Nuclear reactions — Q-value calculation: Visual walkthrough

2.3.20 · D2 · Physics › Modern Physics › Nuclear reactions — Q-value calculation

Step 1 — Pieces ko pehle aur baad mein weigh karo

KYA. Ek nuclear reaction likhi jaati hai $a + X \to Y + b$ . Ise ek recipe ki tarah padho: left side pe do cheezein andar aati hain (ek chhota flying particle $a$ aur ek target nucleus $X$ ), do cheezein bahar aati hain right side pe ( $Y$ , bada leftover nucleus, aur $b$ , ek chhota ejected particle). Hum saare reactants ko ek scale pe rakhte hain, phir saare products ko ek scale pe rakhte hain, aur compare karte hain.

KYUN. $Q$ ke baare mein sab kuch ek shocking experimental fact se shuru hota hai: do scale readings equal nahi hoti. Everyday chemistry mein "before" aur "after" ka weight same hota hai. Nuclear reactions mein aisa nahi hota — aur woh tiny difference hi poori kahaani hai.

PICTURE. Do pans. Left pan = $a + X$ . Right pan = $Y + b$ . Left pan thoda neeche baitha hai — woh zyada bhaari hai. Right side pe jo mass missing hai woh kahin na kahin jaani chahiye.

Step 2 — Missing mass ko naam do: $Δ m$

KYA. Woh mass define karo jo gayab ho gayi: $"after-pile kitni halki hai" Δ m = pehle, left pan (m_{a} + m_{X}) - baad mein, right pan (m_{Y} + m_{b})$

Symbol $Δ$ (Greek "delta") ka matlab hamesha hota hai "mein change" — yahan, before minus after.

KYUN. Hume ek number chahiye jo capture kare "scale kitna unbalanced hai." Woh single number hai $Δ m$ . Agar $Δ m > 0$ toh before-pile zyada bhaari thi (mass gayab ho gayi). Agar $Δ m < 0$ toh after-pile zyada bhaari hai (mass create hui — jo, hum dekhenge, energy lagti hai).

PICTURE. Wohi do pans, ab unke beech ki height difference ek chhote ruler se label ki gayi hai jis par $Δ m$ likha hai. Left side pe positive ruler neeche point kar raha hai = mass lost.

Step 3 — Mass kahan gayi? Einstein ka exchange rate

KYA. Gayi hui mass energy ke roop mein wapas aati hai. Einstein ka mass–energy relation exchange rate deta hai: $energy E = mass m ek fixed number c^{2}$ Yahan $c = 3.0 \times 1 0^{8} m/s$ speed of light hai, aur $c^{2}$ bas ek conversion factor hai jo kilograms ko joules mein convert karta hai.

KYUN. Humne poocha tha "missing mass kahan jaati hai?" Reaction mein aur kuch nahi badalta — protons ki wahi ginti, neutrons ki wahi ginti overall. Extra weight sirf ek jagah chhupi ho sakti hai — flying-apart pieces ki motion energy ke roop mein. $E = m c^{2}$ woh rule hai jo batata hai ki missing mass ki har unit kitni energy ke barabar hai.

PICTURE. Step 2 se mass ka chhota tukda $Δ m$ ek "converter box" mein jaata hai aur doosri taraf se yellow energy $Δ m c^{2}$ ke roop mein bahar aata hai. Kyunki $c^{2}$ ek bahut bada number hai, mass ka ek zarra energy ka ek firework banata hai.

Step 4 — Wohi kahaani kinetic energy se

KYA. Ab mass nahi, energy dekho. Total energy conserved hoti hai: total pehle total baad ke barabar hoti hai. Har side mein do ingredients hain — rest-mass energy ( $m c^{2}$ ) aur kinetic energy $K$ (motion ki energy, idhar udhar fly karne se).

$pehle total energy (m_{a} + m_{X}) c^{2} + (K_{a} + K_{X}) = baad mein total energy (m_{Y} + m_{b}) c^{2} + (K_{Y} + K_{b})$

Har symbol: $m c^{2}$ terms "locked-up" rest energy hain; har $K$ ek moving particle ki energy hai.

KYUN. Hmare paas Step 3 se mass table se $Q$ already hai. Lekin real lab mein tum ek single flying nucleus weigh nahi karte — tum measure karte ho woh kitni tezi se move kar raha hai, yaani uski $K$ . Toh hume $Q$ ka ek doosra roop chahiye jo measurable cheezein mein likha ho. Energy conservation woh bridge hai.

PICTURE. Ek horizontal energy bar. Pehle: rest energy ka ek lamba stack aur kinetic energy ka ek chhota sliver. Baad mein: ek chhota rest-energy stack (mass lost ho gayi) aur ek bada kinetic sliver. Bar ki total height same hai — energy bas "rest" compartment se "motion" compartment mein chali gayi.

Rearrange karo — saari rest masses ek side, saari kinetic energies doosri side: $= Q (Step 3!) [(m_{a} + m_{X}) - (m_{Y} + m_{b})] c^{2} = baad mein motion (K_{Y} + K_{b}) - pehle motion (K_{a} + K_{X})$

Step 5 — Clean-up trick: 931.5 MeV per u

KYA. Masses atomic mass units u mein rehti hain; energies MeV (mega-electron-volts) mein rehti hain. Ek conversion saare physics constants ko absorb kar leta hai: $1 u = 931.5 MeV / c^{2} ⟹ Q (MeV) = Δ m (u) \times 931.5$

KYUN. $c^{2} = (3 \times 1 0^{8})^{2}$ ko haath se carry karna galti-prone hai. Hum woh multiply ek baar karte hain, use number 931.5 mein bake kar dete hain, aur phir kabhi $c$ ko touch nahi karte.

