Radioactive decay — alpha, beta, gamma — mechanisms: Visual walkthrough

2.3.21 · D2 · Physics › Modern Physics › Radioactive decay — alpha, beta, gamma — mechanisms

Yeh Radioactive decay — alpha, beta, gamma — mechanisms ka companion note hai. Agar $Q$ -value ya nucleon number jaisi cheezein shaky lagti hain, toh parent note pehle unhe build karta hai.

Step 1 — "Pehle" ki picture banao: ek nucleus, bilkul still baitha hai

KYA HAI. Hum ek akele unstable parent nucleus se shuru karte hain, space mein, hil nahi raha. Aas paas kuch nahi hai.

YAHAN SE KYUN SHURU KAREIN? Kyunki ek nucleus jo sample mein baitha hai, average pe, at rest hota hai. Agar hum apna viewpoint uske saath chalane wala choose karein, toh decay se pehle zero total motion hai. Yahi ek fact — "pehle ki picture frozen hai" — woh beej hai jis se poori derivation ugti hai.

PICTURE. Parent ek mota, shant ball hai. Hum ek number track karte hain jise momentum kehte hain. Momentum bas "koi cheez kitna motion ka oomph carry karti hai": mass times velocity, $p = m v$ . Ek still object ka $v = 0$ hai, isliye uska momentum $p = 0$ hai. Poore system ka momentum abhi ek bada mota zero hai.

Figure — Radioactive decay — alpha, beta, gamma — mechanisms

Step 2 — "Baad" ki picture banao: do tukde ud jaate hain

KYA HAI. Decay hoti hai. Ab do objects hain:

alpha particle, mass $m_{α}$ , kisi velocity $v_{α}$ se ud raha hai;
bacha hua daughter nucleus, mass $m_{Y}$ , velocity $v_{Y}$ se recoil kar raha hai.

DO KYUN, AUR OPPOSITE KYUN? Kyunki parent exactly inhi do chunks mein toot gaya (parent note ne dikhaya $_{Z}^{A} X \to_{Z - 2}^{A - 4} Y + α$ ). Aur unhe opposite directions mein jaana chahiye — yeh reason aage aata hai.

PICTURE. Ek chhota ball (alpha, coral) right shoot karta hai; ek bada ball (daughter, lavender) left drift karta hai. Unke arrows opposite directions mein point karte hain.

Step 3 — Woh law jo inhe baandh ke rakhta hai: momentum conserved hota hai

KYA HAI. Total momentum after = total momentum before. Pehle zero tha (Step 1). Toh baad mein bhi zero add hona chahiye.

YEH LAW KYUN? Momentum conservation physics ke sabse gehre rules mein se ek hai: bahar se kuch push nahi kar raha, toh total oomph appear ya vanish nahi ho sakta. Nucleus ne alpha ko ek direction mein push kiya; Newton ke "har push ka equal opposite push hota hai" ke according, alpha ne nucleus ko doosri direction mein equal strength se dhakela.

PICTURE. Do arrows, equal length, opposite directions mein point karte hue, kuch nahi milta — jaise do ice skaters ek doosre ko dhakelte hain.

$pehle, Step 1 0 = alpha ka oomph p_{α} - daughter ka oomph p_{Y}$

Minus sign kehta hai woh opposite directions mein hain. Rearrange karne pe:

$p_{α} = p_{Y} \equiv p$

Dono tukde same size ka momentum $p$ carry karte hain. Yahi woh key hai jo sab kuch unlock karti hai — chhota alpha aur bada daughter momentum equally share karte hain, chahe woh energy equally share na karein.

Step 4 — Energy: kinetic energy ko momentum use karke likhna

KYA HAI. Ab hum kinetic energy (KE) laate hain — motion ki energy. Usual form $K E = \frac{1}{2} m v^{2}$ hai, lekin hum ise $v$ ki jagah $p$ mein likhna chahte hain.

ISSE $p$ SE KYUN LIKHEIN? Kyunki Step 3 ne hume $p$ ke baare mein ek fact diya (dono tukde same $p$ share karte hain), $v$ ke baare mein nahi. Woh gift use karne ke liye, hamara energy formula $p$ ki language mein bolna chahiye. Toh hum translate karte hain.

KAISE (translation). $K E = \frac{1}{2} m v^{2}$ se shuru karein. $p = m v$ se $v = p / m$ milta hai. Substitute karein: $K E = \frac{1}{2} m (\frac{p}{m})^{2} = \frac{1}{2} m \cdot \frac{p ^{2}}{m ^{2}} = \frac{p ^{2}}{2 m}$

PICTURE. Same motion, do tareekon se describe ki gayi — ek speedometer view ( $v$ ) aur ek oomph view ( $p$ ). Humne bas dials switch kiye.

