Binding energy — mass defect, BE per nucleon curve: Question bank

2.3.19 · D5 · Physics › Modern Physics › Binding energy — mass defect, BE per nucleon curve

Test se pehle, ek 10-second ka refresher us shape par jo sab kuch thaame hue hai:

Binding-energy-per-nucleon curve mein $\overline{E_{B}}$ (vertical, MeV mein) ko mass number $A$ = total nucleons (horizontal) ke against plot kiya jaata hai.
Yeh left side par tezi se utha hai (halke nuclei), $A \approx 56$ ke paas peak karta hai (iron region, $\approx 8.8$ MeV), phir right side par dheere se girta hai (bhaari nuclei).
Jitna upar = jitna zyada tightly glued = utna zyada stable. Woh reactions jo nucleons ko peak ki taraf move karti hain, energy release karti hain.

Yeh mental picture taiyaar rakho. Neeche har ek trap aslaan mein is curve par tum kahan baithe ho ya mass aur energy kis taraf point karte hain — isi ke baare mein ek sawaal hai.

True or false — justify

Ek nucleus apne free protons aur neutrons ke sum se bhaari hota hai.

False. Yeh halka hota hai — kuch mass woh binding (glue) energy ban gayi jo system se tab nikal gayi jab nucleons clump hue, isliye

E = m c^{2}

se nucleus kam weighs karta hai. Dekho Mass–energy equivalence ($E=mc^2$).

Binding energy woh energy hai jo nucleus ke andar stored hai, jise tum usse intact rakh kar tap kar sakte ho.

False. Binding energy nucleus ke banne par system se nikal chuki thi; yeh woh energy hai jo tum ko nucleus ko todne ke liye wapas daalni padti hai, na ki koi reserve jo bina kuch kiye nikalti ho.

Zyada total binding energy

E_{B}

ka matlab hamesha zyada stable nucleus hota hai.

False. Total

E_{B}

A

ke saath badhta hai kyunki zyada nucleons zyada bonds banate hain;

^{238}

U ki total

E_{B}

bahut badi hai lekin low

\overline{E_{B}} \approx 7.6

MeV hai, isliye woh fission kar sakta hai. Stability ka andaaza

\overline{E_{B}}

, yaani per-nucleon value se lagaya jaata hai.

^{4}

He ki BE per nucleon uske halke neighbours jaise

^{3}

H ya

^{6}

Li se zyada hai.

True.

^{4}

He curve par ek local peak hai — doubly closed shells isse exceptionally tightly glued banate hain (

\approx 7.1

MeV/nucleon), isliye yeh left slope par ek sharp spike ki tarah dikhta hai.

Kyunki fusion energy release karta hai, koi bhi do nuclei ko fuse karne se energy release hoti hai.

False. Sirf woh fusion jo product ko curve par upar le jaaye, energy release karta hai — matlab peak se neeche wale halke nuclei. Agar iron ke paas pahunche do nuclei ko fuse karo toh curve neeche move hoga aur energy cost hogi. Dekho Nuclear fusion.

Iron (

^{56}

Fe) fission aur fusion dono se energy release kar sakta hai.

False. Iron peak par baitha hai, isliye koi bhi movement — todna ya jodna —

\overline{E_{B}}

mein neeche jaata hai, jo energy absorb karta hai. Iron "stellar ash" hai, energy-releasing reactions ka ant.

Mass defect aur binding energy do alag physical quantities hain.

Aadha sach — yeh ek hi baat hai do alag units mein.

E_{B} = Δ m c^{2}

; "missing mass" aur "glue energy" ek hi cheez ke mass aur energy descriptions hain.

Δ m = Z m_{p} + N m_{n} - M

mein tum ko hamesha nuclear masses use karni chahiye.

Practically False. Tables atomic masses (nucleus +

Z

electrons) list karti hain. Agar tum protons ke liye

m_{^{1} H}

aur

M

ke liye atomic mass use karo, toh

Z

electron masses automatically cancel ho jaate hain — aur yeh cleaner aur correct hai.

Spot the error

"Nucleus mein energy add karne se woh bhaari ho jaata hai, isliye ek strongly bound nucleus ek heavy nucleus hota hai."

Error direction mein hai: binding energy nucleus se nikal gayi. Strongly bound nucleus halka hota hai (bada mass defect), aur "heavy" (

A

, mass number) ka "tightly bound" (

\overline{E_{B}}

) se koi lena-dena nahi.

Δ m = Z m_{p} + N m_{n} - M_{nucleus}

, aur main table se atomic mass

M_{nucleus}

ki jagah plug in karunga."

Error hai conventions mix karna:

m_{p}

ek bare proton hai lekin table mass mein

Z

electrons bhi hain. Ya to sab nuclear masses use karo ya sab atomic masses; standard fix mein

m_{^{1} H}

use karte hain taaki electrons cancel ho jayein.

"BE per nucleon hamesha badhti rahti hai jab nuclei bade hote hain, kyunki zyada nucleons zyada bonds banate hain."

Error Coulomb repulsion ko ignore karta hai.

A \approx 56

ke baad, long-range proton–proton repulsion (

Z^{2}

ke saath badhti hai) short-range Strong nuclear force ko peeche chhod deti hai, isliye

\overline{E_{B}}

girta hai — curve palat jaata hai.

"Fission energy release karta hai kyunki ek bada nucleus strong force bonds todta hai."

Ulta hai. Bonds todna energy cost karta hai. Fission energy release karta hai kyunki products per nucleon zyada tightly bound hain (curve par upar); binding mein net gain hi bahar aata hai. Dekho Nuclear fission.

