Radioactive decay — alpha, beta, gamma — mechanisms: Worked examples

2.3.21 · D3 · Physics › Modern Physics › Radioactive decay — alpha, beta, gamma — mechanisms

The scenario matrix

Har decay problem inhi cells mein se ek hota hai. Columns ko kaun sa decay maano aur rows ko kya mushkil hai maano.

Cell	Case class	Nasty kya hai	Example
A1	Alpha — energy split	daughter recoil karta hai, $Q$ share karta hai	Ex 1
A2	Alpha — limiting mass	$A \to \infty$ hone par kya hota hai?	Ex 2
B1	Beta-minus — endpoint	continuous spectrum, max energy	Ex 3
B2	Beta-plus vs capture — threshold	$Q$ zero ya negative ho sakta hai (degenerate)	Ex 4
C1	Gamma — recoil correction	photon ka bhi momentum hota hai	Ex 5
D1	Chain — $Z, A$ track karo	multiple decays, bookkeeping	Ex 6
E1	Real-world word problem	smoke detector ( $^{241}$ Am)	Ex 7
E2	Exam twist — impossible pehchano	ek decay jo ho nahi sakta ( $Q < 0$ )	Ex 8

Hum in constants ko poore time use karte hain. $1 u = 931.5 MeV / c^{2}$ (yeh sirf ek atomic mass unit mein chupi energy hai, $E = m c^{2}$ se). Electron ki rest energy hai $m_{e} c^{2} = 0.511 MeV$ .

Definition Symbols, zero se

$A$ = mass number = protons + neutrons ka total count ("chunk count").
$Z$ = atomic number = protons ki sankhya (isse pata chalta hai kaunsa element hai).
$Q$ = energy released, MeV mein. Positive $Q$ ⇒ decay allowed hai.
$K E$ = kinetic energy = motion ki energy, $K E = \frac{1}{2} m v^{2}$ kam speed par.
$p$ = momentum = $m v$ ("quantity of motion" ka ek measure). Conserved: pehle ka total = baad ka total.
$c$ = speed of light ( $3 \times 1 0^{8} m/s$ ). Yeh sirf $E = m c^{2}$ ke zariye aata hai (mass ↔ energy) aur massless photons ke liye $p = E / c$ (photon ka momentum uski energy divided by $c$ hai). Jab bhi $c$ dekho, padho "mass aur energy ke beech exchange rate".
$m_{α}$ = alpha particle ki mass; $m_{Y}$ = daughter nucleus ki mass (hum kisi bhi daughter ko $Y$ bolte hain, jaise $X \to Y + α$ ); $M$ = ek nucleus ki total mass generally. Teeno ek hi tarah ki cheez hain — rest mass — bas alag objects ke liye.

Ex 1 · Cell A1 — Alpha: energy kise milti hai?

Step 1 — Split draw karo (kyun: momentum conservation dekhne ke liye). Uranium shuru mein rest par hai, toh total momentum zero hai. Decay ke baad alpha ek taraf udta hai aur thorium ulti direction mein recoil karta hai — equal and opposite momentum ke saath. Neeche ke figure mein, orange arrow alpha ka momentum $p_{α}$ hai aur teal arrow thorium ka momentum $p_{T h}$ hai. Dono arrows ki length same hai (momentum ka size equal) lekin opposite directions mein point karte hain — yahi "total momentum zero rehta hai" dikhta hai.

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

Figure: U-238 rest par baitha hai (black dot). Woh ek halke alpha (orange, right) aur ek bhaare thorium daughter (teal, left) mein split hota hai. Dono momentum arrows length mein equal aur direction mein opposite hain, toh woh zero tak cancel ho jaate hain — momentum conserved hai. Kyunki $K E = p^{2} /2 m$ aur alpha ki mass kaafi chhoti hai, alpha lagbhag saari kinetic energy le jaata hai.

Step 2 — Momentum conservation likho. Yahan $p_{T h}$ matlab thorium daughter ka momentum hai (hamara general daughter $Y$ is case mein thorium hai). $0 = p_{α} - p_{T h} \Rightarrow p_{α} = p_{T h} \equiv p$ Yeh step kyun? Kyunki "total momentum zero rehta hai" woh ek hi handle hai jo do unknown speeds ko link karta hai — yeh dono particles ko ek common momentum size $p$ share karne par majboor karta hai.

