Nuclear structure — protons, neutrons, nuclear forces: Worked examples

2.3.18 · D3 · Physics › Modern Physics › Nuclear structure — protons, neutrons, nuclear forces

Shuru karne se pehle, ek symbol reminder taaki kuch bhi bina samjhe use na ho:

$A$ = mass number = protons + neutrons ka total count (the "how many marbles" number).
$Z$ = atomic number = protons ki count (the "which element" number).
$N = A - Z$ = neutrons ki count.
$R_{0} = 1.2 fm$ jahan $1 fm = 1 0^{- 15} m$ (ek femtometre — ek trillionth ke hazarwen hisse ka ek metre).
$ρ$ = density = mass divided by volume, $ρ = \frac{mass}{volume}$ . Yeh jawab deta hai "har cubic metre mein kitna stuff packed hai?"
$c$ = speed of light $= 3.0 \times 1 0^{8} m/s$ — $E = m c^{2}$ mein mass aur energy ke beech fixed conversion rate (dekho Einstein mass-energy equivalence E=mc^2).
$u$ = atomic mass unit, mass ka yardstick jahan $1 u \cdot c^{2} = 931.5 MeV$ . Ise aise padho: "mass ki ek unit, agar $E = m c^{2}$ ke zariye poori tarah energy mein convert ho jaye, toh 931.5 million electron-volts milenge."

The scenario matrix

Nuclear structure ke baare mein har question in cells mein se ek mein aata hai. Last column us example ka naam batata hai jo use cover karta hai.

#	Case class	Kya tricky banata hai	Covered by
C1	Radius, ordinary $A$	seedha $R = R_{0} A^{1/3}$ plug-in	Ex 1
C2	Do nuclei ka Radius ratio	$R_{0}$ cancel ho jata hai — sirf ratio shortcut	Ex 2
C3	Degenerate input: $A = 1$	single proton, $A^{1/3} = 1$	Ex 3
C4	Density ( $A$ algebraically cancel hota hai)	prove karo ki $ρ$ sab $A$ ke liye same hai	Ex 4
C5	Mass defect + binding energy	atomic vs nuclear mass, electron bookkeeping	Ex 5
C6	Binding energy per nucleon	$A$ se divide karo — "stability" number	Ex 6
C7	Range of the force (uncertainty)	Yukawa $r \sim ℏ/ (m_{π} c)$ , unit juggling	Ex 7
C8	Coulomb vs strong tug-of-war	ek distance par do forces compare karo	Ex 8
C9	Real-world word problem	"nucleus ki teaspoon" mass	Ex 9
C10	Exam twist: ratio se unknown $A$ nikalna	cube-root law ko invert karo	Ex 10

Gaur karo ki yahan koi signs ya quadrants nahi hain (trig/vectors ke unlike) — nuclear counts $A, Z, N$ non-negative integers hain, aur $R, ρ, E_{B}$ hamesha positive hote hain. "Edge cases" yahan $A = 1$ par milte hain (ek akela proton, Ex 3) aur cancellations mein (Ex 2, Ex 4). Hum unhe deliberately cover karte hain.

Ex 1 — C1: Ek ordinary nucleus ki Radius

Worked example Iron

$_{26}^{56} Fe$ ki Radius Statement: Iron-56 nucleus ki radius nikalo. $R_{0} = 1.2 fm$ use karo.

Forecast: $A = 56$ . $56$ ka cube root thoda $4$ se kam hai (kyunki $4^{3} = 64$ ). Toh expect karo $R$ roughly $1.2 \times 3.8 \approx 4.6 fm$ . Aage padhne se pehle guess karo.

Step 1 — Law likho. $R = R_{0} A^{1/3}$ . Yeh step kyun? Yeh hamare paas ek-maatra radius formula hai, aur yeh "constant density → volume ∝ $A$ → radius ∝ $A^{1/3}$ " se aaya tha (parent note mein build kiya gaya). Kuch aur chahiye nahi.

