Nuclear structure — protons, neutrons, nuclear forces: Exercises

2.3.18 · D4 · Physics › Modern Physics › Nuclear structure — protons, neutrons, nuclear forces

Inki poori derivations parent note mein hain Nuclear structure — protons, neutrons, nuclear forces. Yahan hum sirf inhe use karte hain.

Level 1 — Recognition

L1·Q1 — Symbol padho

$_{20}^{40} Ca$ ke liye $Z$ , $A$ , aur $N$ (neutrons ki sankhya) batao.

Recall Solution

Symbols ka MATLAB kya hai. $_{Z}^{A} X$ mein neeche wala number $Z$ = protons hai, upar wala number $A$ = kul nucleons hai.

$Z = 20$ (protons).
$A = 40$ (nucleons).
$N = A - Z = 40 - 20 = 20$ (neutrons). Calcium-40 nucleus mein $20$ protons + $20$ neutrons hote hain.

L1·Q2 — Relationship dhundo

Neeche diye pair mein kaun se isotopes hain, aur kaun se isobars? $_{1}^{3} H,_{2}^{3} He,_{1}^{2} H .$

Recall Solution

Isotopes = same $Z$ (same element, alag mass). $_{1}^{3} H$ aur $_{1}^{2} H$ dono ka $Z = 1$ hai → isotopes. Isobars = same $A$ (same mass number, alag element). $_{1}^{3} H$ aur $_{2}^{3} He$ dono ka $A = 3$ hai → isobars.

Level 2 — Application

L2·Q1 — Nucleus ka radius

$_{13}^{27} Al$ ka radius nikalo.

Recall Solution

YE formula kyun. Nuclear matter ki density constant hoti hai, isliye volume ∝ nucleons ki sankhya, jisse $R = R_{0} A^{1/3}$ milta hai. Humein sirf $A$ chahiye. $R = R_{0} A^{1/3} = 1.2 fm \times 2 7^{1/3} .$ Ab $2 7^{1/3} = 3$ (kyunki $3^{3} = 27$ ). To $R = 1.2 \times 3 = 3.6 fm = 3.6 \times 1 0^{- 15} m .$

L2·Q2 — Radii ka ratio

$^{125} Te$ ( $A = 125$ ) ka radius $^{27} Al$ ( $A = 27$ ) se kitne factor bada hai?

Recall Solution

Ratio kyun easy hai. $R_{0}$ cancel ho jaata hai, isliye sirf $A$ bachta hai: $\frac{R _{125}}{R _{27}} = (\frac{125}{27})^{1/3} = \frac{12 5 ^{1/3}}{2 7 ^{1/3}} = \frac{5}{3} \approx 1.67.$ Tellurium nucleus kareeban $1.67 \times$ bada hai — nahi $125/27 \approx 4.6 \times$ , kyunki radius cube root follow karta hai, $A$ directly nahi.

L2·Q3 — $_{2}^{4} He$ ki binding energy

$M (^{4} He) = 4.002603 u$ diya hai, total binding energy nikalo.

Recall Solution

Step 1 — parts ka sum. Helium-4 mein $Z = 2$ , $N = 2$ hai. Atomic masses use karte hue (taaki electrons cancel ho jaayein): $2 m_{p} + 2 m_{n} = 2 (1.007825) + 2 (1.008665) = 2.015650 + 2.017330 = 4.032980 u .$ Step 2 — mass defect. Bound nucleus halka hota hai: $Δ m = 4.032980 - 4.002603 = 0.030377 u .$ Step 3 — energy mein convert karo. $E_{B} = Δ m \times 931.5 MeV$ : $E_{B} = 0.030377 \times 931.5 \approx 28.3 MeV .$ Yahi energy lagti hai ek helium nucleus (alpha particle) ko 4 alag nucleons mein todne ke liye.

Level 3 — Analysis

L3·Q1 — Nuclear density $A$ se independent hai

Algebraically dikhao ki $ρ$ , $A$ par depend nahi karta, phir uski value nikalo.

Recall Solution

Kyun cancel hona chahiye. Mass ∝ $A$ (har nucleon ≈ same mass). Volume ∝ $A$ ( $R^{3} \propto A$ se). Ratio → koi $A$ nahi. $ρ = \frac{mass}{volume} = \frac{A m _{n}}{\frac{4}{3} π R ^{3}} = \frac{A m _{n}}{\frac{4}{3} π R _{0}^{3} A} = \frac{m _{n}}{\frac{4}{3} π R _{0}^{3}} .$ $A$ divide ho jaata hai — har nucleus ke liye same. Numbers daalo ( $m_{n} \approx 1.67 \times 1 0^{- 27}$ kg, $R_{0} = 1.2 \times 1 0^{- 15}$ m): $ρ = \frac{1.67 \times 1 0 ^{- 27}}{\frac{4}{3} π ( 1.2 \times 1 0 ^{- 15} ) ^{3}} \approx 2.3 \times 1 0^{17} kg/m^{3} .$

L3·Q2 — Contact par Coulomb vs strong

Do protons bas touch kar rahe hain, centre-to-centre separation $r = 2.0 fm$ hai. Coulomb repulsion energy MeV mein nikalo. (Yeh dikhata hai ki strong force ko kaun sa "hill" paar karna hai.)

