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

Neeche wala number $Z$ hai (protons), upar wala number $A$ hai (total nucleons).

• $Z=8$
• $A=16$
• $N=A-Z=16-8=8$ Yeh kaisa dikhta hai: 8 positive protons aur 8 neutral neutrons ek clump mein packed — ek accha even-even, doubly magic nucleus jo curve ke early peak par baitha hai.

Atomic masses use karte hain taaki 8 electron masses cancel ho jayein: $\Delta m=[8m_{{}^1\text{H}}+8m_n]-M{(}^{16}\text{O})$ $\Delta m=[8(1.007825)+8(1.008665)]-15.994915$ Atomic masses kyun? Tables mein atoms listed hote hain (nucleus + electrons). Har proton ke liye $m_{{}^1\text{H}}$ (jo ek electron carry karta hai) use karne se left side ke total $Z=8$ electrons right side ke $M{(}^{16}\text{O})$ ke andar ke $Z=8$ electrons se exactly match karte hain — woh cancel ho jaate hain, aur hume $m_e$ ki zarurat hi nahi padti.

Step 1 — mass defect (kyun: yahi $E_B$ ka source hai): $8(1.007825)+8(1.008665)=8.062600+8.069320=16.131920u$ $\Delta m=16.131920-15.994915=0.137005u$ Step 2 — energy (kyun: $E=m{c}^2$ missing mass ko glue energy mein convert karta hai): $E_B=0.137005\times 931.5=127.62\text{MeV}$ Step 3 — per nucleon (kyun: stability ki fair comparison ke liye): $\overline{E_B}=\frac{127.62}{16}=7.98\text{MeV}$ Yeh iron ke $\approx 8.8$ MeV peak se thoda neeche hai — ¹⁶O bahut tightly bound hai, jo curve par iske early sharp peak ke saath consistent hai.

$Z=26$, $N=56-26=30$. Step 1: $26(1.007825)+30(1.008665)=26.203450+30.259950=56.463400u$ $\Delta m=56.463400-55.934939=0.528461u$ Step 2: $E_B=0.528461\times 931.5=492.26\text{MeV}$ Step 3: $\overline{E_B}=\frac{492.26}{56}=8.79\text{MeV}$ Matlab kya hai: Yeh curve ka top hai — sabse tightly bound region. Iron ka koi bhi fusion ya fission energy release nahi karta, kyunki tum aur upar nahi ja sakte.

B ka per-nucleon value calculate karo: $\overline{E_B}{(}^{238}\text{U})=\frac{1802}{238}=7.57\text{MeV}$ Compare karo: $8.79>7.57$, toh iron-56 zyada stable hai. Totals kyun mislead karte hain: Total $E_B$ hamesha $A$ ke saath badhta hai (zyada nucleons → zyada attractive bonds), toh ²³⁸U ka bada total (1802 vs 492) sirf yeh reflect karta hai ki uske paas zyada particles hain, tighter bonding nahi. Stability iss bare mein hai ki ek nucleon hatana kitna mushkil hai, jo exactly $\overline{E_B}$ hai. Uranium ka lower per-nucleon value precisely isliye hai ki woh fission karke energy release kar sakta hai — splitting se woh peak ki taraf chadh jaata hai.

Yeh kyun kaam karta hai: Har nucleon ka glue $\Delta \overline{E_B}=8.5-7.6=0.9$ MeV improve hota hai. Binding energy jo increase hoti hai woh energy released hoti hai (products energy well mein gehra baithte hain). $Q\approx A\times \Delta \overline{E_B}=236\times 0.9=212.4\text{MeV}$ Yeh kaisa dikhta hai: Curve par, reaction peak ki taraf right slope par upar jaati hai. Vertical rise, times carried nucleons ki sankhya, energy budget hai. Yeh ~200 MeV per fission uranium ka textbook figure hai.

