Binding energy — mass defect, BE per nucleon curve: Worked examples

2.3.19 · D3 · Physics › Modern Physics › Binding energy — mass defect, BE per nucleon curve

Yahan sab kuch atomic masses (neutral atoms ki masses, jisme electrons shaamil hain) use karta hai taaki electron masses cancel ho jaayein — yahi fix parent note ne stress ki thi. Constants jo hum baar baar use karenge:

Yahan $Z$ = protons, $N = A - Z$ = neutrons, $A$ = total nucleons — sab parent ki tarah. Har symbol wahan se aaya hai; yeh page sirf unhe apply karta hai.

Scenario matrix

Binding energy par har problem in case classes mein se ek (ya mix) hoti hai. Table mein har ek listed hai, aur neeche konsa worked example use cover karta hai.

Cell	Case class	Kya alag banata hai	Covered by
A	Sabse chhota real nucleus	Sirf 2 nucleons → left slope pe bahut neeche	Ex 1 (deuteron)
B	Local-peak light nucleus	Doubly-magic, unusually high $\overline{E_{B}}$	Ex 2 ( $^{4}$ He)
C	Iron-region maximum	Curve ka top, reference stability	Ex 3 ( $^{56}$ Fe)
D	Heavy nucleus	Bada total $E_{B}$ lekin low $\overline{E_{B}}$	Ex 4 ( $^{235}$ U)
E	Reaction — energy released	$Q > 0$ , products zyada bound (fusion)	Ex 5 (D–T)
F	Reaction — energy absorbed	$Q < 0$ , energy supply karni padti hai (photodisintegration)	Ex 6 (break deuteron)
G	Degenerate / zero input	1 nucleon: koi partner nahi → $E_{B} = 0$	Ex 7 ( $^{1}$ H, free $n$ )
H	Limiting behaviour	Per-nucleon kyun flatten hota hai phir girta hai	Ex 8 (compare Fe vs U)
I	Word / exam twist	Reactor energy, hidden per-nucleon trap	Ex 9 (fission power)

Hum nine examples mein A–I cover karte hain. Signs explicitly E, F, aur G mein handle ki gayi hain, taaki "energy kis taraf flow karti hai" wala question kabhi ambiguous na rahe.

Case A — sabse chhota nucleus (deuteron)

Worked example Example 1 —

$_{1}^{2}$ H ki Binding energy Data: $M (^{2} H) = 2.014102 u$ . $E_{B}$ aur $\overline{E_{B}}$ nikalo.

Forecast: Sirf 2 nucleons hain, toh guess karo — kya $\overline{E_{B}}$ iron ki value $8.8$ MeV ke paas hogi, ya bahut neeche? (Aage padhne se pehle curve ke left slope ke baare mein socho.)

Nucleons count karo. $Z = 1$ , $N = 1$ , $A = 2$ . Yeh step kyun? Har formula ko pehle $Z$ aur $N$ chahiye — yeh decide karte hain ki kitne protons aur neutrons add karne hain.
Mass defect. $Δ m = [1 (1.007825) + 1 (1.008665)] - 2.014102 = 0.002388 u$ Yeh step kyun? $Δ m$ "missing mass" hai — binding energy ka akela source.
Energy mein convert karo. $E_{B} = 0.002388 \times 931.5 = 2.224 MeV$ Yeh step kyun? $E = m c^{2}$ missing mass ko glue energy mein badhata hai; $931.5$ ready-made $u \to$ MeV bridge hai.
Per nucleon. $\overline{E_{B}} = 2.224/2 = 1.11 MeV$ . Yeh step kyun? Per-nucleon fair stability yardstick hai.

Verify: $\overline{E_{B}} = 1.11$ MeV bahut chhota hai $8.8$ MeV ke comparison mein — yeh consistent hai kyunki deuteron curve ke bilkul bottom-left par baitha hai. Units: $u \times (MeV / u) = MeV$ . ✓ Tumhara forecast hona chahiye tha "iron se bahut neeche."

Case B — ek local-peak light nucleus

Worked example Example 2 — Helium-4,

$M (^{4} He) = 4.002602 u$ $\overline{E_{B}}$ nikalo aur explain karo ki yeh apne neighbours se better kyun hai.

Forecast: $^{4}$ He mein sirf 4 nucleons hain — kya yeh deuteron ki tarah low hoga, ya "doubly magic" hone se surprisingly high?