Figure — Nuclear structure — protons, neutrons, nuclear forces

Recall Solution

Yeh energy kyun. $+ e$ charge ke do particles jo $r$ door hain, unki electrostatic potential energy $U = \frac{k e ^{2}}{r}$ hoti hai (Coulomb's law and electrostatic repulsion se). Hum ise MeV mein chahte hain taaki binding energies (~MeV scale) se compare kar sakein. $U = \frac{( 8.99 \times 1 0 ^{9} ) ( 1.602 \times 1 0 ^{- 19} ) ^{2}}{2.0 \times 1 0 ^{- 15}} J .$ Numerator: $(1.602 \times 1 0^{- 19})^{2} = 2.566 \times 1 0^{- 38}$ , times $8.99 \times 1 0^{9} = 2.307 \times 1 0^{- 28}$ . $2.0 \times 1 0^{- 15}$ se divide karo: $U = 1.15 \times 1 0^{- 13} J$ . MeV mein convert karo ( $1 MeV = 1.602 \times 1 0^{- 13} J$ ): $U = \frac{1.15 \times 1 0 ^{- 13}}{1.602 \times 1 0 ^{- 13}} \approx 0.72 MeV .$ To contact par repulsion ek MeV se kam hai, jabki strong attraction har bond mein kai MeV deta hai — strong force jeetta hai, bilkul jaisa parent note ne claim kiya tha.

L3·Q3 — Mediator mass se Yukawa range

Strong force ek pion exchange karke kaam karta hai jiska mass $m_{π} c^{2} = 140 MeV$ hai. Range $r \sim \frac{ℏ}{m _{π} c}$ estimate karo aur check karo ki yeh kuch fm ke paas aata hai ya nahi.

Recall Solution

Yeh estimate kyun. Heisenberg uncertainty principle se, ek mediator sirf $Δ t \sim ℏ/Δ E$ time ke liye energy $Δ E \sim m_{π} c^{2}$ "borrow" kar sakta hai. Us time mein, ≈ $c$ speed se chalta hua, yeh $r \sim c Δ t = ℏ/ (m_{π} c)$ cover karta hai. Mass ko SI energy mein convert karo: $140 MeV = 140 \times 1.602 \times 1 0^{- 13} = 2.243 \times 1 0^{- 11} J$ . Phir $m_{π} c = \frac{m _{π} c ^{2}}{c} = \frac{2.243 \times 1 0 ^{- 11}}{3.0 \times 1 0 ^{8}} = 7.48 \times 1 0^{- 20} kg\cdotm/s$ . $r = \frac{ℏ}{m _{π} c} = \frac{1.055 \times 1 0 ^{- 34}}{7.48 \times 1 0 ^{- 20}} \approx 1.41 \times 1 0^{- 15} m \approx 1.4 fm .$ Zyaada bhaari mediator shorter range deta; massless mediator (photon) infinite range deta — isliye EM hamesha reach karta hai lekin strong force kuch fm pe cut off ho jaata hai.

Level 4 — Synthesis

L4·Q1 — Neutrons glue hain, deadweight nahi

Quantitatively-flavoured reasoning explain karo ki heavy stable nuclei mein $N > Z$ kyun hota hai. Phir check karo: $_{82}^{208} Pb$ mein $N > Z$ hai? Kitna zyaada?

Recall Solution

Physics. Coulomb repulsion har proton pair par sum hota hai, isliye yeh $Z^{2}$ ki tarah badhta hai (fast). Strong attraction saturate hoti hai (har nucleon sirf neighbours se bond karta hai), isliye binding sirf ~ $A$ ki tarah badhti hai (slow). Jaise $Z$ badhta hai, repulsion attraction ko outpace karne lagti hai. Extra neutrons strong attraction add karte hain bina charge add kiye, balance restore karte hain — isliye heavy stable nuclei $N > Z$ ki taraf drift karti hain. Lead-208 check karo: $Z = 82$ , $A = 208$ , to $N = A - Z = 208 - 82 = 126.$ To $N - Z = 126 - 82 = 44$ . Sach mein $N > Z$ hai, $44$ se — ek bada neutron excess, bilkul jaisa predict kiya tha.

L4·Q2 — Binding energy per nucleon comparison

Deuteron $^{2} H$ ( $E_{B} = 2.22 MeV$ ) aur $^{4} He$ ( $E_{B} = 28.3 MeV$ L2·Q3 se) ke liye $E_{B} / A$ compute karo. Kaun zyaada tightly bound hai, aur fusion ke liye yeh kyun matter karta hai?

