Binding energy — mass defect, BE per nucleon curve: Visual walkthrough

2.3.19 · D2 · Physics › Modern Physics › Binding energy — mass defect, BE per nucleon curve

Hum ek hi sawaal follow karte hain poore raste: "Agar main nucleons ko glue karta hoon, toh weight kahaan jaata hai?"

Step 1 — Nucleon kya hota hai, aur "mass" kya hoti hai?

KYA HAI. Ek nucleus do tarah ki choti balls ka ek tight clump hai:

protons (positively charged), inhe hum letter $Z$ se count karte hain,
neutrons (koi charge nahi), inhe hum letter $N$ se count karte hain.

Saath milkar inhe nucleons kehte hain. Total count hai $A = Z + N$ , jise mass number kehte hain.

YEH LETTERS KYUN. Kuch bhi weighing karne se pehle humein names chahiye. $Z$ = "kitne protons", $N$ = "kitne neutrons", $A$ = "total kitni balls". Bas itna hi — abhi tak koi physics nahi, sirf labels.

PICTURE. Left par, balls free aur bahut door float kar rahi hain. Right par, woh ek clump mein snap ho gayi hain. Same balls, alag arrangement — yahi difference poori kahaani hai.

Figure — Binding energy — mass defect, BE per nucleon curve

Step 2 — Pehle loose parts ko weigh karo

KYA HAI. Gluing se pehle har loose ball ko scale par rakho. Ek free proton ki mass hoti hai $m_{p}$ ; ek free neutron ki mass hoti hai $m_{n}$ . Inhe sab add karo:

$M_{parts} = kitne protons Z ek proton ki mass m_{p} + kitne neutrons N ek neutron ki mass m_{n}$

KYUN. Yeh jaanne ke liye ki gluing se weight change hota hai ya nahi, humein "pehle" ka number chahiye. $M_{parts}$ alag-alag rakhi gayi ingredients ki total mass hai.

PICTURE. Ek chalk balance: left pan mein saari alag balls hain, aur pointer $M_{parts}$ read kar raha hai — har ingredient ka honest sum.

Step 3 — Taiyaar nucleus ko weigh karo — yeh lighter hai

KYA HAI. Ab balls ko actual nucleus mein glue karo aur usi ko weigh karo: ise $M_{nucleus}$ kehte hain. Experiment ek surprising baat kehta hai:

$M_{nucleus} < M_{parts}$

Assembled clump kam weighs karta hai compared to same balls ke alag-alag weight se.

KYUN GALAT LAGTA HAI PAR SAHI HAI. Koi ball gaayab nahi hui — aap count kar sakte ho, abhi bhi $Z$ protons aur $N$ neutrons hain. Phir bhi scale kam read karta hai. Missing amount ka ek naam hai, mass defect:

$Δ m = parts alag M_{parts} - clump saath M_{nucleus} = [Z m_{p} + N m_{n}] - M_{nucleus}$

Yahan $Δ$ (Greek "delta") ka matlab sirf "mein fark" hai — yeh ek subtraction hai, koi nayi cheez nahi.

PICTURE. Do balances side by side. Left pan: loose balls (bhaari, pointer neeche jhukta hai). Right pan: glued nucleus (halka, pointer upar uthta hai). Dono pointers ke beech ka chota sa gap $Δ m$ hai.

Step 4 — Kyun work + energy conservation se mass drop hona zaruri hai

KYA HAI. Socho ulta kaam: glued nucleus ko wapas free balls mein pull karo. Strong nuclear force tumse ladta hai — woh balls ko wapas kheenchta hai. Isliye tumhe kaam karna padega, system mein energy daalni padegi, inhe alag karne ke liye.

KYUN YAHI MASS KO DETERMINE KARTA HAI. Energy ko honestly track karo:

Free balls alag mein kuch total energy hoti hai $E_{parts}$ .
Bound clump ki total energy hoti hai $E_{nucleus}$ .
Clump → alag jaane ke liye tumhe energy $E_{B}$ add karni padi.

Energy conservation: $E_{nucleus} + E_{B} = E_{parts}$ , isliye $E_{B} = E_{parts} - E_{nucleus} > 0.$ Bound clump energy mein neeche baith jaata hai — woh energy well mein rehta hai. Kam energy hi wajah hai ki woh bound hai. Yeh $E_{B}$ hi binding energy hai.

PICTURE. Chalkboard par ek valley. Free nucleons upar rim par khade hain; bound nucleus well ke neeche rest kar raha hai. Vertical drop $E_{B}$ hai — neeche jaane par release hoti energy, ya woh energy jo wapas chadhne ke liye supply karni padegi.

Step 5 — Bridge: $E = m c^{2}$ energy drop ko mass drop mein convert karta hai

KYA HAI. Einstein ka relation kehta hai ki energy aur mass do alag units mein same currency hain: $E = m c^{2},$ jahan $c$ speed of light hai (ek fixed conversion number). Energy mein koi bhi change automatically mass mein change hai: $Δ E = Δ m c^{2} .$

KYUN YAHI MISSING LINK HAI. Step 3 mein mass $Δ m$ se neeche gayi. Step 4 mein energy $E_{B}$ se neeche gayi jab clump bana. Yeh same drop hai, do units mein dekhi gayi. Inhe equal karo:

$E_{B} = Δ m c^{2}$

Term by term:

$E_{B}$ = binding energy (woh energy jo system se bahar gayi),
$Δ m$ = Step 3 se mass defect (woh weight jo gaayab hua),
$c^{2}$ = exchange rate jo missing mass ke kilograms ko released energy ke joules mein convert karta hai.