Step 5 — Released energy kahan jaati hai? Q-value aata hai

KYA HAI. Do ud rahe tukdon ki total kinetic energy decay se release hone wali energy ke barabar hai, yani Q-value: $K E_{α} + K E_{Y} = Q$

KYUN? Parent (rest-mass-energy mein) daughter + alpha se bhaari tha. Woh missing mass motion energy mein badal gayi: $Q = (m_{parent} - m_{products}) c^{2}$ . Decay se pehle kuch nahi hil raha tha, isliye $Q$ ka sab kuch baad mein dono tukdon ke KE ke roop mein aata hai. Energy conserved hai.

PICTURE. $Q$ label wala tank do cups mein khaali hota hai — ek chhota alpha cup aur ek daughter cup. Cups unequally bharte hain; hamara kaam split dhundhna hai.

Step 4 ka formula use karke har KE likhein, yaad karke ki dono tukde same $p$ share karte hain: $K E_{α} \frac{p ^{2}}{2 m _{α}} + K E_{Y} \frac{p ^{2}}{2 m _{Y}} = Q$

Step 6 — Shared momentum $p$ solve karo

KYA HAI. Hamare paas ek equation hai ek unknown $p$ ke saath (kyunki $Q$ , $m_{α}$ , $m_{Y}$ known hain). Ise solve karo.

KYUN? Jab ek baar $p$ pata chal jaaye, $K E_{α} = p^{2} /2 m_{α}$ mein daalne se alpha ki energy seedha milti hai — yahi goal hai.

KAISE. $p^{2}$ factor out karo: $p^{2} (\frac{1}{2 m _{α}} + \frac{1}{2 m _{Y}}) = Q$

Fractions ko common bottom $2 m_{α} m_{Y}$ pe combine karo: $\frac{1}{2 m _{α}} + \frac{1}{2 m _{Y}} = \frac{m _{Y} + m _{α}}{2 m _{α} m _{Y}}$

Toh: $p^{2} \cdot \frac{m _{Y} + m _{α}}{2 m _{α} m _{Y}} = Q ⟹ p^{2} = \frac{2 Q m _{α} m _{Y}}{m _{α} + m _{Y}}$

PICTURE. Do "motion ke resistance" terms ( $1/2 m_{α}$ aur $1/2 m_{Y}$ ) row mein do springs ki tarah add hote hain; moti daughter bahut chhoti springiness contribute karti hai kyunki uska $1/ m_{Y}$ chhota hai.

KYA HAI. $p^{2}$ ko $K E_{α} = \frac{p ^{2}}{2 m _{α}}$ mein daalo: $K E_{α} = \frac{1}{2 m _{α}} \cdot \frac{2 Q m _{α} m _{Y}}{m _{α} + m _{Y}}$

$2$ cancel hota hai $2$ se, aur ek $m_{α}$ cancel hota hai $m_{α}$ se: $K E_{α} = Q \cdot \frac{m _{Y}}{m _{α} + m _{Y}}$

YEH SAHI KYUN LAGTA HAI. Fraction $\frac{m _{Y}}{m _{α} + m _{Y}}$ almost 1 hai kyunki $m_{Y}$ (bada daughter) $m_{α}$ se bahut zyada bhaari hai. Toh alpha $Q$ ka almost sab kuch le jaata hai — lekin bilkul nahi, kyunki daughter phir bhi recoil karta hai aur ek chhota sa hissa apne paas rakhta hai.

Handy approximation. Mass roughly nucleon number $A$ ke proportional hoti hai: daughter ke $A - 4$ nucleons hain, alpha ke $4$ hain, parent ke $A$ the. Toh $m_{Y} \approx A - 4$ aur $m_{α} + m_{Y} \approx A$ : $K E_{α} \approx Q \cdot \frac{A - 4}{A}$

PICTURE. Ek see-saw: bada daughter pivot ke paas baitha hai (kam energy leta hai), chhota alpha door baitha hai aur zyaadatar energy ke saath upar ud jaata hai.

Step 8 — Edge aur limiting cases (koi gap mat chhodna)

KYA HAI / KYUN. Ek formula tabhi trustworthy hota hai jab extreme situations mein sensible behave kare. Aaiye ise check karein.

Case A — daughter infinitely heavier ( $m_{Y} \to \infty$ ). Toh $\frac{m _{Y}}{m _{α} + m _{Y}} \to 1$ , isliye $K E_{α} \to Q$ . Ek truly immovable wall alpha ko sari energy rakhne degi. Reassuring: hamara formula smoothly "alpha sab kuch le jaata hai" myth ko limit ke roop mein approach karta hai, real nuclei ke liye kabhi isko reach kiye bina.

Case B — equal masses ( $m_{Y} = m_{α}$ ). Toh fraction $\frac{1}{2}$ hai: woh $Q$ equally split karenge. Yeh alpha decay mein kabhi nahi hota (daughter hamesha bahut bhaari hoti hai), lekin yeh dikhata hai ki formula honest hai — equal masses equally share karte hain.

