Worked examples — Nuclear reactions — Q-value calculation

2.3.20 · D3 · Physics › Modern Physics › Nuclear reactions — Q-value calculation

Kuch bhi shuru karne se pehle, teen quick reminders jo HAARE example neeche kaam aate hain:

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Scenario matrix

Har Q-value problem jo tumhare haath aaye woh in cells mein se kisi ek mein fit hogi. Aane wale examples mein har example ka [Cell N] tag hai taaki tum dekh sako ki poora grid cover ho raha hai.

Cell	Scenario class	Isme kya tricky hai	Example
1	$Q>0$, exothermic, mass data	seedha sign check	Ex 1 (fusion)
2	$Q<0$, endothermic + threshold	products rest mein nahi ho sakte	Ex 2
3	$Q$ sirf kinetic energies se (masses nahi)	kinetic face use karo	Ex 3
4	Splitting $Q$ between fragments (momentum)	daughter recoils, mass se share karo	Ex 4
5	$Q=0$ / degenerate (elastic, rest-mass change nahi)	limiting case, $\Delta m=0$	Ex 5
6	Positron (${\beta }^{+}$) decay — atomic-mass trap	extra $2m_e{c}^2$ correction	Ex 6
7	${\beta }^{-}$ decay — atomic masses cancel cleanly	cell 6 se contrast	Ex 7
8	Threshold ka limiting behaviour ($m_X\to \infty$, $m_X\to 0$)	formula ka sanity-check	Ex 8
9	Exam twist / real-world: reactor fission energy per event → per kg	scale-up word problem	Ex 9

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Ex 1 — Exothermic, mass data [Cell 1]

Forecast: do bahut tightly-bound helium nuclei niklenge — guess: strongly positive $Q$, kai MeV. Apna guess likh lo.

1. Pehle: $7.016004+1.007825=8.023829$ u. Yeh step kyun? Reactant rest masses ko add karo — yeh woh total mass hai jisse hum shuru karte hain.
2. Baad mein: $4.002603+4.002603=8.005206$ u. Yeh step kyun? Product rest masses ko add karo.
3. $\Delta m$: $8.023829-8.005206=0.018623$ u. Yeh step kyun? Jo mass gayi. Positive ⇒ mass gaayab hui ⇒ energy nikli.
4. Convert: $Q=0.018623\times 931.5\approx 17.35\mathbf{MeV}$. Yeh step kyun? $\Delta m$ in u times unit-bridge $931.5$ MeV/u energy deta hai.

Verify: Sign $+$ hai (products halke hain) → exothermic, forecast se match. Units: $\text{u}\times (\text{MeV}/\text{u})=\text{MeV}$. ✓ Do alphas sabse zyada bound nuclei mein se hain, toh itna bada release physically sensible hai.

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Ex 2 — Endothermic + threshold [Cell 2]

Forecast: ek halke fragment ko loose karne mein aam taur par energy lagti hai → guess $Q<0$, aur threshold thodi $|Q|$ se badi.

Template $a+X\to Y+b$ se match karte hue: yahan neutron projectile $a$ hai (toh $m_a=m_n=1.008665$ u), oxygen target $X$ hai (toh $m_X=15.994915$ u), ${}^{13}$C is $Y$ aur ${}^4$He is $b$.