PICTURE. Ek money-changer ki desk: "missing mass" $Δ m$ slot mein daalo, machine $c^{2}$ se multiply karti hai, aur bahar aati hai energy $E_{B}$ . Same value, alag currency.

Step 6 — Edge case: deuteron (sirf 2 balls)

KYA HAI. Sabse chota bound nucleus hai deuteron $_{1}^{2}$ H: ek proton, ek neutron. Chaliye machine ko sabse chota possible input dekar chalate hain.

$Δ m = m_{p}, u (1.007825 + m_{n}, u 1.008665) - M (^{2} H), u 2.014102 = 0.002388 u$ $E_{B} = 0.002388 \times 931.5 = 2.224 MeV$

KYUN EDGE CASE KE TAUR PAR MATTER KARTA HAI. Do balls matlab sirf ek bond. Ek real mass defect hai (well exist karti hai), par woh shallow hai: sirf $2.224$ MeV, aur per nucleon toh sirf $2.224/2 = 1.11$ MeV. Yeh curve ka far-left, barely-bound end hai — sabse chota non-trivial case, aur phir bhi $E_{B} = Δ m c^{2}$ follow karta hai.

PICTURE. Ek bahut shallow chalk well jisme sirf ek proton aur ek neutron hai, drop labelled $2.224$ MeV — aage aane wali gehri wells ke comparison mein bahut choti.

Step 7 — Fair comparison: $A$ se divide karo depth per ball paane ke liye

KYA HAI. Total $E_{B}$ hamesha $A$ ke saath badhta hai (zyada balls, zyada bonds), isliye woh stability rank nahi kar sakta. Iske bajaye poochho: "well kitni gehri hai per nucleon?" $\overline{E_{B}} = \frac{E _{B}}{A} (average par, e k ball hataane ki energy)$

Helium-4 chalao ( $Z = 2, N = 2, M = 4.002602 u$ ): $Δ m = 2 (1.007825) + 2 (1.008665) - 4.002602 = 0.030378 u$ $E_{B} = 0.030378 \times 931.5 = 28.30 MeV, \overline{E_{B}} = \frac{28.30}{4} = 7.07 MeV .$

KYUN. Deuteron ne score kiya $1.11$ MeV/nucleon; helium-4 score karta hai $7.07$ MeV/nucleon — per ball bahut gehri well. $A$ se divide karna woh fair judge hai jo reveal karta hai ki medium nuclei winners hain, $^{56}$ Fe ke paas $\approx 8.8$ MeV par peak karte hue.

PICTURE. Do wells same "per ball" ruler par draw ki gayi hain: deuteron shallow ( $1.11$ ), helium deep ( $7.07$ ) — same physics, par ab per nucleon fairly compare kiya gaya.

Recall Kyun dono fusion aur fission energy release karte hain (ek line mein)

Dono nucleons ko iron ke paas sabse gehri per-ball well ki taraf push karte hain; jo extra depth milti hai woh energy ke roop mein bahar aati hai — fusion steep left ko climb karta hai, fission gentle right ko climb karta hai, aur released amount Q-value hota hai.

Ek picture mein summary

Upar sab kuch ek single chain mein compress ho jaata hai: loose balls (bhaari) → inhe glue karo → clump (halka) → missing mass $Δ m$ released energy $E_{B} = Δ m c^{2}$ ban jaati hai → depth compare karne ke liye $A$ se divide karo, iron-peaked curve milti hai.

Recall Feynman retelling — poora walkthrough seedhe shabd mein

Ek scale par magnet-balls ka pile rakho aur weight note karo — yeh tumhara parts list hai. Ab inhe ek clump mein snap hone do; lock hote waqt woh ek choti si click ki energy release karte hain. Clump ko weigh karo: yeh thoda halka hai. Kuch gira nahi — count karo, sab abhi bhi wahan hain — par weight ka ek chota sa tukda ( $Δ m$ ) us energy mein convert ho gaya jo click ke roop mein bahar nikla. Einstein ka exchange rate $c^{2}$ batata hai exactly kitni energy woh tukda worth tha ( $E_{B}$ ). Clump todne ke liye tumhe woh saari energy wapas deni padegi, isliye clump "bound" hai: woh ek energy valley ke neeche park kiya hua hai. Kuch clumps doosron ke comparison mein per ball zyada gehre park karte hain — iron ke paas wale medium-sized ones sabse gehre hain — isliye chote clumps milna pasand karte hain aur bade clumps todna pasand karte hain, har ek comfortable iron-deep valley ki taraf slide karta hua aur difference ko energy ke roop mein bahar spitting karta hua.

Connections

Mass–energy equivalence ($E=mc^2$)
Atomic mass unit (u)
Strong nuclear force
Nuclear fusion
Nuclear fission
Q-value of nuclear reactions
Stability of nuclei & N-Z curve