Step 3 — $K E = p^{2} /2 m$ se kinetic energy likho. Hum $K E = p^{2} /2 m$ use karte hain (na ki $\frac{1}{2} m v^{2}$ ) kyunki hum shared momentum $p$ jaante hain, velocities nahi. (Yeh form seedha $p = m v$ se aata hai: $v = p / m$ ko $\frac{1}{2} m v^{2}$ mein substitute karo toh $\frac{1}{2} m (p / m)^{2} = p^{2} /2 m$ milta hai.) Dono ke liye same $p$ , toh: $K E_{α} = \frac{p ^{2}}{2 m _{α}}, K E_{T h} = \frac{p ^{2}}{2 m _{T h}}$ Yeh step kyun? Is form mein mass denominator mein hai, toh halka particle (chhota $m$ ) bada $K E$ lega. Alpha jeet jaata hai.

Step 4 — Share formula derive karo (koi formula hawaai nahi). Dono kinetic energies add karo; unhe released energy $Q$ ke barabar hona chahiye: $Q = K E_{α} + K E_{T h} = \frac{p ^{2}}{2 m _{α}} + \frac{p ^{2}}{2 m _{T h}} = \frac{p ^{2}}{2} (\frac{1}{m _{α}} + \frac{1}{m _{T h}}) = \frac{p ^{2}}{2} \cdot \frac{m _{α} + m _{T h}}{m _{α} m _{T h}}$ Yeh step kyun? Hum $K E_{α}$ sirf $Q$ ke terms mein chahte hain, toh pehle $Q$ use karke $p^{2}$ express karo. Rearrange karo: $p^{2} = \frac{2 Q m _{α} m _{T h}}{m _{α} + m _{T h}}$ Ab ise $K E_{α} = p^{2} /2 m_{α}$ mein daalo; $m_{α}$ ek baar cancel ho jaata hai: $K E_{α} = \frac{p ^{2}}{2 m _{α}} = Q \cdot \frac{m _{T h}}{m _{α} + m _{T h}}$ Yeh step kyun? Yahi woh formula hai jo parent note ne quote kiya tha — lekin ab tumne ise momentum + energy conservation se girte dekha hai, thama nahi gaya. Kyunki $m \approx A u$ (upar wala ratio-safe box dekho), $m_{T h} / (m_{α} + m_{T h}) \approx 234/ (4 + 234) = 234/238 = (A - 4) / A$ . Daalo: $K E_{α} = 4.27 \times \frac{234}{238} = 4.198 MeV$ Phir recoil baaki leta hai: $K E_{T h} = Q - K E_{α} = 4.27 - 4.198 = 0.072 MeV$

Verify: $K E_{α} + K E_{T h} = 4.198 + 0.072 = 4.27 = Q$ . ✓ Energy poori account ho gayi. Alpha ko ~98% mila — halka particle jeeta, jaisa Step 3 ne forecast kiya tha.

Ex 2 · Cell A2 — Alpha: limiting case

Step 1 — Limit lo. $lim_{A \to \infty} \frac{A - 4}{A} = lim_{A \to \infty} (1 - \frac{4}{A}) = 1$ Yeh step kyun? Ek limit is sawaal ka jawaab deta hai "jab $A$ bez-had badhta hai toh yeh expression kis value ki taraf jaata hai?" — long-run behaviour padhne ka sabse saaf tarika. Jab $A \to \infty$ toh term $4/ A$ zero tak simit jaata hai, toh fraction $1$ ki taraf badhta hai. Ek super-heavy nucleus barely recoil karta hai, toh alpha essentially sabhi $Q$ le jaata hai.

Step 2 — Chhote end check karo (degenerate input). Formula ko ek real daughter chahiye, toh $A - 4 \geq 1$ , yaani $A \geq 5$ . Sabse chhote sensible case $A = 5$ par: fraction $= (5 - 4) /5 = 0.2$ . Alpha sirf 20% leta hai — chhoti daughter zor se recoil karti hai. Yeh step kyun? Extreme small input test karna tab pata chalta hai jab formula toota ya oddly behave kiya; yahan yeh dikhata hai ki alpha ka share utna hi kam ho sakta hai jitna ek-paanchwa, Step 1 ke opposite extreme par.