Case C — $Q \leq 0$ . Koi decay nahi. Agar products parent se halke nahi the, toh baantne ke liye koi energy nahi hai, aur Step 6 mein $p^{2}$ zero ya "negative" (unphysical) hoga. Yahi wajah hai ki alpha decay sirf bhaare nuclei ke liye hoti hai jahan $Q > 0$ ho.

Case D — daughter excited chhuta hai. Kabhi kabhi daughter excited state mein land karti hai, isliye $Q$ ka thoda hissa internal excitation ke roop mein store ho jaata hai aur baad mein gamma ke roop mein nikalta hai. Tab kinetic sharing ke liye available $Q$ chhota hota hai, jo thoda lower $K E_{α}$ line deta hai. Yahi wajah hai ki alpha spectra kai discrete lines dikhate hain (daughter state ke hisaab se ek), sirf ek nahi — har ek phir bhi apne reduced $Q$ ke saath hamare split formula ka palan karta hai.

Worked check — Uranium-238

$_{92}^{238} U \to_{90}^{234} Th + α$ ke liye $Q = 4.27$ MeV ke saath: $K E_{α} \approx 4.27 \times \frac{238 - 4}{238} = 4.27 \times \frac{234}{238} = 4.20 MeV$ Daughter thorium quietly remaining $4.27 - 4.20 = 0.07$ MeV rakhti hai. Alpha ne budget ka 98% le liya — exactly wahi jo Step 7 ka see-saw predict karta hai.

Reveal-lines:

U-238 ka alpha kitna KE carry karta hai?

Lagbhag

4.20

MeV (≈98% of

Q = 4.27

MeV).

Isse poore 4.27 MeV kyun nahi milte?

Bhaari Th-234 daughter ko recoil karna padta hai aur woh ≈0.07 MeV rakhti hai.

Ek-picture summary

Poori chain ek image mein: still parent (p=0) → equal-and-opposite $p$ wale do tukde → same $p$ , isliye $K E = p^{2} /2 m$ mass ke inversely split hoti hai → halka alpha bada share leta hai, bhaari daughter sliver leti hai → woh $Q$ mein sum hote hain.

Recall Feynman retelling — poora walk simple shabdon mein

Socho ek bhaari nucleus bilkul dead still float kar raha hai. Woh achanak ek chhote alpha aur ek bade leftover chunk mein crack hota hai. Kyunki pehle kuch nahi hil raha tha, dono tukdon ko equal push ke saath alag udna hai — jaise do skaters ek doosre ko dhakelte hain, ek chhota aur zippy, ek bada aur slow. Ab, "motion ki energy" push-language mein (push squared) ÷ (twice the mass) hai. Dono ke liye same push, lekin bade chunk ke bottom pe badi mass hai — isliye use tiny energy milti hai, aur chhote alpha ko almost sab kuch milta hai. Woh sari energy nucleus ke weight se aayi jo decay hone pe usne khoyi — yahi Q-value hai. Dono shares wapas add karo aur exactly Q milta hai. Alpha ka share $Q$ times (daughter mass)/(total mass) hai, jo bhaare nucleus ke liye almost $Q$ hai lekin kabhi bilkul nahi — kyunki daughter hamesha thoda recoil karti hai. Extremes pocho: infinitely heavy daughter alpha ko sab kuch de degi; equal masses evenly split karenge; aur agar nucleus decay hone se actually halka nahi hota, toh bilkul bhi energy nahi hai aur kuch nahi hota.

Recall Khud test karo

Equal-momentum fact aur kyun batao. ::: Parent at rest → pehle total $p = 0$ → baad mein, $p_{α} = p_{Y}$ (equal aur opposite), isliye dono same magnitude $p$ share karte hain. $K E = p^{2} /2 m$ kyun likhein $\frac{1}{2} m v^{2}$ ki jagah? ::: Kyunki conservation ne hume equal $p$ diya, equal $v$ nahi; $p$ -form hume woh fact use karne deta hai aur dikhata hai ki KE mass badhne ke saath girti hai. $K E_{α} = Q m_{Y} / (m_{α} + m_{Y})$ kya ban jaata hai jab $m_{Y} \to \infty$ ? ::: Yeh $Q$ approach karta hai — immovable-daughter limit jahan alpha sari energy le jaata hai. Alpha spectra mein kai discrete lines kyun dikhti hain? ::: Daughter alag alag excited states mein land kar sakti hai, har ek split ke liye alag available $Q$ chhodti hai, isliye har state ke liye ek distinct fixed $K E_{α}$ hota hai.

Related builds: Quantum tunnelling (alpha barrier se escape kaise karta hai), Nuclear binding energy and mass defect (jahan se $Q$ aata hai), Nuclear energy levels and shell model (Case D ki discrete daughter states), Law of radioactive decay (time ke saath tunnelling kaise add hoti hai).