai — yeh reactor ka normal operating point hai.

Verify: $0.98^{200}=0.01758$, giving $N_{200}=17,580$ — factor $\approx 57$ ki drop, yani subcritical khatam ho jaata hai. $k=1$ ke liye, $1^{200}=1$, count unchanged. ✓

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Example 4 — Cell E: supercritical growth aur doubling time

Step 1 (a). $k^n=1000$ chahiye. Logs lo (isi liye hum logarithms use karte hain — yeh ek exponent ko extract karte hain): $n=\frac{\ln 1000}{\ln k}=\frac{6.908}{\ln 1.05}=\frac{6.908}{0.04879}\approx \mathrm{141.6.}$

Yeh step kyun? $n$ ek exponent mein trapped hai; use nikaalane ka ek hi tool hai logarithm — woh function jo exponentiation ko undo karta hai.

Step 2 (b). Real time $=n\tau =141.6\times 10^{-8}\text{s}\approx 1.42\times 10^{-6}$ s.

Yeh step kyun? Generations × time-per-generation $=$ elapsed time. Lagbhag ek microsecond — koi bhi insaan ya machine intervene nahi kar sakta.

Step 3 — contrast. Ek reactor mein effective $\tau$ delayed neutrons ki wajah se $\sim 0.1$ s tak stretch ho jaata hai, isliye same growth mein $\sim 14$ s lagenge: controllable. (Dekho Nuclear Reactor.)

Yeh step kyun? Yeh dikhata hai ki same math ek alag $\tau$ ke saath bomb ko power plant se alag kar deta hai.

Verify: $\ln 1000/\ln 1.05=141.6$; $141.6\times 10^{-8}=1.416\times 10^{-6}$ s. Microsecond regime confirmed. ✓

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Example 5 — Cell F: degenerate case $k=0$

Step 1. Generation-0 neutrons $500$ tak fissions cause karte hain, lekin $k=0$ matlab generation 1 mein $N_1=N_0\times k=500\times 0=0.$

Yeh step kyun? $k=0$ formula $N_n=N_0{k}^n$ ki boundary hai: chain ek step ke baad cut off ho jaati hai. Yeh no-chain limit hai.

Step 2. Total fissions $\approx N_0=500$ (sirf ek generation), phir yeh ruk jaata hai.

Yeh step kyun? Koi surviving neutron nahi hai to generation 1 nahi hai — reaction ek single flash hai, chain nahi. Yeh hai jo "subcritical" apni extreme par degenerate ho jaata hai.

Verify: $500\times 0^1=0$ aur ${\sum }_{n\ge 1}{N}_0{k}^n=0$ — ratio $0$ waali geometric series generation 0 ke baad kuch nahi sum karti. Lump inert hai. ✓

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Example 6 — Cells G & H: critical radius aur dono limits (figure ke saath)

Step 1 (a). Critical matlab $k=1$. $\frac{a}{b}{R}_c=1$ set karo: $R_c=\frac{b}{a}=\frac{12}{1}=12\text{cm}.$

Yeh step kyun? $k=1$ growing aur dying ke beech ka razor's edge hai; wahan $R$ ke liye solve karne par threshold milta hai — critical mass ki geometric origin.

Step 2 (b). $k(8)=\frac{1}{12}\times 8=0.667<1$ → subcritical (figure mein dashed line ke baaye blue point dekho). $k(20)=\frac{1}{12}\times 20=1.667>1$ → supercritical (green point, line ke daaye).

Yeh step kyun? $k-1$ ka sign hi poori classification hai. $R_c$ ke baaye: marega. Daaye: phateega.

Step 3 (c) — limits.

• Jab $R\to \infty$: $k=\frac{a}{b}R\to \infty$. Leakage fraction $\propto R^2/R^3=1/R\to 0$; production totally jeet jaata hai. (Reality mein $k$ saturate ho jaata hai jab absorption use cap karti hai, lekin yeh kabhi 1 se neeche nahi jaata.)
• Jab $R\to 0$: $k\to 0$. Saari surface, koi volume nahi — har neutron instantly escape kar jaata hai. Yeh Ex 5 ka degenerate $k=0$ case hai, geometry ki limit ke roop mein recover hua.