Step 2 — Mass number ka cube-root lo. $5 6^{1/3} = 3.826$ . Yeh step kyun? Cube root poori physics hai — volume linearly $A$ ke saath badhta hai, lekin radius volume ka cube root hai, isliye yeh dheere badhta hai.

Step 3 — $R_{0}$ se multiply karo. $R = 1.2 \times 3.826 = 4.59 fm$ . Yeh step kyun? $R_{0}$ single-nucleon blob ki radius hai; ise $A^{1/3}$ se scale karne par yeh poore nucleus tak inflate ho jaata hai.

Verify: $4.59 fm = 4.59 \times 1 0^{- 15} m$ — femtometre-scale, jaise har nucleus hona chahiye. Yeh hamare forecast ( $\approx 4.6$ ) se match karta hai. ✔

Ex 2 — C2: Radius ratio ( $R_{0}$ cancel ho jaata hai)

Worked example

$^{216}$ Rn, $_{13}^{27}$ Al se kitna bada hai? Statement: $A = 216$ nucleus aur $A = 27$ nucleus ki radii compare karo.

Forecast: Yeh parent note ka forecast-verify problem hai. $216/27 = 8$ , aur $8 = 2^{3}$ , toh cube root exactly $2$ hai. Guess: bada wala exactly do guna wide hai.

Step 1 — Dono radii likho. $R_{216} = R_{0} (216)^{1/3}$ aur $R_{27} = R_{0} (27)^{1/3}$ . Yeh step kyun? Inhe side-by-side likhne se shared $R_{0}$ reveal hota hai.

Step 2 — Divide karo. $\frac{R _{216}}{R _{27}} = \frac{R _{0} ( 216 ) ^{1/3}}{R _{0} ( 27 ) ^{1/3}} = (\frac{216}{27})^{1/3} .$ Yeh step kyun? Unknown-ish constant $R_{0}$ cancel ho jaata hai — toh hume kabhi uski numerical value chahiye hi nahi. Isliye ratio questions absolute questions se aasaan hote hain.

Step 3 — Evaluate karo. $(\frac{216}{27})^{1/3} = 8^{1/3} = 2$ . Yeh step kyun? Ratio ko pehle single number $8$ mein convert karna, phir cube-root lena, messy decimals se bachata hai.

Verify: Radius $A^{1/3}$ ke saath scale karti hai, toh $8 \times$ bhaari nucleus $8^{1/3} = 2 \times$ wider hota hai. Volume ratio $8$ hoga, jo "8× nucleons 8× volume mein pack hote hain" se match karta hai. ✔

Ex 3 — C3: Degenerate case $A = 1$ (ek akela proton)

Worked example Ek single proton (hydrogen-1 nucleus) ki Radius

Statement: $R = R_{0} A^{1/3}$ $_{1}^{1} H$ ke liye kya predict karta hai, jiska nucleus ek single proton hai?

Forecast: $A = 1$ , aur $1^{1/3} = 1$ , toh formula bas $R = R_{0}$ dega. Guess: lagbhag $1.2 fm$ .

Step 1 — $A = 1$ substitute karo. $R = R_{0} \cdot 1^{1/3} = R_{0} = 1.2 fm$ . Yeh step kyun? Yeh degenerate limit hai — sabse chhota possible nucleus. Yeh dikhata hai ki $R_{0}$ literally ek nucleon ki effective radius hai.

Step 2 — Meaning ka reality check karo. Measured proton "charge radius" $\approx 0.84 fm$ hai. Yeh step kyun? Hamara formula ek packing model hai — yeh ek akele particle ke liye thoda overshoot karta hai kyunki ek akela marble neighbours ke against "pack" nahi karta. Yeh law bohot-saare-nucleon nuclei ke liye hai; $A = 1$ par yeh sirf ek rough guide hai.