Step 1 — mass balance (kyun: $Q=(\text{reactant mass}-\text{product mass})c^2$): $M_{\text{react}}=2.014102+3.016049=5.030151u$ $M_{\text{prod}}=4.002602+1.008665=5.011267u$ $\Delta m=5.030151-5.011267=0.018884u$ Step 2 — energy: $Q=0.018884\times 931.5=17.59\text{MeV}$ Electron check: Left side mein $1+1=2$ electrons hain (do hydrogen atoms se), right side mein $2$ hain (helium se). Woh balance karte hain, toh atomic masses yahan safe hain. Dekho Q-value of nuclear reactions. Positive kyun hai: ⁴He curve par high baitha hai (≈7.07 MeV/nucleon vs fuel ki low values) — glue improve hua, energy nikli. Yahi Sun mein fusion ko power karta hai.

BE bookkeeping kyun kaam karta hai: Free neutron ki binding energy zero hai, toh $Q=E_B(\text{products})-E_B(\text{reactants})=E_B{(}^4\text{He})-[E_B{(}^2\text{H})+E_B{(}^3\text{H})]$ $Q=28.30-(2.224+8.482)=28.30-10.706=17.59\text{MeV}$ Match: L4.1 se identical (17.59 MeV) rounding tak. Do independent routes — raw masses aur binding energies — agree karte hain, kyunki dono same $E=m{c}^2$ accounting encode karte hain.

(a) Energy per fission (kyun: nucleons peak ki taraf chadh rahe hain): $Q\approx 236\times (8.45-7.59)=236\times 0.86=202.96\text{MeV}$ Joules mein: $Q=202.96\times 1.602\times 10^{-13}=3.251\times 10^{-11}\text{J}$.

(b) Fission rate (kyun: electrical power / efficiency = thermal power, phir har fission ki energy se divide karo): Thermal power needed $=\frac{1000\text{MW}}{0.33}=3.030\times 10^9\text{W}$. $\text{fissions/s}=\frac{3.030\times 10^9}{3.251\times 10^{-11}}=9.32\times 10^{19}{\text{s}}^{-1}$

(c) Daily mass (kyun: ek din mein nuclei count karo, moles phir grams mein convert karo): Fissions/day $=9.32\times 10^{19}\times 86400=8.05\times 10^{24}$. Moles $=\frac{8.05\times 10^{24}}{6.022\times 10^{23}}=13.37\text{mol}$. Mass $=13.37\times 235=3142\text{g}\approx 3.14\text{kg/day}$.

Matlab kya hai: Kuch kilogram fuel per day ek gigawatt city ko power karta hai — yeh curve par nucleons ko upar le jaane ka fayda hai. Chemical fuel se compare karo (coal: hazaron tonnes/day), jo nuclear vs chemical energy scales ka numeric signature hai.

Argument: Fusion energy tabhi release karta hai jab product $\overline{E_B}$ curve par reactants se upar baitha ho. Iron (8.79 MeV) maximum hai. Gold (7.91 MeV) descending right slope par hai — woh iron se neeche hai. Toh lighter nuclei ko gold tak fuse karne ka matlab hoga peak se aage jaana aur waapas neeche aana, jo energy release karne ki jagah cost karega. Ek star khud ko tabhi power kar sakta hai jab woh peak ki taraf upar jaye; jaise hi woh iron pahunchta hai waahan aur koi jagah nahi hai, energy generation ruk jaata hai, aur core collapse ho jaata hai. Gold aur heavy elements ordinary fusion se nahi balki supernovae mein rapid neutron capture se bante hain. Dekho Stability of nuclei & N-Z curve aur Strong nuclear force.

Jab nucleons higher $\overline{E_B}$ ki taraf move karte hain tab released energy equals ==$A\times \Delta \overline{E_B}$== (nucleons times per-nucleon gain). Fission of a heavy nucleus releases about ::: ~200 MeV D–T fusion releases about ::: ~17.6 MeV