Nuclear reactions — Q-value calculation: Exercises

2.3.20 · D4 · Physics › Modern Physics › Nuclear reactions — Q-value calculation

Constants jo poore mein use hote hain: $1 u = 931.5 MeV / c^{2}$ . Atomic/particle masses (u): $^{1}$ H $= 1.007825$ , n $= 1.008665$ , $^{2}$ H $= 2.014102$ , $^{3}$ H $= 3.016049$ , $^{3}$ He $= 3.016029$ , $^{4}$ He $= 4.002603$ , $^{6}$ Li $= 6.015123$ , $^{7}$ Li $= 7.016004$ , $^{9}$ Be $= 9.012182$ , $^{12}$ C $= 12.000000$ , $^{14}$ N $= 14.003074$ , $^{17}$ O $= 16.999132$ , $^{235}$ U $= 235.043930$ , $^{236}$ U $= 236.045568$ , $m_{e} = 0.000549$ .

Level 1 — Recognition

L1.1 — Sign of Q

Neeche di gayi har reaction ke liye batao ki woh exothermic ( $Q > 0$ ) hai ya endothermic ( $Q < 0$ ), sirf mass change $Δ m = m_{before} - m_{after}$ se: (a) $Δ m = + 0.0189$ u (b) $Δ m = - 0.0013$ u (c) $Δ m = 0$ .

Recall Solution

Rule (mnemonic BAM): $Q = (Before - After) \times 931.5$ . Positive mass loss ⇒ energy bahar aati hai.

(a) $Δ m > 0 \Rightarrow Q > 0 \Rightarrow$ exothermic. Products halke hain; missing mass kinetic energy ban ke nikli.
(b) $Δ m < 0 \Rightarrow Q < 0 \Rightarrow$ endothermic. Products bhaari hain; aapko energy andar dhakelnI padi.
(c) $Δ m = 0 \Rightarrow Q = 0 \Rightarrow$ elastic — koi rest mass exchange nahi hua.

L1.2 — One-line conversion

Ek reaction mein $Δ m = 0.0189$ u hai. Ise MeV mein $Q$ mein convert karo.

Recall Solution

$Q = 0.0189 \times 931.5 = 17.6 MeV .$ 931.5 se kyun multiply karte hain? Itne MeV ek atomic mass unit ki vanished mass ke andar hote hain — wahi energy nuclear-physics ke kapdon mein likhi hoti hai.

Level 2 — Application

L2.1 — Direct Q from a mass table

D–D fusion branch ke liye $Q$ calculate karo: $_{1}^{2} H +_{1}^{2} H \to_{2}^{3} He +_{0}^{1} n .$

Recall Solution

Before: $2.014102 + 2.014102 = 4.028204$ u. After: $3.016029 + 1.008665 = 4.024694$ u. $Δ m$ : $4.028204 - 4.024694 = 0.003510$ u. Convert: $Q = 0.003510 \times 931.5 = 3.27 MeV$ . Exothermic (fusion energy release karta hai).

L2.2 — Neutron-induced, atomic masses cancel

$Q$ calculate karo: $_{3}^{6} Li +_{0}^{1} n \to_{1}^{3} H +_{2}^{4} He .$

Recall Solution

Pehle electrons check karo: left mein $Z = 3$ (Li) $+ 0$ (n) $= 3$ ; right mein $Z = 1 + 2 = 3$ . Balanced hai ⇒ atomic masses electrons ko safely cancel kar dete hain. Before: $6.015123 + 1.008665 = 7.023788$ u. After: $3.016049 + 4.002603 = 7.018652$ u. $Δ m$ : $7.023788 - 7.018652 = 0.005136$ u. $Q = 0.005136 \times 931.5 = 4.78 MeV$ (exothermic; yahi reaction tritium breed karti hai).