Count. $Z = 2$ , $N = 2$ , $A = 4$ . Kyun? Yeh decide karta hai ki kitne $m_{^{1} H}$ aur $m_{n}$ add karne hain.
Mass defect. $Δ m = 2 (1.007825) + 2 (1.008665) - 4.002602 = 0.030378 u$ Kyun? Same source-of-energy logic; electrons cancel ho jaate hain kyunki humne poori jagah atomic masses use ki hain.
Energy. $E_{B} = 0.030378 \times 931.5 = 28.30 MeV$ . Kyun? $E = m c^{2}$ missing mass ko glue energy mein convert karta hai; $931.5$ $u \to$ MeV bridge hai.
Per nucleon. $\overline{E_{B}} = 28.30/4 = 7.07 MeV$ . Kyun? Har particle ke liye fair comparison.

Verify: $7.07$ MeV bahut zyada hai deuteron ke $1.11$ se, chahe dono "light" hain. Yahi famous local spike hai $^{4}$ He par — closed proton shell aur closed neutron shell (doubly magic) use bahut strong bind karte hain. ✓

Case C — iron maximum

Worked example Example 3 — Iron-56,

$M (^{56} Fe) = 55.934939 u$ , $Z = 26$ $\overline{E_{B}}$ nikalo aur confirm karo ki yeh curve ka peak hai.

Forecast: Parent ne peak ke liye $\approx 8.8$ MeV quote ki thi. Kya humara computed number wahan land karega?

Count. $Z = 26$ , $N = 56 - 26 = 30$ , $A = 56$ . Kyun? $N = A - Z$ ; tumhe sahi split karna hoga warna neutron count galat ho jaayega.
Mass defect. $Δ m = [26 (1.007825) + 30 (1.008665)] - 55.934939 = 0.528461 u$ Kyun? 26 hydrogen-atom masses + 30 neutron masses, minus actual atom.
Energy. $E_{B} = 0.528461 \times 931.5 = 492.3 MeV$ . Kyun? $E = m c^{2}$ mass defect ko binding energy mein badhata hai; $931.5$ factor humein $u$ se MeV le jaata hai.
Per nucleon. $\overline{E_{B}} = 492.3/56 = 8.79 MeV$ . Kyun? $A$ se divide karne par fair per-particle glue strength milti hai — nuclei mein stability compare karne ka yahi ekmaatra fair tarika hai.

Verify: $8.79 \approx 8.8$ MeV — bilkul quoted peak par. Total $E_{B}$ yahan ( $492$ MeV) enormous hai, phir bhi per-nucleon value hi iron ko most stable crown deti hai. ✓

Case D — ek heavy nucleus (total-vs-per-nucleon trap)

Worked example Example 4 — Uranium-235,

$M (^{235} U) = 235.043924 u$ , $Z = 92$ Total $E_{B}$ aur $\overline{E_{B}}$ nikalo. Kiska total zyada bada hai, uska ya iron ka?

Forecast: $^{235}$ U bahut bada hai. Uski total binding energy zaroor iron ko dwarf karegi — lekin kya uska per-nucleon value iron ke $8.79$ se zyada hoga ya kam?

Count. $Z = 92$ , $N = 235 - 92 = 143$ , $A = 235$ . Kyun? Bada $Z$ matlab baad mein bahut Coulomb repulsion — lekin pehle sirf count karo.
Mass defect. $Δ m = [92 (1.007825) + 143 (1.008665)] - 235.043924 = 1.867521 u$ Kyun? 92 hydrogen-atom masses aur 143 neutron masses ka sum, minus actual atom — woh missing mass jo binding energy ban gayi.
Energy. $E_{B} = 1.867521 \times 931.5 = 1739.6 MeV$ . Kyun? $E = m c^{2}$ mass defect ko glue energy mein convert karta hai; $931.5$ humein $u$ se MeV le jaata hai.
Per nucleon. $\overline{E_{B}} = 1739.6/235 = 7.40 MeV$ . Kyun? Sirf per-nucleon value fair stability measure hai — total $E_{B}$ hamesha $A$ ke saath badhta hai aur mislead karta.

Verify: Total $E_{B} \approx 1740$ MeV iron ke $492$ MeV ko beat karta hai — phir bhi per nucleon $7.40 < 8.79$ . Yahi parent ki "stability ke liye total use mat karo" wali warning hai: $^{235}$ U kam tightly bound hai per particle, isliye yeh split (fission) ho sakta hai aur peak ki taraf chadh sakta hai. ✓

Agla figure in charon nuclei (A–D) ko curve par place karta hai taaki tum story dekh sako.

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

Notice karo: deuteron (Case A) bilkul bottom par hai, $^{4}$ He (B) early spike karta hai, $^{56}$ Fe (C) crest par baitha hai, aur $^{235}$ U (D) right side par neeche slide kar gaya hai.

Case E — ek reaction jo energy RELEASE karta hai (sign $Q > 0$ )

Worked example Example 5 — D–T fusion,

$^{2} H +^{3} H \to^{4} He + n$ Diye gaye total $E_{B}$ values: $E_{B} (^{2} H) = 2.22$ , $E_{B} (^{3} H) = 8.48$ , $E_{B} (^{4} He) = 28.30$ MeV. Neutron ek single nucleon hai, $E_{B} = 0$ . $Q$ nikalo.