Verify: monotonic aur bounded — $(A - 4) / A$ $0.2$ se (at $A = 5$ ) $1$ ki taraf (at $A = \infty$ ) badhta hai aur kabhi $1$ se zyada nahi hota. Physically sane: alpha kabhi $Q$ ka 100% se zyada nahi le sakta. ✓

Ex 3 · Cell B1 — Beta-minus: endpoint energy

Step 1 — Mass difference compute karo. $Δ m = 14.003242 - 14.003074 = 0.000168 u$ Yeh step kyun? Decay energy gayi hui mass se aati hai, toh pehla kadam hamesha yeh dhundhna hai ki kitni mass gayi. $β^{-}$ ke liye, atomic masses use karne se electron masses automatically sahi tarike se bundle ho jaati hain, toh $m_{e}$ alag se add nahi karna.

Step 2 — Mass ko energy mein convert karo. $Q = Δ m \times 931.5 MeV/u = 0.000168 \times 931.5 = 0.1565 MeV$ Yeh step kyun? $E = m c^{2}$ ( $c$ yahan mass-energy exchange rate hai): jo mass gayi woh released energy $Q$ ke roop mein wapas aati hai. Factor $931.5 MeV/u$ exactly $c^{2}$ hai in units mein likha hua.

Step 3 — Spectrum interpret karo. Yeh ek three-body decay hai ( $N + e^{-} + \overset{ν}{ˉ}_{e}$ ). Electron aur antineutrino $Q$ continuously share karte hain. Toh electron ki energy $0$ se ek maximum tak jaati hai (the endpoint), jab antineutrino ~zero energy leta hai: $K E_{e, m a x} \approx Q = 0.156 MeV$ Yeh step kyun? Forecast ka jawaab: two-body split (ek fixed energy) ke unlike, teen bodies $Q$ ko anginat tareekon se divide kar sakti hain, ek smooth band deti hain — neutrino ka historical fingerprint.

Verify: parent note ke stated $Q \approx 0.156$ MeV se match karta hai. ✓ Continuous spread (Forecast) wahi reason tha kyun neutrino ko exist karna pada — dekho Neutrino and lepton number conservation.

Ex 4 · Cell B2 — Beta-plus vs electron capture: ek threshold aur ek degenerate case

Step 1 — Atomic masses ke saath $β^{+}$ Q-value likho. $Q_{β^{+}} = (m_{parent}^{atom} - m_{daughter}^{atom} - 2 m_{e}) c^{2}$ Yeh step kyun? Atomic masses orbital electrons count karti hain, aur $β^{+}$ ek orbital electron bhi khota hai (daughter mein ek kam proton hai) aur ek positron create karta hai; $Q$ is tarah likhne se woh double cost clearly zahir ho jaata hai. Daughter mein ek kam proton hota hai, toh ek neutral daughter atom mein ek kam orbital electron hota hai; upar se humne ek positron create kiya. Do extra electron-masses ki energy pay karni padti hai — ek real threshold, bookkeeping fluff nahi.

Step 2 — Threshold set karo. $Q_{β^{+}} > 0$ ke liye zaroori hai $(m_{parent}^{atom} - m_{daughter}^{atom}) c^{2} > 2 m_{e} c^{2} = 1.022 MeV$ Yeh step kyun? "Decay allowed" matlab $Q > 0$ ; us inequality ko rearrange karne se exact mass gap isolate hota hai jo parent ko beat karna hai, jo woh number hai jो examiner chahta hai.

Step 3 — Electron capture Q-value. $Q_{E C} = (m_{parent}^{atom} - m_{daughter}^{atom}) c^{2} > 0$ Yeh step kyun? Dono channels side by side compare karne se poora difference dikhta hai — missing $2 m_{e}$ : capture ek existing electron absorb karta hai nayi banane ki bajaye, toh positron-creation penalty nahi pay karta.

Step 4 — Degenerate window. Agar $0 < (m_{parent} - m_{daughter}) c^{2} < 1.022 MeV$ , toh $Q_{β^{+}} < 0$ (forbidden) lekin $Q_{E C} > 0$ (allowed). Exactly $0.5$ MeV gap par sirf electron capture hota hai — yahi Forecast ka jawaab hai. Yeh step kyun? Yeh "zero/degenerate input" cell hai: yeh masses ka poora ek band dikhata hai jahan ek channel dead hai aur doosra alive, woh subtlety jo matrix ne promise ki thi.