L2.3 — Q straight from kinetic energies

$^{210}$ Po rest mein $α$ emission se decay karta hai. $α$ $K_{α} = 5.30$ MeV ke saath nikalta hai, $^{206}$ Pb daughter $K_{D} = 0.10$ MeV ke saath recoil karti hai. Mass table ke bina $Q$ nikalo.

Recall Solution

$Q = K_{final} - K_{initial} = (5.30 + 0.10) - 0 = 5.40 MeV .$ Ye kyun kaam karta hai: vanished rest mass saare fragments ki total kinetic energy ban ke wapas aati hai — daughter ka recoil bhi count hota hai.

Figure — Nuclear reactions — Q-value calculation

Level 3 — Analysis

L3.1 — Endothermic + threshold

Rutherford ki transmutation ke liye $_{7}^{14} N +_{2}^{4} He \to_{8}^{17} O +_{1}^{1} H,$ $Q$ aur threshold $α$ kinetic energy nikalo (N target rest mein hai).

Recall Solution

Before: $14.003074 + 4.002603 = 18.005677$ u. After: $16.999132 + 1.007825 = 18.006957$ u. $Δ m = - 0.001280$ u ⇒ bhaare products ⇒ endothermic. $Q = - 0.001280 \times 931.5 = - 1.19 MeV$ . Threshold (projectile $a = α$ , target $X =$ N): $K_{th} = ∣ Q ∣ (1 + \frac{m _{a}}{m _{X}}) = 1.19 (1 + \frac{4.0026}{14.0031}) = 1.19 \times 1.286 = 1.53 MeV .$ $∣ Q ∣$ se zyaada kyun? Products ek dam nahi ruk sakte — center-of-mass momentum conserve karne ke liye aage drift karta rehta hai, aur woh bulk KE reaction ke liye available nahi hoti.

L3.2 — Split of Q in $α$ -decay

$^{210}$ Po $\to^{206}$ Pb $+ α$ ka $Q = 5.40$ MeV hai. Momentum conservation use karke $K_{α}$ aur $K_{D}$ alag-alag nikalo. $m_{α} \approx 4$ u, $m_{D} \approx 206$ u lo.

Recall Solution

Parent rest mein hai ⇒ fragments equal, opposite momentum $p$ ke saath alag jaate hain: $p_{α} = p_{D} = p$ . Kinetic energy $K = p^{2} / (2 m)$ , isliye $K \propto 1/ m$ : halka $α$ bada hissa le jaata hai. $K_{α} = Q \frac{m _{D}}{m _{D} + m _{α}} = 5.40 \times \frac{206}{210} = 5.30 MeV,$ $K_{D} = Q \frac{m _{α}}{m _{D} + m _{α}} = 5.40 \times \frac{4}{210} = 0.103 MeV .$ Check: $5.30 + 0.103 = 5.40 = Q$ . ✓ L2.3 ke saath consistent hai.

Level 4 — Synthesis

L4.1 — Positron-emission correction

$^{?}$ ... ek $β^{+}$ decay $X \to Y + e^{+} + ν$ consider karo jahan atomic masses $m_{X} - m_{Y} = 0.00220$ u satisfy karti hain. True $Q$ nikalo, electron-count mismatch yaad rakhte hue. ( $m_{e} = 0.000549$ u lo; neutrino massless hai.)

Recall Solution

$β^{+}$ decay mein nucleus ek proton kho deta hai, daughter ke atom count mismatch mein ek electron gain karte hue — atomic masses use karne par aapko $2 m_{e} c^{2}$ subtract karna padta hai: $Q = [(m_{X} - m_{Y}) - 2 m_{e}] c^{2} .$ $Δ m = 0.00220 - 2 (0.000549) = 0.00220 - 0.001098 = 0.001102 u .$ $Q = 0.001102 \times 931.5 = 1.03 MeV .$ $2 m_{e}$ kyun? Atomic-mass tables mein har atom ke saath $Z$ electrons shaamil hote hain. $β^{+}$ mein daughter ke paas ek kam proton (ek kam atomic electron) hota hai, AUR ek positron create hota hai — do electron masses unaccounted hain, isliye unhe haath se hatana padta hai. Ye caveat Beta decay mein flag ki gayi hai.