Verify: Formula apne sabse chhote input par sensibly behave karta hai — yeh $R_{0}$ return karta hai, zero ya infinity nahi — aur measured value ( $\sim 1 fm$ ) ke ballpark mein aata hai. Koi blow-up nahi, koi nonsense nahi. ✔

Ex 4 — C4: Density — jahan $A$ algebraically cancel ho jaata hai

Worked example Dikhao ki har nucleus ki density same hoti hai

Statement: Nuclear density $ρ = \frac{mass}{volume}$ compute karo aur prove karo ki yeh $A$ par depend nahi karta.

Forecast: Mass $A$ ki tarah badhti hai; volume $R^{3} \propto A$ ki tarah badhta hai. $A$ ki same power upar aur neeche hai, toh $A$ divide out ho jaata hai (yeh cancel hota hai — yeh zero nahi jaata) → density ek constant hai. Number guess karo: around $2 \times 1 0^{17} kg/m^{3}$ .

Step 1 — Mass aur volume ko $A$ ke terms mein likho. $mass \approx A m_{n}, V = \frac{4}{3} π R^{3} = \frac{4}{3} π (R_{0} A^{1/3})^{3} = \frac{4}{3} π R_{0}^{3} A .$ Yeh step kyun? Hum jaanbhoojhkar $A$ ko dono mein visible rakhte hain taaki hum use cancel hote dekh sakein. Note karo $(A^{1/3})^{3} = A$ — cube, cube root ko undo karta hai.

Step 2 — Density banao. $ρ = \frac{A m _{n}}{\frac{4}{3} π R _{0}^{3} A} = \frac{m _{n}}{\frac{4}{3} π R _{0}^{3}} .$ Yeh step kyun? Numerator aur denominator mein $A$ divide out ho jaata hai — yahi punchline hai. Bada ya chhota nucleus, same density.

Step 3 — Numbers plug karo. $m_{n} = 1.675 \times 1 0^{- 27} kg$ , $R_{0} = 1.2 \times 1 0^{- 15} m$ ke saath: $ρ = \frac{1.675 \times 1 0 ^{- 27}}{\frac{4}{3} π ( 1.2 \times 1 0 ^{- 15} ) ^{3}} \approx 2.31 \times 1 0^{17} kg/m^{3} .$ Yeh step kyun? Symbols ko ek number mein badalna humein feel karaata hai ki nuclear matter kitna absurdly dense hai.

Verify: Units: $kg / m^{3}$ . ✔ Value $\sim 2.3 \times 1 0^{17}$ parent note se match karta hai. Aur sabse important — koi $A$ survive nahi kiya, constant-density claim confirm ho gayi. ✔

Neeche ki figure cancellation ko visual banati hai. Yeh $ρ$ ko real nuclei ki poori range ke liye $A$ ke against plot karti hai: blue line bilkul flat hai (hydrogen, iron aur lead sab ke liye same $2.31 \times 1 0^{17} kg/m^{3}$ ), aur $^{2}$ H, $^{56}$ Fe aur $^{208}$ Pb ke orange dots sab us ek horizontal line par aate hain. Ek flat line wahi hai jo " $A$ se independent" dikhta hai — agar density $A$ ke saath badhti, yeh line upar slope karti; yeh nahi karti.

Ex 5 — C5: Mass defect & binding energy (electron bookkeeping)

Worked example Helium-4,

$_{2}^{4} He$ ki Binding Energy Statement: Atomic masses diye gaye hain $m (^{1} H) = 1.007825 u$ , $m_{n} = 1.008665 u$ , $M (^{4} He) = 4.002603 u$ , helium-4 nucleus ki binding energy nikalo.

Forecast: Helium-4 (ek alpha particle) mashoor roop se bahut tightly bound hai. Guess: total ek bada number hai, tens of MeV — shayad $28 MeV$ ke paas.