Verify: threshold $2 m_{e} c^{2} = 2 \times 0.511 = 1.022 MeV$ . ✓ Aur $0.5 < 1.022$ , toh $Q_{β^{+}} = 0.5 - 1.022 = - 0.522 < 0$ (forbidden) jabki $Q_{E C} = 0.5 > 0$ (allowed). ✓

Ex 5 · Cell C1 — Gamma: recoil correction

Step 1 — Photons momentum carry karte hain. Energy $E_{γ}$ wala photon momentum $p_{γ} = E_{γ} / c$ carry karta hai (yaad karo $c$ = speed of light; massless particle ke liye yeh $p = m v$ ki jagah leti hai). Momentum conservation — nucleus shuru mein rest par tha — nucleus ko equal, opposite recoil momentum deta hai $p = E_{γ} / c$ . Yeh step kyun? Photon ki rest mass nahi hoti, toh $p = m v$ bekaar hai; $p = E / c$ sahi momentum tool hai, aur momentum conservation photon ko recoiling nucleus se couple karti hai.

Step 2 — Recoil kinetic energy. $K E_{recoil} = \frac{p ^{2}}{2 M} = \frac{E _{γ}^{2}}{2 M c ^{2}}$ Energy conservation: $E_{0} = E_{γ} + K E_{recoil}$ . Kyunki recoil tiny hai, $E_{γ} \approx E_{0}$ andar set karo: $\frac{K E _{recoil}}{E _{0}} \approx \frac{E _{0}}{2 M c ^{2}}$ Yeh step kyun? Hum $K E = p^{2} /2 M$ reuse karte hain (Ex 1 jaisa tool) kyunki hum recoil momentum jaante hain, recoil speed nahi; energies ka ratio banana result ko dimensionless banata hai aur crude $M \approx A u$ mass ko 3 figures tak cancel hone deta hai.

Step 3 — Numbers daalo. $M c^{2} = 57 \times 931.5 MeV = 53, 095 MeV = 5.31 \times 1 0^{10} eV$ , aur $E_{0} = 1.44 \times 1 0^{4} eV$ : $\frac{K E _{recoil}}{E _{0}} = \frac{1.44 \times 1 0 ^{4}}{2 \times 5.31 \times 1 0 ^{10}} \approx 1.36 \times 1 0^{- 7}$ Yeh step kyun? Real numbers daalne se symbolic ratio concrete "kitna tiny hai tiny" ban jaata hai jो Forecast ne poochha tha.

Verify: fraction $\sim 1 0^{- 7}$ hai — bilkul tiny (Forecast: tiny). ✓ Isliye gamma lines sharp dikhti hain; famous Mössbauer effect exactly is near-zero recoil ko exploit karta hai.

Ex 6 · Cell D1 — Decay chain: $Z$ aur $A$ ki bookkeeping

Step 1 — Har decay ke changes yaad karo. $α$ : $A - 4, Z - 2$ . $β^{-}$ : $A$ same, $Z + 1$ . (Parent ki comparison table se.) Yeh step kyun? Chain sirf do numbers par repeated arithmetic hai, toh single-step rule nail karna zaroori hai pehle iterate karne se.

Step 2 — $A$ track karo. Do alphas $4$ each remove karte hain; betas $A$ ko akela chhodte hain. $A = 238 - 4 - 4 = 230$ Yeh step kyun? $A$ aur $Z$ independently change hote hain, toh dono ledgers alag-alag chalana zyada clean hai taaki mix-up na ho.

Step 3 — $Z$ track karo. Do alphas $2$ each remove karte hain ( $- 4$ ); do betas $1$ each add karte hain ( $+ 2$ ). $Z = 92 - 2 + 1 + 1 - 2 = 90$ Yeh step kyun? Ab doosra ledger; signed contributions ko sum karne se final proton count milta hai.

Step 4 — Naam lo. $Z = 90$ matlab thorium hai: $_{90}^{230} Th$ . Yeh step kyun? Sirf $Z$ element fix karta hai, toh periodic table se naam padhna numbers ko ek real isotope mein translate karne ka final step hai.

Verify: total $Δ A = - 8$ (chaar missing chunks = do alphas ✓), total $Δ Z = - 2$ . Order totals ke liye matter nahi kiya — ek achha sanity check. ✓

Ex 7 · Cell E1 — Real-world: smoke-detector alpha

Step 1 — Decay balance karo. $α$ removes $A - 4$ , $Z - 2$ : $A = 241, Z = 95$ se $A = 237, Z = 93$ = neptunium milta hai. $_{95}^{241} Am \to_{93}^{237} Np + α$ Yeh step kyun? Energy split karne se pehle daughter ka mass number chahiye, aur $A, Z$ balance karne se daughter ka naam bhi milta hai.