L4.2 — Binding-energy route to Q

D–T fusion $^{2}$ H $+^{3}$ H $\to^{4}$ He $+$ n ke liye masses ki jagah binding energies use karke $Q$ compute karo. Total binding energies diye hain: $B (^{2} H) = 2.224$ MeV, $B (^{3} H) = 8.482$ MeV, $B (^{4} He) = 28.296$ MeV, $B (n) = 0$ .

Recall Solution

$Q = (final total B) - (initial total B)$ — reaction tab energy release karta hai jab products zyaada tightly bound hote hain (Binding energy and mass defect). $Q = B (^{4} He) + B (n) - B (^{2} H) - B (^{3} H)$ $= 28.296 + 0 - 2.224 - 8.482 = 17.59 MeV .$ Ye mass-table answer ( $\approx 17.6$ MeV) se match karta hai — ek hi kamre ke do darwaaze.

Level 5 — Mastery

L5.1 — Fission energy budget

Compound nucleus $^{236}$ U banate neutron-induced fission step ke liye $Q$ estimate karo: $^{235} U + n \to^{236} U^{*} .$ Phir comment karo ki full fission $\sim 200$ MeV kyun release karta hai jabki yeh capture step chhota hai.

Recall Solution

Capture step: Before: $235.043930 + 1.008665 = 236.052595$ u. After: $236.045568$ u. $Δ m = 236.052595 - 236.045568 = 0.007027$ u. $Q = 0.007027 \times 931.5 = 6.55 MeV$ — woh excitation energy jo $^{236}$ U $^{*}$ ko deform aur split hone deti hai. Overall $\sim 200$ MeV kyun: itni badi release actual splitting se aati hai do mid-mass fragments mein, jo binding-energy-per-nucleon curve par $^{236}$ U se kaafi zyaada upar hain (Nuclear fission). $\sim 236$ nucleons ko rearrange karke binding mein $\sim 0.85$ MeV/nucleon gain karne se $\sim 200$ MeV milta hai. Capture $Q$ yahan sirf trigger hai, payoff nahi.

L5.2 — Full synthesis: reaction, sign, threshold, split

$^{9}$ Be $(α, n)^{12}$ C consider karo, yaani $_{2}^{4} He +_{4}^{9} Be \to_{6}^{12} C +_{0}^{1} n .$ (a) $Q$ nikalo. (b) Exo- ya endothermic hai? (c) Agar endothermic ho toh threshold energy do; agar exothermic ho toh explain karo ki threshold kyun nahi. (d) Batao konse conservation laws use kiye.

Recall Solution

(a) Electron check: $Z$ left $= 2 + 4 = 6$ ; right $= 6 + 0 = 6$ . Balanced. Before: $4.002603 + 9.012182 = 13.014785$ u. After: $12.000000 + 1.008665 = 13.008665$ u. $Δ m = 13.014785 - 13.008665 = 0.006120$ u. $Q = 0.006120 \times 931.5 = 5.70 MeV$ . (b) $Q > 0$ ⇒ exothermic (ye classic lab neutron source hai). (c) Koi threshold nahi: energy release hoti hai, isliye Coulomb barrier cross karne wala ek slow $α$ bhi react kar sakta hai. Threshold energy purely endothermic concept hai ( $K_{th} = ∣ Q ∣ (1 + m_{a} / m_{X})$ tab hi sense karta hai jab $Q < 0$ ). (d) Humne use kiya: charge aur nucleon number ka conservation (equation balance karne ke liye), mass-energy ka conservation ( $Q$ nikalne ke liye), aur implicitly momentum (tab hi zaroori hota hai jab hum KE split karein, jaise L3.2 mein).