Nuclear reactions — Q-value calculation — Hinglish notes

2.3.20 · Physics › Modern Physics

Q-value KYA hota hai?

Masses nuclei ki rest masses ho sakti hain, YA (zyada convenient tarike se) atomic masses — kyunki electrons cancel ho jaate hain (hum yeh neeche check karenge).

Mass energy mein KYU convert hoti hai?

Q ko first principles se KAISE derive karein

Total energy ke conservation se shuru karo (rest + kinetic), kyunki nucleon number aur charge conserved hain lekin rest mass nahi hoti:

$before (m_{a} + m_{X}) c^{2} + (K_{a} + K_{X}) = after (m_{Y} + m_{b}) c^{2} + (K_{Y} + K_{b})$

Rest masses ko ek taraf aur kinetic energies ko doosri taraf group karne ke liye rearrange karo:

$\equiv Q [(m_{a} + m_{X}) - (m_{Y} + m_{b})] c^{2} = (K_{Y} + K_{b}) - (K_{a} + K_{X})$

Kaam aane wala unit fact (ek baar derive karo, hamesha use karo)

$1 u = 931.494 MeV / c^{2}$ . Toh agar $Δ m$ atomic mass units u mein hai: $Q (MeV) = Δ m (u) \times 931.5$

Atomic masses use karna safe kyun hai

$a + X \to Y + b$ mein total proton number (electrons) conserved hota hai: $Z_{a} + Z_{X} = Z_{Y} + Z_{b}$ . Har atomic mass apne $Z$ electrons saath laata hai. Jab tum difference lete ho, same number ki electron masses dono taraf aati hain aur cancel ho jaati hain. (Caveat: yeh positron emission / electron capture ke liye fail hota hai, jahan electron count match nahi hoti — wahan tumhe correction terms add karni padengi.)

Figure — Nuclear reactions — Q-value calculation

Endothermic reactions: THRESHOLD energy

$Q < 0$ ke liye tumhe kinetic energy supply karni padti hai. Lekin tum sirf $∣ Q ∣$ supply nahi kar sakte — momentum bhi conserved hona chahiye, isliye products ko rest mein nahi chhoda ja sakta. Threshold kinetic energy (projectile $a$ stationary target $X$ se takraye) yeh hai:

$K_{th} = ∣ Q ∣ (1 + \frac{m _{a}}{m _{X}})$

Worked Example 1 — Exothermic (classic D-T fusion)

Reaction: $_{1}^{2} H +_{1}^{3} H \to_{2}^{4} He +_{0}^{1} n$ Atomic/particle masses (u): $^{2}$ H = 2.014102, $^{3}$ H = 3.016049, $^{4}$ He = 4.002603, n = 1.008665.

Step 1 — mass before. $2.014102 + 3.016049 = 5.030151$ u. Kyun? Reactant rest masses ka sum.

Step 2 — mass after. $4.002603 + 1.008665 = 5.011268$ u. Kyun? Product rest masses ka sum.

Step 3 — $Δ m$ . $5.030151 - 5.011268 = 0.018883$ u. Kyun? Jo mass gayab hui.

Step 4 — convert. $Q = 0.018883 \times 931.5 \approx 17.6 MeV$ . Kyun? $Δ m > 0 \Rightarrow Q > 0$ , exothermic — isliye fusion stars ko power karti hai.

Worked Example 2 — Endothermic + threshold

Reaction: $_{7}^{14} N +_{2}^{4} He \to_{8}^{17} O +_{1}^{1} H$ (Rutherford ka pehla transmutation). Masses (u): $^{14}$ N = 14.003074, $^{4}$ He = 4.002603, $^{17}$ O = 16.999132, $^{1}$ H = 1.007825.

Step 1 — before: $14.003074 + 4.002603 = 18.005677$ u. Step 2 — after: $16.999132 + 1.007825 = 18.006957$ u. Step 3 — $Δ m = 18.005677 - 18.006957 = - 0.001280$ u. Negative kyun? Products bhaari hain → energy absorb hoti hai. Step 4 — $Q = - 0.001280 \times 931.5 = - 1.19 MeV$ . Endothermic. Step 5 — threshold (α projectile, N target): $K_{th} = 1.19 (1 + \frac{4.0026}{14.0031}) = 1.19 \times 1.286 \approx 1.53 MeV$ . Yeh step kyun? Tumhe kam se kam itni α kinetic energy chahiye warna reaction ho hi nahi sakti.

Worked Example 3 — Q kinetic energies se (bina mass table ke!)