Step 1 — Parts sum karo. Helium mein $Z = 2$ protons aur $N = 2$ neutrons hain: $parts = 2 m (^{1} H) + 2 m_{n} = 2 (1.007825) + 2 (1.008665) = 4.032980 u .$ Yeh step kyun? Hum atomic hydrogen mass (proton + 1 electron) do baar use karte hain, jo 2 electrons supply karta hai jo helium ke neutral atom mein bhi hain. Toh jab hum neutral helium atom mass subtract karte hain toh electron masses cancel ho jaate hain. Saaf bookkeeping.

Step 2 — Mass defect. $Δ m = 4.032980 - 4.002603 = 0.030377 u .$ Yeh step kyun? Bound nucleus apne parts se halka hota hai; missing mass binding energy ban gayi.

Step 3 — Energy mein convert karo. $E_{B} = Δ m \cdot 931.5 = 0.030377 \times 931.5 \approx 28.30 MeV .$ Yeh step kyun? Factor $931.5 MeV/u$ hai hi $c^{2}$ convenient units mein dressed — yeh "u of mass" ko "MeV of energy" mein badalta hai $E = Δ m c^{2}$ ke zariye, jahan $c = 3.0 \times 1 0^{8} m/s$ speed of light hai.

Verify: $28.3 MeV$ hamare forecast aur $^{4}$ He ki accepted value se match karta hai. Positive hai (ise tod ne ke liye energy lagti hai), aur parent note ke deuteron ke $2.22 MeV$ se comfortably bada hai — helium-4 far more tightly bound hai. ✔

Ex 6 — C6: Binding energy per nucleon

Worked example Zyada stable kaun hai: helium-4 ya deuteron?

Statement: Stability compare karo binding energy per nucleon, $E_{B} / A$ use karke. $E_{B} (^{4} He) = 28.30 MeV$ aur $E_{B} (^{2} H) = 2.22 MeV$ use karo.

Forecast: Total binding energy compare karna fair nahi hai — helium mein zyada nucleons hain, toh obviously uska total bada hoga. Fair measure per-nucleon hai. Guess: helium badi jeet se aage hai.

Step 1 — Har ek ke liye $A$ se divide karo. $\frac{E _{B}}{A}_{^{4} He} = \frac{28.30}{4} = 7.07 MeV/nucleon, \frac{E _{B}}{A}_{^{2} H} = \frac{2.22}{2} = 1.11 MeV/nucleon .$ Yeh step kyun? $A$ se divide karna jawab deta hai "ek typical single nucleon kitna tightly held hai?" — yahi real stability yardstick hai (yeh Mass defect and binding energy curve ka $y$ -axis hai).

Step 2 — Compare karo. $7.07 > 1.11$ , toh helium-4 per nucleon far more tightly bound hai. Yeh step kyun? Per-nucleon binding unfair size advantage hata deta hai; ab comparison apples-to-apples hai.

Verify: $7.07 MeV/nucleon$ bilkul wahan baithta hai jahan binding-energy curve kehti hai ki $^{4}$ He jaise light-but-magic nuclei hote hain (apne early peak ke paas). Deuteron ka $1.11$ mashoor roop se low hai — yeh ek fragile nucleus hai. ✔

Ex 7 — C7: Nuclear force ki Range (Yukawa + uncertainty)

Worked example Pion mass se force range estimate karo

Statement: Ek pion ki rest energy $m_{π} c^{2} \approx 140 MeV$ hai. Force ki range $r \sim ℏ/ (m_{π} c)$ estimate karo jo yeh mediate karta hai.

Forecast: Parent note ne $\approx 1.4 fm$ quote kiya tha. Dekhte hain kya hum scratch se wahan pahunch sakte hain. Guess: lagbhag $1 - 2 fm$ .