PICTURE. Ek currency exchange counter: "1 u" dete ho, "931.5 MeV" milta hai. Teller $c^{2}$ hai.

Step 6 — Sign padho: $Q$ ki teen duniyaan

KYA. $Q$ positive, negative, ya zero ho sakta hai. Har ek ek alag physical world hai.

Sign of $Q$	Mass	Matlab
$Q > 0$	after-pile halki	exothermic — energy bahar nikalti hai
$Q < 0$	after-pile bhaari	endothermic — energy andar dhekelni padti hai
$Q = 0$	barabar	elastic — kuch release nahi

KYUN. Step 4 ne kaha tha $Q = K_{final} - K_{initial}$ . Agar $Q > 0$ toh pieces tez ho jaati hain — reaction kinetic energy muft de deti hai. Agar $Q < 0$ toh pieces ko zero se bhi neeche slow down karna padta, jo impossible hai jab tak bahar se energy supply na ki jaaye. Woh impossibility hi exactly woh reason hai kyun endothermic reactions ko ek push chahiye hota hai (agla step).

PICTURE. Teen scale-and-bar panels side by side: left pan neeche ( $Q > 0$ , green), right pan neeche ( $Q < 0$ , red), balanced ( $Q = 0$ , grey).

Step 7 — Degenerate case: kyun endothermic ko $∣ Q ∣$ se zyada chahiye

KYA. $Q < 0$ ke liye tumhe kinetic energy supply karni padti hai — lekin exactly $∣ Q ∣$ supply karna kaafi nahi hai. Agar projectile $a$ ek stationary target $X$ se takraata hai, toh reaction hone ke liye minimum kinetic energy hai: $K_{th} = ∣ Q ∣ (1 + \frac{m _{a}}{m _{X}})$ Har symbol: $K_{th}$ threshold (minimum) kinetic energy hai $a$ ki; factor $(1 + m_{a} / m_{X})$ hamesha 1 se bada hota hai.

KYUN. Momentum conserved hota hai. Incoming $a$ forward momentum carry karta hai, toh products baad mein sab milke still nahi baith sakte — kuch forward move karte rehna chahiye. Woh forced forward-motion energy (woh "center-of-mass" energy) nucleus ko todne ke liye available nahi hai. Sirf woh energy jo us momentum tax ke baad bachi hai woh reaction kar sakti hai — isliye tumhe extra supply karna padta hai.

PICTURE. Upar: $a$ andar fly karta hai, $X$ still baitha hai. Neeche: threshold par bhi products milke aage drift karte hain (ruk nahi sakte). Ek yellow bracket "wasted" forward energy mark karta hai; ek green bracket woh part mark karta hai jo actually reaction mein kaam aaya.

Step 8 — Payout ko split karna: sab chhote particle ko nahi milta

KYA. Jab $Q > 0$ motion ke roop mein release hota hai, toh share kaise hota hai? Rest pe ek decay ke liye ( $X$ rest pe, $Y$ aur $b$ mein split hota hai), momentum conservation force karta hai: $K_{b} = Q \cdot \frac{m _{Y}}{m _{Y} + m _{b}}, K_{Y} = Q \cdot \frac{m _{b}}{m _{Y} + m _{b}}$ Light particle $b$ ko lion's share milta hai (kyunki upar $m_{Y}$ bada hai).

KYUN. Do fragments equal aur opposite momentum se alag fly karte hain. Equal momentum lekin unequal mass ⇒ halka wala tezi se move karta hai (bahut zyada $K$ ), bhaari wala barely recoil karta hai. Toh $Q$ hai $K_{b} + K_{Y}$ , dono terms — kabhi sirf fast particle ki energy nahi.

PICTURE. α-decay cartoon: bhaari daughter slowly left recoil karti hai (chhota arrow), light α right tezi se jaata hai (lamba arrow), momenta equal aur opposite. Energy pie mostly α ko milta hai.

Ek-picture summary

Sab ek saath: ek scale (mass lost, Steps 1–2) ek converter box ko feed karta hai ( $E = m c^{2}$ , Step 3) jiska output products ki kinetic energies mein split hota hai (Step 4), corners mein 931.5 exchange rate (Step 5) aur sign key (Step 6) ke saath.

Recall Feynman retelling — poora walkthrough simple words mein

Ingredients ko ek scale pe rakho aur number likh lo. Reaction karo. Leftovers ko scale pe rakho — number ek baal chhoti hai. Woh missing baal kahi leak nahi hui; Einstein ke rule $E = m c^{2}$ ne use motion ka ek burst bana diya, aur kyunki $c^{2}$ bahut bada hai, weight ka ek zarra ek firework ban jaata hai. Woh firework of energy hi Q-value hai. Tum ise do tareekon se nikal sakte ho: woh mass weigh karo jo gayab hui, ya measure karo ki pieces kitni tezi se fly apart hoti hain — dono same $Q$ dete hain. Agar leftovers bhaare nikalte hain, toh tumhe unhe banane ke liye energy dhakkelni padti hai — aur us rule ki wajah se ki momentum cheezein chalti rakhta hai, tumhe mass gap se bhi zyada dhakkelna padta hai, kyunki products ko standing still nahi chhoda ja sakta. Aakhir mein, jab energy bahar nikalti hai, chhota fast piece uska zyada hissa le jaata hai jabki bada piece barely hilta hai — lekin $Q$ sahi paane ke liye tumhe dono nudges count karne honge.