1. Pehle: $15.994915+1.008665=17.003580$ u. Kyun? Reactant masses $m_a+m_X$.
2. Baad mein: $13.003355+4.002603=17.005958$ u. Kyun? Product masses $m_Y+m_b$.
3. $\Delta m$: $17.003580-17.005958=-0.002378$ u. Negative kyun? Products bhaari hain — mass create karni padi, isliye energy absorb hui.
4. $Q$: $-0.002378\times 931.5=-2.215\mathbf{MeV}$. Endothermic.
5. Threshold — poori algebraic derivation. Symbols yaad karo: $m_a$ = projectile mass, $m_X$ = (stationary) target mass, $K_a$ = projectile kinetic energy lab mein, $K_{\text{th}}$ = $K_a$ ki minimum value jis par reaction ho sake. Hum formula char honest steps mein banate hain.
   • (i) Momentum in. Projectile lab momentum $p_a$ carry karta hai; target rest mein hai, toh total momentum $p_a$ hai. Non-relativistically $K_a=\frac{p_a^2}{2m_a}$, toh $p_a=\sqrt{2m_a{K}_a}$.
   • (ii) Products ko chalte rehna hai. Total momentum zero nahi ho sakta, toh saare products rest mein nahi ho sakte. Threshold par (minimum energy) products barely milke chalte hue create hote hain — ek single lump of mass $(m_a+m_X)$ poora momentum $p_a$ carry karta hai. Uski kinetic energy unavoidable centre-of-mass energy hai: $K_{\text{cm}}=\frac{p_a^2}{2(m_a+m_X)}=\frac{2m_a{K}_a}{2(m_a+m_X)}=\frac{m_a}{m_a+m_X}{K}_a.$
   • (iii) Jo energy actually available hai. Sirf isi bulk-motion floor ke upar wali kinetic energy reaction ke mass deficit $|Q|$ par kharch ho sakti hai. Toh available energy hai: $K_a-K_{\text{cm}}=K_a\left(1-\frac{m_a}{m_a+m_X}\right)=K_a\cdot \frac{m_X}{m_a+m_X}.$
   • (iv) Available energy $=|Q|$ set karo aur $K_a$ solve karo. Threshold par $K_a=K_{\text{th}}$: $K_{\text{th}}\cdot \frac{m_X}{m_a+m_X}=|Q|⟹K_{\text{th}}=|Q|\frac{m_a+m_X}{m_X}=|Q|\left(1+\frac{m_a}{m_X}\right).$ Ab numbers daalo ($m_a=1.008665$, $m_X=15.994915$, $|Q|=2.215$ MeV): $K_{\text{th}}=2.215\left(1+\frac{1.008665}{15.994915}\right)=2.215\times 1.06306=2.355\mathbf{MeV}.$ Yeh step kyun? Bracket exactly woh extra energy hai jo unavoidable bulk motion mein phasi hai; tumhe $|Q|$ ke upar yeh bhi supply karni hogi.

Verify: $K_{\text{th}}>|Q|$ ✓ (jaisa hona chahiye). Correction factor $1.063$ close to $1$ hai kyunki projectile target ke muqable halka hai — Ex 8 mein limits dekho.

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Ex 3 — Q seedha kinetic energies se [Cell 3]

Forecast: parent rest mein hai toh kuch andar nahi ja raha; $Q$ bas woh hai jo bahar aata hai. Guess $\approx 4.87$ MeV.

1. $K_{\text{initial}}$: parent rest mein hai $\Rightarrow K_{\text{initial}}=0$. Kyun? Pehle kuch bhi move nahi kar raha tha.
2. $K_{\text{final}}$: $4.78+0.086=4.866$ MeV. Kyun? SAARE fragments ki total kinetic energy — daughter recoil bhi count hoti hai.
3. $Q$: $Q=K_{\text{final}}-K_{\text{initial}}=4.866-0=4.866\mathbf{MeV}$. Yeh step kyun? $Q$ ka kinetic face: gaayab rest mass fragment kinetic energy ke roop mein wapas aati hai. Humhe masses ki zaroorat hi nahi padi.

Verify: $Q>0$ (spontaneous decay hamesha exothermic hoti hai — warna hoti hi nahi). $0.086$ MeV recoil bhoolna $Q$ ko kam count karega — yehi exact common mistake hai jo yeh example rokne ke liye banaya gaya hai.

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Ex 4 — Q ko fragments mein split karna [Cell 4]

Figure s01 (neeche reference hai): yeh poora event ek horizontal axis par dikhata hai (recoil line ke along position). Parent ek single horizontal axis ke center par rest mein baitha hai (black dot, momentum $=0$); phir woh split hota hai fast light alpha (red, lamba arrow, right pointing) aur slow heavy radon daughter (black, chota arrow, left pointing) mein. Do labelled arrows momenta $p_{\alpha }$ aur $p_{\text{Rn}}$ hain — length mein equal but opposite direction mein drawn, equal and opposite momenta ka key visual. Alpha kaafi zyada right mein end hota hai kyunki woh faster move karta hai.

Forecast: halka alpha tezi se bhagega; bhaari daughter mushkil se hilegi. Guess: alpha almost saara $Q$ le jaata hai.