Step 2 — Energy split karo. $K E_{α} = Q \cdot \frac{m _{N p}}{m _{α} + m _{N p}} \approx Q \cdot \frac{A - 4}{A} = 5.64 \times \frac{237}{241} = 5.546 MeV$ Yeh step kyun? Hum Ex 1 se derived share formula reuse karte hain (daughter ab neptunium hai ke saath); mass ratio $\approx (A - 4) / A$ 3 figures tak valid hai jaisa ratio box mein justify kiya gaya.

Step 3 — "Bahar safe, andar useful" ki physics. Alphas ki penetration low hoti hai (paper ya skin rok deti hai). Bahar: harmless — skin ki dead layer unhe block kar deti hai. Andar: alpha ek chhote chamber mein air ionise karta hai, ek tiny current flow karne deta hai; smoke particles is current ko disrupt karte hain, alarm trigger karta hai. Ionising power high hai, range chhoti hai — ek sealed sensor ke liye perfect. Yeh step kyun? Part (c) ek number nahi, physical consequence maangta hai; penetration ko real behaviour se link karna kinematics ko engineering answer mein badalta hai.

Verify: $K E_{α} = 5.546$ MeV, $Q$ ka $98.3%$ hai (Forecast: $Q$ ke kareeb). ✓ Recoil $N p$ leta hai $5.64 - 5.546 = 0.094$ MeV, aur $5.546 + 0.094 = 5.64 = Q$ . ✓

Ex 8 · Cell E2 — Exam twist: impossible decay pehchano

Step 1 — Spontaneity rule yaad karo. Ek decay spontaneously tabhi hota hai agar $Q > 0$ ho (products parent se halke, energy release hoti hai). Yeh master conservation rule ek yes/no test mein convert ho jaata hai. Yeh step kyun? Ek number ( $Q$ ) sab kuch decide karta hai, toh pehle uska sign check karna tumhe aagey ki kisi bhi calculation se bachata hai.

Step 2 — Sign padho. $Q_{α} = - 7.6 MeV < 0$ . Negative $Q$ matlab products parent se bhaare hain — tumhe ise force karne ke liye $7.6$ MeV add karna padega. Toh yeh apne aap ho nahi sakta. Yeh step kyun? Sign ko physically interpret karna (negative ⇒ energy supply karni padegi) ek bare number ko verdict mein convert karta hai.

Step 3 — Iron kyun especially. $^{56} Fe$ binding-energy-per-nucleon curve ke peak ke paas baitha hai (dekho Nuclear binding energy and mass defect) — yeh sabse zyada tightly bound nuclei mein se ek hai. Toote se kuch gain nahi; bahar ka har rasta energy costa hai. Yeh step kyun? Yeh explain karna kyun iron ke liye sign negative nikla (maximal binding) dikhata hai ki result koi fluke nahi balki mass curve ki ek deep feature hai.

Verify: $Q = - 7.6 < 0 \Rightarrow$ forbidden. Student galat hai. ✓ Sanity: iron ka maximally stable hona exactly wahi reason hai kyun stars iron se aage fuse karke energy nahi nikaal sakte — consistent physics.

Recall Self-test — answers chhupao

Alpha decay mein kaunsa particle zyada KE leta hai, aur kyun? ::: Alpha (halka); shared momentum $p$ ke saath, $K E = p^{2} /2 m$ chhote $m$ ke liye bada hota hai. $A \to \infty$ par alpha ka $Q$ mein share kya hota hai? ::: $1$ (100%) — bhaari daughter barely recoil karti hai. Beta electron ki energy continuous band kyun hoti hai? ::: Three-body decay; antineutrino electron ke saath $Q$ continuously share karta hai. $β^{+}$ ko jo extra energy pay karni padti hai woh electron capture nahi karta — woh kya hai? ::: $2 m_{e} c^{2} = 1.022$ MeV (ek positron create karna + ek orbital electron khona). Heavy nucleus ke liye gamma recoil energy fraction kitni badi hoti hai? ::: $1 0^{- 7}$ ka order — negligible; lines sharp rehti hain. Ek number decide karta hai ki koi bhi decay allowed hai ya nahi — woh kya hai? ::: Q-value; decay spontaneous tabhi hota hai jab $Q > 0$ .

Parent par wapas jao: Radioactive decay mechanisms.