Step 1 — Energy-friendly form mein rewrite karo. Upar aur neeche $c$ se multiply karo: $r \sim \frac{ℏ}{m _{π} c} = \frac{ℏ c}{m _{π} c ^{2}} .$ Yeh step kyun? Hume $m_{π} c^{2}$ MeV mein diya gaya hai, aur constant $ℏ c = 197.3 MeV \cdot fm$ ek famous "handy number" hai. Ise is tarah likhne se MeV cancel hota hai aur fm milta hai — koi messy SI conversions nahi. Yeh Heisenberg uncertainty principle kaam kar raha hai: ek bhaari borrowed particle sirf ek chhoti distance tak pahunch sakta hai.

Step 2 — Plug in karo. $r \sim \frac{197.3 MeV \cdot fm}{140 MeV} = 1.41 fm .$ Yeh step kyun? MeV units divide out ho jaate hain, fm reh jaata hai — exactly ek length. Mediator jitna bhaari, $r$ utna chhota; ek massless photon ( $m = 0$ ) $r = \infty$ deta, isliye electromagnetism ki infinite range hai.

Verify: $1.41 fm$ parent note ke $\approx 1.4 fm$ aur strong force ke observed "few-fm" cutoff se match karta hai. Units fm mein check out. ✔

Ex 8 — C8: Ek fixed distance par Coulomb vs strong tug-of-war

Worked example

$r = 1 fm$ par do protons — repulsion kitna strong hai? Statement: $r = 1.0 fm$ se alag hue do protons ke beech Coulomb repulsion compute karo. $k = 8.99 \times 1 0^{9} N\cdotm^{2} / C^{2}$ , $e = 1.602 \times 1 0^{- 19} C$ use karo. Phir comment karo ki nucleus isko kyun survive karta hai.

Forecast: Aam distances par proton–proton repulsion tiny hoti hai, lekin $1 fm$ microscopically close hai aur force $\propto 1/ r^{2}$ blow up karta hai. Guess: ek shockingly badi force — ek single proton par hundreds of newtons.

Step 1 — Coulomb's law likho. Coulomb's law and electrostatic repulsion se: $F = \frac{k e ^{2}}{r ^{2}} .$ Yeh step kyun? Dono particles charge $+ e$ carry karte hain, toh charges ka product $e^{2}$ hai; yeh law repulsive push deta hai jise hume overcome karna hai.

Step 2 — Numbers plug karo ( $r = 1.0 \times 1 0^{- 15} m$ ). $F = \frac{( 8.99 \times 1 0 ^{9} ) ( 1.602 \times 1 0 ^{- 19} ) ^{2}}{( 1.0 \times 1 0 ^{- 15} ) ^{2}} \approx 231 N .$ Yeh step kyun? Tiny $r$ ko square karne se denominator $1 0^{- 30}$ ban jaata hai — enormous division — toh force huge ho jaati hai tiny charges ke bawajood.

Step 3 — Interpret karo. $231 N$ jo $1.67 \times 1 0^{- 27} kg$ mass ke proton par act kare, yeh ek unimaginable push hai. Phir bhi nucleus hold karta hai — kyunki strong force is range par $\sim 100 \times$ stronger hai aur attractive hai. Yeh step kyun? Yahi poora "tug-of-war" numeric ban gaya: repulsion real aur large hai, lekin yeh harti hai.

Verify: $\approx 231 N$ — comparison ke liye yeh ek proton par ~23 kg ke weight jaisa hai. Force positive (repulsive) hai jaise like charges ke liye expected hai. Strong force, $\sim 100 \times$ badi aur short-ranged hoti hai, $\sim 2 fm$ ke andar jeet jaati hai. ✔

Ex 9 — C9: Real-world word problem

Worked example Pure nuclear matter ki ek teaspoon

Statement: Ek teaspoon lagbhag $5 cm^{3} = 5 \times 1 0^{- 6} m^{3}$ hota hai. Agar yeh pure nuclear matter (density $ρ = 2.31 \times 1 0^{17} kg/m^{3}$ Ex 4 se) se bhara hota, toh iska wazan kitna hota?