1. Momentum conservation. Parent rest mein hai ⇒ total momentum $=0$, toh dono fragments equal and opposite momenta ke saath alag bhaagte hain (Figure s01 mein red alpha arrow aur black recoil arrow dekho — same length, opposite direction): $p_{\alpha }=p_{\text{Rn}}\equiv p.$ Yeh step kyun? Pehle kuch move nahi kar raha tha; momentum kahin se aa nahi sakta, toh momenta cancel out hone chahiye.
2. Momentum ke through kinetic energy likho. Non-relativistic fragment ke liye (speed $\ll c$, jo yahan valid hai kyunki alpha/daughter energies few MeV hoti hain aur speed $c$ ke sirf kuch percent hoti hai), kinetic energy hai $K=\frac{p^2}{2m}$. Yeh step kyun? Dono fragments same $p$ share karte hain, toh $K$ ko $p$ ke terms mein likhne se hum unhe directly compare kar sakte hain. (MeV-scale nuclear fragments ke liye yeh classical formula $1\%$ se bhi zyada accurate hai; sirf hundreds of MeV par relativistic version chahiye hoga.)
3. Share ratio. $\frac{K_{\alpha }}{K_{\text{Rn}}}=\frac{p^2/2m_{\alpha }}{p^2/2m_{\text{Rn}}}=\frac{m_{\text{Rn}}}{m_{\alpha }}.$ Yeh step kyun? Halka fragment zyada bada share leta hai — energy mass ke inversely split hoti hai.
4. $K_{\alpha }$ solve karo. Kyunki $Q=K_{\alpha }+K_{\text{Rn}}$: $K_{\alpha }=Q\cdot \frac{m_{\text{Rn}}}{m_{\text{Rn}}+m_{\alpha }}=4.87\times \frac{222}{226}=4.784\mathbf{MeV}.$ Yeh step kyun? Yahi exact split rule hai jo parent note ke mistake box mein quoted tha, ab derive kiya.

Verify: $K_{\text{Rn}}=Q-K_{\alpha }=4.87-4.784=0.086$ MeV — yehi exactly Ex 3 mein di gayi recoil energy hai! ✓ Dono examples consistent hain. Alpha $Q$ ka $98.2\%$ rakhta hai, forecast se match karta hai.

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Ex 5 — Degenerate case: $Q=0$ (elastic) [Cell 5]

Forecast: "pehle ka pile" aur "baad ka pile" same particles hain — kuch create ya destroy nahi hua. Guess $Q=0$.

1. Pehle: $m_n+m_{\text{C}}$. Baad mein: $m_{\text{C}}+m_n$. Kyun? List karo — literally identical hain.
2. $\Delta m$: $(m_n+m_{\text{C}})-(m_{\text{C}}+m_n)=0$. Kyun? Same terms term-by-term cancel ho jaate hain.
3. $Q=0\times 931.5=0$. Yeh step kyun? Rest-mass change nahi ⇒ koi rest-mass energy release ya absorb nahi hui. Total kinetic energy sirf dono ke beech redistribute hoti hai, change nahi hoti.

Verify: Yeh sign axis ka limiting middle hai: $Q>0$ (Ex 1), $Q<0$ (Ex 2), $Q=0$ yahan. Parent ke teesre bullet se match karta hai: "$Q=0$ → elastic." Kinetic energy conserve hoti hai lekin repackage hoti hai — reactor neutrons ko moderate (slow down) karne ke liye carbon use karne ka yehi poora point hai.

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Ex 6 — Positron decay: atomic-mass trap [Cell 6]

Forecast: correction $2m_e{c}^2=2\times 0.511=1.022$ MeV ek real chunk hai. Agar tum ise bhool gaye toh tumhara $Q$ almost $1$ MeV zyada aayega.

1. Nuclear-mass truth. Nuclear masses $M$ ke terms mein, $Q=[M_{\text{C}}-M_{\text{B}}-m_e]c^2.$ Yeh step kyun? Ek positron ($+m_e$) genuinely create hoti hai, aur neutrino massless hai — toh hum explicitly ek electron-mass ki rest energy kho dete hain.
2. Nuclear se atomic masses mein convert karo. Atomic mass $=$ nuclear mass $+Z{m}_e$ (electron binding ignore karte hue), jahan $Z$ (atomic number) proton/electron count hai. Toh $M_{\text{C}}=m{(}^{11}\text{C})-6m_e$ (carbon ka $Z=6$) aur $M_{\text{B}}=m{(}^{11}\text{B})-5m_e$ (boron ka $Z=5$). Yeh step kyun? Humhe atomic masses di gayi thi; substitute karo taaki unhe use kar sakein.
3. Substitute karo aur simplify karo. $Q=[(m_{\text{C}}-6m_e)-(m_{\text{B}}-5m_e)-m_e]c^2=[m_{\text{C}}-m_{\text{B}}-2m_e]c^2.$ Yeh step kyun? Electron bookkeeping collapse hokar ek single $-2m_e$ correction mein aa jaati hai — woh extra term jiske baare mein parent ne warn kiya tha.
4. Numbers. $m_{\text{C}}-m_{\text{B}}=11.011434-11.009305=0.002129$ u. $2m_e=0.001098$ u subtract karo: $\Delta m=0.001031$ u. $Q=0.001031\times 931.5=0.960\mathbf{MeV}.$