Forecast: Sun isi par chalta hai — toh $Q$ positive hona chahiye. Lekin kitna bada, aur kya free neutron bookkeeping mein count hota hai?

Reactants ka total BE. $E_{B} (reactants) = E_{B} (^{2} H) + E_{B} (^{3} H) = 2.22 + 8.48 = 10.70 MeV$ Kyun? $Q$ -value formula compare karta hai ki "before" aur "after" states kitne tightly bound hain.
Products ka total BE. $E_{B} (products) = E_{B} (^{4} He) + E_{B} (n) = 28.30 + 0 = 28.30 MeV$ Kyun? Ek akele neutron ke paas bind karne ke liye koi partner nahi, isliye yeh $0$ contribute karta hai — ise include karne ka sahi tarika.
Released energy — $Q$ -value formula apply karo. $Q = E_{B} (products) - E_{B} (reactants) = 28.30 - 10.70 = 17.6 MeV$ Yeh step kyun? Agar products zyada bound hain, toh woh extra glue energy fragments ki kinetic energy ke roop mein release hoti hai.

Verify: $Q > 0$ ✓ — Sun ke apne aap power karne se match karta hai. Sign rule confirm hua: products curve par zyada upar ⇒ energy bahar. ✓

Case F — ek reaction jo energy ABSORB karta hai (sign $Q < 0$ )

Worked example Example 6 — Photodisintegration: deuteron tod do,

$γ +^{2} H \to p + n$ Ek gamma photon ek deuteron ko ek free proton aur ek free neutron mein tod deta hai. Minimum photon energy kitni chahiye?

Forecast: Hum ek bound nucleus ko un-glue kar rahe hain — kya energy bahar aayegi ya humein energy daalni padegi? Compute karne se pehle $Q$ ka sign predict karo.

States identify karo. Pehle: bound deuteron ( $E_{B} = 2.224$ MeV). Baad mein: ek free proton aur free neutron ( $E_{B} = 0$ each). Kyun? Ek bound state todna Example 5 ki logic ka reverse hai.
$Q$ -value formula apply karo. $Q = E_{B} (products) - E_{B} (reactants) = 0 - 2.224 = - 2.224 MeV$ Yeh step kyun? Products kam bound hain, isliye $Q$ negative hai — yeh reaction free mein nahi ho sakta.
Minimum photon energy. Photon ko kam se kam $∣ Q ∣$ supply karna hoga: $E_{γ}^{m i n} = 2.224 MeV$ Kyun? Tumhe woh binding energy wapas add karni hogi jo deuteron ko ek saath pakde thi.

Verify: $∣ Q ∣$ Example 1 se deuteron ki binding energy ke bilkul equal hai — kyunki ek nucleus todna iska binding energy chukana hai. Sign $Q < 0$ ⇒ energy absorbed, jaise ek unbinding process ke liye forecast kiya tha. ✓

Case G — degenerate inputs (ek single nucleon)

Worked example Example 7 — Lone

$_{1}^{1}$ H aur ek free neutron ki Binding energy Ordinary hydrogen $^{1} H$ ( $Z = 1, N = 0$ ) aur ek free neutron ka $E_{B}$ kya hai?

Forecast: Bind karne ke liye kuch bhi nahi toh mass defect aur binding energy kya honi chahiye?

$^{1}$ H ka mass defect. $Z = 1$ , $N = 0$ : $Δ m = [1 (1.007825) + 0] - 1.007825 = 0 u$ Kyun? "Nucleus" yahan ek single proton hai — parts ka sum whole ke barabar hai. Koi mass missing nahi jaata.
Binding energy. $E_{B} = 0 \times 931.5 = 0 MeV$ . Kyun? Ek nucleon ke paas koi partner nahi, isliye koi glue account karne ke liye nahi.
Free neutron. Same logic: yeh already ek single free nucleon hai, $E_{B} = 0$ . Kyun? Toolkit apply karne par, iska "mass defect" apni mass minus khud $= 0$ hai, isliye $E_{B} = Δ m c^{2} = 0$ — ek akele particle ki koi binding energy nahi ho sakti.

Verify: Kisi bhi single nucleon ke liye $E_{B} = 0$ . Yeh sahi zero/degenerate case hai aur yahi justify karta hai ki Examples 5–6 mein free neutron ne $0$ contribute kiya. Isliye BE-per-nucleon curve $A = 1$ par 0 se shuru hoti hai. ✓

Case H — limiting behaviour (curve kyun flatten hoti hai phir girti hai)

Worked example Example 8 —

Marginal nucleon compare karo: Fe vs U Examples 3 aur 4 use karke quantify karo ki uranium, iron se per nucleon kitna kam bound hai, aur ise curve ke downhill slope se connect karo.