Recall Solution

Per-nucleon kyun. Total $E_{B}$ sirf zyaada parts hone ki wajah se bade nuclei ko favour karta hai. Stability fairly compare karne ke liye, $A$ se divide karo (dekho Mass defect and binding energy curve).

Deuteron: $E_{B} / A = 2.22/2 = 1.11 MeV per nucleon$ .
Helium-4: $E_{B} / A = 28.3/4 = 7.07 MeV per nucleon$ . Helium-4 per nucleon bahut zyaada tightly bound hai (~ $7$ vs ~ $1$ MeV). Kyun matter karta hai: light nuclei (jaise deuterons) ko $^{4} He$ ki taraf fuse karna nucleons ko ek gehra binding well mein le jaata hai, difference energy ke roop mein release hoti hai — yahi Nuclear fission and fusion (stellar fusion) ka basis hai.

Level 5 — Mastery

L5·Q1 — Radius se ek testable prediction tak

Ek nucleus ka radius $R = 4.8 fm$ hai. (a) Uska mass number $A$ estimate karo. (b) Predict karo ki yeh $N = Z$ ke kis side par likely hai, aur ek plausible $Z$ do. (c) Uski density ka sanity-check karo.

Recall Solution

(a) Radius law invert karo. $R = R_{0} A^{1/3}$ se, $A = (\frac{R}{R _{0}})^{3} = (\frac{4.8}{1.2})^{3} = 4^{3} = 64.$ To $A \approx 64$ . (b) $N = Z$ ke kis side? $A = 64$ ek mid-heavy nucleus hai, jahan Coulomb repulsion significant hai, isliye hum expect karte hain ek modest neutron excess $N ≳ Z$ . Ek real example hai $_{30}^{64} Zn$ : $Z = 30$ , $N = 34$ — sach mein $N > Z$ hai $4$ se, jo "bhaare nuclei ko spacer neutrons chahiye" reasoning se match karta hai. (c) Density check. Density same $\sim 2.3 \times 1 0^{17} kg/m^{3}$ honi chahiye jaise har nucleus ki — kyunki $ρ$ , $A$ se independent hai (L3·Q1). Koi nayi computation nahi chahiye; agar tumne actually $\frac{A m _{n}}{\frac{4}{3} π R ^{3}}$ compute kiya to wahi number milega.

L5·Q2 — Poori chain: mass defect → energy → mass-energy consistency

$_{3}^{7} Li$ ke liye, $M (^{7} Li) = 7.016004 u$ diya hai: (a) mass defect nikalo, (b) total binding energy, (c) $E_{B} / A$ , aur (d) mass defect ko kilograms mein convert karo aur $E = m c^{2}$ se confirm karo ki joules mein wahi energy milti hai.

Recall Solution

Lithium-7: $Z = 3$ , $N = 7 - 3 = 4$ . (a) Parts ka sum, phir defect. $3 m_{p} + 4 m_{n} = 3 (1.007825) + 4 (1.008665) = 3.023475 + 4.034660 = 7.058135 u .$ $Δ m = 7.058135 - 7.016004 = 0.042131 u .$ (b) Binding energy. $E_{B} = 0.042131 \times 931.5 \approx 39.2 MeV .$ (c) Per nucleon. $\frac{E _{B}}{A} = \frac{39.2}{7} \approx 5.6 MeV per nucleon .$ (d) $E = m c^{2}$ se independent check (dekho Einstein mass-energy equivalence E=mc^2). Defect ko kg mein convert karo ( $1 u = 1.6605 \times 1 0^{- 27}$ kg): $Δ m = 0.042131 \times 1.6605 \times 1 0^{- 27} = 6.996 \times 1 0^{- 29} kg .$ Phir $E_{B} = Δ m c^{2} = 6.996 \times 1 0^{- 29} \times (3.0 \times 1 0^{8})^{2} = 6.30 \times 1 0^{- 12} J .$ Wapas convert karo: $6.30 \times 1 0^{- 12} / (1.602 \times 1 0^{- 13}) \approx 39.3 MeV$ ✔ — part (b) se rounding ke andar match karta hai. Dono routes agree karte hain, yeh confirm karta hai ki $931.5 MeV/u$ shortcut actually sirf $E = m c^{2}$ hi hai disguise mein.

Recall Self-test: yeh memory se bharo

Radius law aur $R_{0}$ ki value ::: $R = R_{0} A^{1/3}$ , $R_{0} = 1.2$ fm Nuclear density constant kyun hai ::: mass ∝ $A$ aur volume ∝ $A$ , isliye $A$ cancel ho jaata hai Mass defect ki definition ::: $Δ m = [Z m_{p} + N m_{n}] - M_{nucleus}$ Conversion factor ::: $1 u \cdot c^{2} = 931.5$ MeV Heavy nuclei ko $N > Z$ kyun chahiye ::: Coulomb $Z^{2}$ ki tarah badhta hai, strong force saturate hoti hai (∝ $A$ ), isliye extra neutrons binding add karte hain bina charge ke