Verify: ${}^{11}$C ${\beta }^{+}$ decay ka known experimental $Q$ $\approx 0.96$ MeV hai ✓. Agar tumne (galti se) $-2m_e$ skip kar diya hota: $0.002129\times 931.5=1.983$ MeV — lagbhag double, aur exactly $1.022$ MeV off, woh $2m_e{c}^2$ jo tumne drop kiya. Yehi trap hai.

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Ex 7 — ${\beta }^{-}$ decay: atomic masses cleanly cancel [Cell 7]

Forecast: Ex 6 se contrast — yahan $Z$ ek badh jaata hai, aur emitted electron ki mass pehle se daughter ki atomic mass ke andar account hai. Guess: koi correction term nahi.

1. Nuclear truth: $Q=[M_{\text{C}}-M_{\text{N}}-m_e]c^2$ (create hue electron ke liye $-m_e$). Kyun? Ek electron genuinely emit hota hai.
2. Atomic substitution: $M_{\text{C}}=m_{\text{C}}-6m_e$ (carbon $Z=6$), $M_{\text{N}}=m_{\text{N}}-7m_e$ (nitrogen $Z=7$). Kyun? Jo atomic masses di gayi hain unhe use karne ke liye convert karo.
3. Simplify karo: $Q=[(m_{\text{C}}-6m_e)-(m_{\text{N}}-7m_e)-m_e]c^2=[m_{\text{C}}-m_{\text{N}}]c^2.$ Yeh step kyun? $-6m_e$, $+7m_e$, aur $-m_e$ exactly cancel ho jaate hain — emitted electron neatly daughter ke electron cloud mein slot ho jaata hai. Koi correction nahi bachti, Ex 6 ke unlike.
4. Numbers: $\Delta m=14.003242-14.003074=0.000168$ u. $Q=0.000168\times 931.5=0.156\mathbf{MeV}.$ Yeh step kyun? $\Delta m$ in u times $931.5$ MeV/u released energy deta hai.

Verify: ${}^{14}$C beta spectrum ki textbook endpoint energy $0.156$ MeV hai ✓. Symmetry dekho: ${\beta }^{-}$ mein koi correction nahi (Ex 7), ${\beta }^{+}$ mein $-2m_e{c}^2$ correction hai (Ex 6). Same idea, opposite bookkeeping.

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Ex 8 — Threshold ka limiting behaviour [Cell 8]

Figure s02 (neeche reference hai) threshold formula ka key ratio $K_{\text{th}}/|Q|$ (vertical axis) plot karta hai mass ratio $m_a/m_X$ = projectile-over-target (horizontal axis) ke against. Red curve formula $K_{\text{th}}/|Q|=1+m_a/m_X$ hai; dashed black line height $1$ par woh floor hai jis par woh heavy targets ke liye approach karta hai; black dot equal-mass case $m_a=m_X$ mark karta hai jahan ratio exactly $2$ hai.

Forecast: ek brick wall se takrana (heavy target) recoil mein almost koi energy waste nahi karni chahiye; ek pankh se takrana (light target) almost sab kuch waste kar dena chahiye.

1. Heavy target, $m_X\to \infty$: $\frac{m_a}{m_X}\to 0$, toh $K_{\text{th}}\to |Q|$. Yeh step kyun? Ek massive target barely recoil karta hai, toh bulk motion mein almost koi energy waste nahi hoti — tumhe bas $|Q|$ chahiye. Isliye Ex 2 ka factor sirf $1.063$ tha: oxygen neutron se $16\times$ bhaari hai. Graph par red curve dashed floor par flatten ho jaati hai.
2. Light target, $m_X\to 0$: $\frac{m_a}{m_X}\to \infty$, toh $K_{\text{th}}\to \infty$. Yeh step kyun? Ek tiny target enormous momentum lekar ud jaata hai; almost saari incoming energy wasted bulk motion ban jaati hai, toh threshold blow up ho jaata hai. Graph par red curve right par upar shoot karti hai.
3. Middle check, $m_X=m_a$: $K_{\text{th}}=2|Q|$. Yeh step kyun? Equal masses recoil symmetrically share karte hain — exactly aadhi energy unavailable hoti hai, toh tumhe $|Q|$ ka double chahiye. Curve marked black dot se cleanly guzarti hai.