α-decay $^{210}$ Po $\to$ $^{206}$ Pb $+ α$ . Maan lo α $K_{α} = 5.30$ MeV ke saath nikalta hai aur daughter $K_{Pb} = 0.10$ MeV ke saath recoil karti hai; parent rest mein hai.

$Q = K_{final} - K_{initial} = (5.30 + 0.10) - 0 = 5.40 MeV$ Kyun? $Q$ ka KE wala form — released rest mass fragments ki total kinetic energy ke roop mein dikhta hai. Masses lookup karne ki zaroorat nahi.

Common mistake Common errors ko steel-man karke dekho

(1) " $Q$ = sirf emitted particle ki kinetic energy." Sahi kyun lagta hai: α (ya fast particle) zyaatar energy le jaata hai. Fix: Momentum conservation heavy daughter ko bhi recoil karne par majboor karti hai. $Q = K_{α} + K_{daughter}$ . α-decay ke liye, $K_{α} = Q \cdot \frac{m _{D}}{m _{D} + m _{α}}$ (thoda $Q$ se kam).

(2) "Nuclear masses use kar raha hoon, lekin positron emission ke liye atomic masses plug in kar diye." Sahi kyun lagta hai: atomic masses usually electrons ko cleanly cancel kar dete hain. Fix: $β^{+}$ decay mein electron count mismatched hoti hai; atomic masses use karne par extra $2 m_{e} c^{2}$ subtract karna padta hai.

(3) "Threshold energy = |Q|." Sahi kyun lagta hai: tumhe "bas" missing energy replace karni hai. Fix: Products rest mein nahi ho sakte (momentum!). $K_{th} = ∣ Q ∣ (1 + m_{a} / m_{X})$ use karo.

(4) Sign confusion. $Q = m_{before} - m_{after}$ . Agar tum ise ulta karo, exothermic endothermic lagegi. Anchor: halke products ⇒ positive $Q$ ⇒ energy out.

Recall Feynman: 12-saal ke bachche ko samjhao

Socho LEGO blocks ek saath snap kiye hue hain — unhe alag karna ya unhe ek zyada tight shape mein re-snap karna thodi si "click" energy release kar sakta hai. Nuclei bhi aise hi hain, lekin "click" bahut badi hoti hai kyunki $E = m c^{2}$ thode se gayab hue wazan ko bahut badi energy mein badal deta hai. Q-value bas yeh hai: pehle sab kuch tolo, baad mein sab kuch tolo, aur jo grams missing hain woh energy ban jaate hain. Agar kuch missing nahi hua, koi energy nahi. Agar baad wali pile bhaari hai, tumhe energy andar push karni padi hai usse banane ke liye.

Ek nuclear reaction ki Q-value define karo.

Q = [(m_{a} + m_{X}) - (m_{Y} + m_{b})] c^{2}

, release hone wali energy; total kinetic energy ke change

K_{f ina l} - K_{ini t ia l}

ke barabar.

Q>0 vs Q<0 ka kya matlab hai?

Q > 0

exothermic (energy release, products halke);

Q < 0

endothermic (energy absorb, products bhaari, threshold hota hai).

u aur MeV ke beech conversion factor?

1 u = 931.5 MeV / c^{2}

, isliye

Q (MeV) = Δ m (u) \times 931.5

Q mein nuclear masses ki jagah atomic masses kyun use kar sakte hain?

Charge (electron number) conserved hota hai, isliye dono sides par equal electron masses aati hain aur cancel ho jaati hain —

β^{+}

/ electron capture ke liye fail hota hai.

Endothermic reaction ke liye threshold kinetic energy (projectile a, target X)?

K_{t h} = ∣ Q ∣ (1 + \frac{m _{a}}{m _{X}})

Threshold energy |Q| se zyada kyun hoti hai?

Momentum conservation products ko chalte rehne par majboor karti hai; CM kinetic energy available nahi hoti, isliye extra energy chahiye.

D-T fusion

^{2}

H+

^{3}

H→

^{4}

He+n ke liye approximate Q?

Lagbhag 17.6 MeV (exothermic).

α-decay mein Q kaise split hota hai?

Q = K_{α} + K_{d a ug h t er}

;

K_{α} = Q m_{D} / (m_{D} + m_{α})

, daughter baaki ke saath recoil karti hai.

Connections

Mass-energy equivalence E=mc^2
Binding energy and mass defect
Nuclear fission aur Nuclear fusion
Conservation of momentum in decays
Alpha decay / Beta decay energetics
Threshold energy in particle reactions