Forecast: Hamare paas already $8.79$ (Fe) aur $7.40$ (U) hain. Kya woh ~1.4 MeV gap, 235 nucleons mein spread hoke, ek badi ya chhoti total energy hogi?

U vs Fe ka per-nucleon deficit. $Δ \overline{E_{B}} = 8.79 - 7.40 = 1.39 MeV per nucleon$ Kyun? Yeh curve par peak se uranium tak vertical drop hai.
Nucleon count se multiply karo. Agar $^{235}$ U ke nucleons mein se har ek iron ki tarah tightly bound hota, toh extra binding hoti $1.39 \times 235 = 326.7 MeV$ Yeh step kyun? Yeh estimate karta hai ki peak ki taraf rearrange karne se total energy "available" kitni hai — fission energy ka origin.
Limit interpret karo. Jaise $A$ iron ke baad badhta hai, Coulomb repulsion ( $\propto Z^{2}$ , long-range) accumulate hota hai jabki strong force ( $short-range$ , sirf nearest neighbours) saturate hoti hai. Isliye $\overline{E_{B}}$ sirf decline kar sakti hai — yeh kabhi recover nahi hoti. Curve ka ek single broad maximum hota hai aur uske baad hamesha dheere dheere girta hai. Yeh step kyun? Humein explain karna hoga ki numbers is tarah kyun aaye — do competing forces ki physics hi uranium ko iron se neeche force karti hai aur poori curve ko shape deti hai.

Verify: $^{235}$ U ke ek fission mein release hone wali real energy $\approx 200$ MeV hai — hamare $327$ MeV estimate ke same order mein (estimate ek upper bound hai kyunki fission fragments peak par nahi, peak ke paas land karte hain). Order-of-magnitude sanity ✓. Direction: U below Fe ⇒ fission favourable ✓.

Case I — word problem / exam twist

Worked example Example 9 — Reactor bookkeeping

Ek reactor $^{235}$ U fission karta hai, har fission event mein $\approx 200$ MeV release karta hai. (a) $200$ MeV joules mein kitni energy hai? (b) Roughly kitne fissions per second $1 W$ power denge? Use $1 MeV = 1.602 \times 1 0^{- 13} J$ (toolkit se).

Forecast: $1$ watt chhota lagta hai — lekin ek single fission sirf $\sim 1 0^{- 11}$ J deta hai. Toh kya humein hazaron fissions per second chahiye, ya trillions?

Per-event energy ko joules mein convert karo. $E_{1} = 200 \times 1.602 \times 1 0^{- 13} = 3.204 \times 1 0^{- 11} J$ Yeh step kyun? Power watts mein measure hoti hai (joules per second), isliye divide karne se pehle MeV chhodni padegi aur SI units mein aana padega.
1 W ke liye fissions per second. Power $=$ energy per event $\times$ events per second, yani $P = E_{1} \cdot n$ . Rate $n$ ke liye solve karo: $n = \frac{P}{E _{1}} = \frac{1 W}{3.204 \times 1 0 ^{- 11} J} = 3.12 \times 1 0^{10} fissions/s$ Yeh step kyun? $P = E_{1} \cdot n$ rearrange karne se woh count $n$ isolate hota hai jo humse maanga gaya tha.

Verify: Sirf $1$ W ke liye $\sim 3 \times 1 0^{10}$ fissions per second — reassuringly huge, confirm karta hai ki nuclei individually tiny (joules mein) lekin collectively vast energy release karte hain. Units: $J/s \div J = 1/ s$ ✓. Exam "twist" yahan tha ki " $200$ MeV per event" ko galti se per-nucleon number padhna — yeh per fission hai (saare 235 nucleons milke). Woh trap dekho.

Recall Self-test: har question kaunsa cell hai?

Ek gamma ray $^{4}$ He tod deti hai — energy released hui ya absorbed? ::: Absorbed ( $Q < 0$ , Case F logic): tumhe $^{4}$ He ki binding energy wapas supply karni padegi. $^{56}$ Fe ya $^{235}$ U mein se kiska total binding energy bada hai? ::: $^{235}$ U ka ( $1740$ vs $492$ MeV) — lekin per nucleon iron jeet ta hai (Case D trap). Ek free neutron ka $E_{B}$ kya hai? ::: Exactly $0$ — bind karne ke liye koi partner nahi (Case G). Iron ke upar koi bhi nucleus per-nucleon binding zyada kyun nahi rakh sakta? ::: Coulomb repulsion $Z^{2}$ ke saath badhti hai jabki strong force saturate ho jaati hai (Case H limit).

Connections

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