Verify: Teeno limits monotonic aur sensible hain: bhaara target ⇒ sasta threshold. Formula kabhi $|Q|$ se neeche nahi jaata (bracket hamesha $\ge 1$ positive masses ke liye) ✓, $K_{\text{th}}\ge |Q|$ hamesha guarantee karta hai — parent note ka promise.

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Ex 9 — Exam twist / real-world: fission energy per kilogram [Cell 9]

Forecast: nuclear energy densities famously "ek million times" chemical hain. Guess: answer $\sim 10^{13}$–$10^{14}$ J around hoga 1 kg se, aur coal se ratio near a million.

1. Energy per fission in joules: $200\text{MeV}\times 1.602\times 10^{-13}\text{J/MeV}=3.204\times 10^{-11}$ J. Yeh step kyun? Q-value ko SI joules mein convert karo taaki hum ise scale up kar sakein. (MeV khud $\Delta m{c}^2$ se aaya tha, toh $c^2$ pehle se baked in hai.)
2. 1 kg mein nuclei ki sankhya: $N=\frac{1000\text{g}}{235\text{g/mol}}\times N_A=4.2553\text{mol}\times 6.022\times 10^{23}=2.5626\times 10^{24}$ nuclei. Yeh step kyun? Total energy $=$ (energy per fission) $\times$ (kitne nuclei hain), toh pehle nuclei count karo.
3. Total energy: $E=N\times (3.204\times 10^{-11}\text{J})=2.5626\times 10^{24}\times 3.204\times 10^{-11}=8.21\times 10^{13}\mathbf{J}$. Yeh step kyun? Seedha multiplication — payoff.
4. Coal se compare karo: $\frac{8.21\times 10^{13}\text{J}}{3\times 10^7\text{J}}\approx 2.7\times 10^6$. Yeh step kyun? Poora point yehi hai: 1 kg nuclear fuel 1 kg coal se 2.7 million times zyada energy release karta hai.

Verify: Units: $\text{nuclei}\times \text{J/nucleus}=\text{J}$ ✓. Order of magnitude $10^{13}$ J forecast se match karta hai, aur ratio $\sim 10^6$ "a million times" se match karta hai. $8.2\times 10^{13}$ J $\approx 23$ GWh — roughly ek chote power plant ka daily output ek akele kilogram se, woh famous headline number. ✓

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Recall Quick self-test (answers cover karo)

Har ek kahan fit hota hai? Match karo, sirf recite mat karo.

Reaction mein products reactants se bhaari hain — Q ka sign kya aur kya threshold hai? ::: $Q<0$ endothermic, haan threshold hai $K_{\text{th}}=|Q|(1+m_a/m_X)$ [Cell 2]. Tumhe sirf $K_{\alpha }$ aur $K_{\text{daughter}}$ pata hai, masses nahi — Q ka kaunsa face? ::: Kinetic face, $Q=K_{\text{final}}-K_{\text{initial}}$ [Cell 3]. Atomic masses ke saath ${\beta }^{+}$ decay — kaunsa extra term? ::: $2m_e{c}^2$ subtract karo [Cell 6]; ${\beta }^{-}$ ko koi correction nahi chahiye [Cell 7]. Jab target mass $m_X\to \infty$, threshold kya approach karta hai? ::: $K_{\text{th}}\to |Q|$ [Cell 8]. Alpha decay mein kaunsa fragment zyaada $Q$ carry karta hai aur kyun? ::: halka alpha, share $\propto m_{\text{daughter}}/(m_{\text{daughter}}+m_{\alpha })$ momentum conservation se [Cell 4].

Prerequisites & neighbours: Mass-energy equivalence E=mc^2 · Binding energy and mass defect · Conservation of momentum in decays · Alpha decay · Beta decay · Threshold energy in particle reactions · Nuclear fission · Nuclear fusion