Yeh step kyun? Dono extremes check karne se prove hota hai ki model $R$ mein monotonic hai: $R_c$ par exactly ek crossing hai, isliye critical mass unique hai — koi magic nahi, bas geometry.

Verify: $R_c=12$; $k(8)=0.667$, $k(20)=1.667$; ${\lim }_{R\to 0}k=0$ Ex 5 se match karta hai. ✓

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Example 7 — Cell I: real-world word problem (burn-up)

Step 1. Fissions per second $=\frac{150\times 10^6}{3.2\times 10^{-11}}=4.6875\times 10^{18}$/s.

Yeh step kyun? Same power-÷-energy logic jaise Ex 2 mein; "steady" matlab $k=1$, isliye power constant hai.

Step 2. Ek saal mein seconds $=3.156\times 10^7$. Total fissions $=4.6875\times 10^{18}\times 3.156\times 10^7=1.479\times 10^{26}$.

Yeh step kyun? Rate × total time $=$ total count. "Continuous" key word hai — koi downtime factor nahi.

Step 3. Mass $=1.479\times 10^{26}\times 3.90\times 10^{-25}\text{kg}\approx 57.7$ kg.

Yeh step kyun? Count × mass-per-nucleus. To 60 kg se kam uranium ek submarine ko ek saal tak power deta hai — nuclear energy density ka poora point yahi hai.

Verify: $4.6875\times 10^{18}\times 3.156\times 10^7\times 3.90\times 10^{-25}\approx 57.7$ kg. Order of magnitude (tens of kg/year) standard textbook result hai. ✓

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Example 8 — Cell J: exam twist (compression aur $M_c\propto 1/{\rho }^2$)

Step 1. Kyunki $M_c\propto {\rho }^{-2}$ hai, critical masses ka ratio hai $\frac{M_c^{\prime }}{M_c}={\left(\frac{\rho }{{\rho }^{\prime }}\right)}^2={\left(\frac{1}{2.5}\right)}^2=\mathrm{0.16.}$

Yeh step kyun? Compress karne se $R$ shrink hota hai (kam leakage) aur nuclei paas pack hote hain (neutrons jaldi hit karte hain). Dono effects threshold ko neeche push karte hain; combined scaling parent mein quoted inverse-square law hai.

Step 2. $M_c^{\prime }=0.16\times 10\text{kg}=1.6$ kg.

Yeh step kyun? Compressed metal ko critical hone ke liye sirf $1.6$ kg chahiye.

Step 3. Lump mein abhi bhi poora $10$ kg plutonium hai — ab apne naye $1.6$ kg threshold se kaafi upar. To yeh massively supercritical hai ($k>1$): exactly aise hi ek implosion device ek sub-critical mass ko detonate karta hai.

Yeh step kyun? Same atoms, same total mass — sirf geometry/density badli. Critical mass arrangement ke baare mein hai, jo Ex 6 ke geometric threshold ko echo karta hai.

Verify: $(1/2.5{)}^2=0.16$; $0.16\times 10=1.6$ kg $<10$ kg → supercritical. ✓

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Recall Self-check: har ek ko uski cell se match karo

Ek single split 200 MeV release karta hai — kaunsa cell? ::: Cell A (Ex 1) Woh number jo decide karta hai bomb vs reactor vs dud ::: $k$; $k>1$, $=1$, $<1$ Jab $R\to 0$, $k$ kya banta hai aur yeh pehle ke kaunse example ko reproduce karta hai? ::: $k\to 0$, Ex 5 ka degenerate one-shot (Cell F) Plutonium compress karne se subcritical lump kyun explode karta hai? ::: $M_c\propto 1/{\rho }^2$, isliye zyada density threshold ko present mass se neeche le jaati hai (Cell J) Ek reactor ${}^{235}$U roughly kitna per day burn karta hai 3 GW par? ::: Lagbhag 3 kg (Ex 2)

Related: Nuclear Reactor · Neutron Cross-section · Radioactive Decay and Half-life · Nuclear Fusion · Binding Energy per Nucleon Curve