"D+T→He+n mein neutron koi energy carry nahi karta, isliye saari

17.6

MeV helium ko jaati hai."

Error momentum conservation skip karne mein hai: halka neutron actually kinetic energy ka zyaada hissa le jaata hai.

17.6

MeV total

Q

-value hai, shared hoti hai, na ki sirf helium ka share. Dekho Q-value of nuclear reactions.

1 u = 931.5

MeV, isliye mass aur energy ke same units hain."

Sloppy: yeh likhna chahiye

1 u \cdot c^{2} = 931.5

MeV. Mass aur energy mein

c^{2}

ka factor hota hai;

c^{2}

drop karne se unit conversion chhupp jaata hai.

"Strong force poore nucleus mein phaila rehta hai, isliye ek giant nucleus har jagah equally glued hai."

Error range mein hai. Strong force short-range hai (sirf nearest-neighbour tak), isliye nucleus ke andar ka ek nucleon sirf apne neighbours ko feel karta hai, jabki har proton poore nucleus mein har doosre proton ko repel karta hai — yahi imbalance heavy-nucleus stability ko barbaad karta hai.

Why questions

Hum total binding energies compare karne ki jagah

A

se kyun divide karte hain?

Total

E_{B}

bade nuclei ko unfairly reward karta hai (zyada particles = zyada bonds).

A

se divide karne par sawaal hota hai "ek average nucleon ko nikaalna kitna mushkil hai?" — ek fair, size-independent stability comparison.

Curve ki left slope itni steeply kyun uthti hai?

Ek tiny nucleus mein ek nucleon add karne par ussse achanak bahut saare naye attractive neighbours mil jaate hain, isliye binding per particle bahut zyada jump karta hai. Jab nucleus bada hota hai, surface par naye nucleons ko kam naye neighbours milte hain, isliye payoff flatten ho jaata hai.

Ek stable bound nucleus ka mass defect hamesha positive kyun hota hai?

Ek bound state energy well mein hoti hai, free nucleons se energy mein neeche; kam energy ka matlab kam mass hai, isliye alag-alag pieces hamesha whole se zyada weight karte hain — ek positive

Δ m

Fusion aur fission dono ulte processes hone ke bawajood energy kyun release karte hain?

Dono nucleons ko curve ke peak ki taraf le jaate hain. Halke nuclei fuse hokar upar chahdte hain; bhaari nuclei split hokar upar chahdte hain.

\overline{E_{B}}

mein rise hi dono directions mein release hone wali energy hai. Dekho Nuclear fusion aur Nuclear fission.

Curve iron ke aas-paas peak kyun karta hai, say, lead ke paas nahi?

Iron woh sweet spot hai jahan short-range attraction lagbhag saturate ho chuki hai lekin Coulomb repulsion abhi dominant nahi hui. Iron ke baad,

Z^{2}

repulsion jeet jaati hai aur

\overline{E_{B}}

girta hai, isliye lead peak se neeche hai.

Agar hum dono sides par atomic masses use karein toh electron masses ignore kyun kar sakte hain?

Atomic masses apne electrons saath laati hain:

Z

hydrogen atoms ke

Z

electrons neutral product atom ke

Z

electrons se match karte hain, isliye exactly cancel ho jaate hain (electron binding energies tiny hain aur neglect ki jaati hain).

^{4}

He ka BE per nucleon halka nucleus hone ke bawajood zyada kyun hai?

Iske neutrons aur protons closed shells fill karte hain ("doubly magic"), jo ek especially stable arrangement hai, aur iron se bahut left mein ek sharp local peak produce karta hai. Yeh smooth-curve trend ka ek exception hai.

Edge cases

Ek single free proton (hydrogen-1 nucleus) ki binding energy kya hai?

Yeh zero hai — kuch bhi hold karne wala nahi, koi partner nucleon nahi, isliye na mass defect aur na glue energy. Yeh

\overline{E_{B}} = 0

par baitha hai, curve ki bilkul shuruaat mein.

Ek lone neutron ke liye, kya binding energy hoti hai?

Nahi — ek single nucleon ke koi bonds nahi hote, isliye

E_{B} = 0

. (Ek free neutron unstable hai aur decay karta hai, lekin yeh beta decay hai, nuclear binding se unrelated.)

Curve ke exact peak par, ek reaction kitni energy release karta hai?

Kisi bhi chhhoti rearrangement ke liye essentially zero net energy, kyunki kisi bhi taraf move karna — fission ya fusion — peak chhod ke

\overline{E_{B}}

girata hai, isliye woh directions energy cost karti hain — peak products ke liye energy floor hai.

Agar ek "nucleus" ka positive mass defect doosre tarike se define kiya jaaye (nucleus parts se bhaari ho), toh iska kya matlab hoga?

Yeh unbound hoga — tukde khud apni marzi se udenge, kyunki assembled state free nucleons se zyada energy mein hai. Real stable nuclei aisa kabhi nahi karte.

Δ m = 0

physically kis cheez ko correspond karta hai?

Ek bilkul unbound collection: whole aur parts ka sum equal hai, koi glue nahi, koi released energy nahi — zero binding energy per nucleon ke saath borderline case.

Kya bahut-halke region mein curve kabhi dip karta hai smooth rise ki jagah?

Haan — curve tiny

A

ke liye jagged hota hai:

^{4}

He upar spike karta hai, lekin

^{6}

Li aur

^{7}

Li usse neeche dip karte hain, aur

^{8}

Be unstable hai. Smooth "rise" sirf overall trend hai, har point nahi.

Connections

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