Forecast: Nuclear matter insanely dense hai, toh ek chamach bhi preposterous amount weigh karna chahiye — parent note ne "billions of tonnes" kaha tha. Guess: around $1 0^{12} kg$ .

Step 1 — Mass = density × volume. $m = ρ V = (2.31 \times 1 0^{17}) (5 \times 1 0^{- 6}) .$ Yeh step kyun? Density hai hi mass per volume ( $ρ = mass / volume$ ), toh volume se multiply karne par mass milti hai — definition ulti chalayi.

Step 2 — Multiply karo. $m = 1.155 \times 1 0^{12} kg \approx 1.16 \times 1 0^{9} tonnes .$ Yeh step kyun? $1 0^{17} \times 1 0^{- 6} = 1 0^{11}$ , coefficients se multiply karo ( $2.31 \times 5 \approx 11.6$ ) toh milta hai $\sim 1.16 \times 1 0^{12} kg$ . $1000$ se divide karne par kg se tonnes mein convert hota hai.

Verify: Ek teaspoon ke liye $\sim 1.16$ billion tonnes — parent note ke "billions of tonnes" se match. Units: $(kg/m^{3}) (m^{3}) = kg$ ✔. Yeh ek chhote pahaad ki mass hai, jo dikhata hai ki ek normal atom mein kitna khali space hota hai.

Ex 10 — C10: Exam twist — radius ratio se $A$ nikalna

Worked example Unknown nucleus, carbon-12 ki radius ka 3 guna hai

Statement: Ek nucleus $X$ ki radius $_{6}^{12} C$ se exactly $3$ guni hai. Uska mass number $A_{X}$ nikalo.

Forecast: Radius $A^{1/3}$ ke saath scale karti hai, toh $3 \times$ radius ka matlab hai $3^{3} = 27 \times$ mass number. Guess: $A_{X} = 27 \times 12 = 324$ .

Step 1 — Ratio likho. $\frac{R _{X}}{R _{C}} = (\frac{A _{X}}{12})^{1/3} = 3.$ Yeh step kyun? $R_{0}$ ratio mein cancel ho jaata hai (jaise Ex 2 mein), ek clean cube-root equation reh jaata hai jise invert karna hai.

Step 2 — Cube root undo karne ke liye dono sides cube karo. $\frac{A _{X}}{12} = 3^{3} = 27.$ Yeh step kyun? Cubing, cube root lene ka inverse operation hai — yeh $A_{X} /12$ isolate karta hai. Yahi "law ko invert karo" move hai jo exam twists ko pasand hai.

Step 3 — Solve karo. $A_{X} = 27 \times 12 = 324$ . Yeh step kyun? Simple multiplication se khatam hota hai.

Verify: Forward check: $(324/12)^{1/3} = 2 7^{1/3} = 3$ ✔ — diye gaye ratio aur hamare forecast se match karta hai. (Real nuclei $A \approx 250$ ke paas khatam hoti hain, toh $324$ hypothetical hai — ek accha reminder ki ek formula ko wahan se aage extrapolate kiya ja sakta hai jahan nature rukti hai.) ✔

Recall Quick self-test

$^{125}$ Te ki radius agar $R_{0} = 1.2 fm$ ho? ::: $1.2 \times 12 5^{1/3} = 1.2 \times 5 = 6.0 fm$ . Radius ratio mein $R_{0}$ kyun gayab ho jaata hai? ::: Yeh dono radii ko multiply karta hai, toh division mein cancel ho jaata hai. $^{4}$ He ki binding energy per nucleon? ::: $28.3/4 \approx 7.07 MeV/nucleon$ . Force range ke liye $ℏ c = 197.3 MeV\cdotfm$ kyun use karte hain? ::: Taaki MeV pion ki rest energy ko cancel kare aur answer seedha fm mein aaye.