Radioactive decay — alpha, beta, gamma — mechanisms: Exercises

2.3.21 · D4 · Physics › Modern Physics › Radioactive decay — alpha, beta, gamma — mechanisms

Yeh page ek self-testing ladder hai. Har problem parent mechanisms note pe build hoti hai. Har ek ko Solution callout collapsed rakhtey hue try karo, phir kholo. Rungs "sirf naam lo" (L1) se lekar "sab kuch combine karo" (L5) tak jaate hain.

Shuru karne se pehle, yeh ek picture hai jis par har problem tiki hai: decay ke liye "conserved" ka matlab kya hota hai.

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

Isse balance scale ki tarah padho. Woh do numbers jo hum dekhte hain:

$Z$ = the number of protons (yeh charge hai, aur element ka naam batata hai).
$A$ = the number of nucleons = protons + neutrons (mass number, top-left number).

Parent note ka rule: arrow ke dono taraf, $Z$ ke totals match karne chahiye, aur $A$ ke totals bhi match karne chahiye. "Conservation of charge and nucleon number" ka bas itna hi matlab hai. Poori page ke liye yeh scale apne dimag mein rakho.

Level 1 — Recognition

Problem 1.1 (L1)

Har change ke liye decay type ka naam batao:

(a) $A$ 4 se girta hai, $Z$ 2 se girta hai.
(b) $A$ same rehta hai, $Z$ 1 se badhta hai.
(c) $A$ aur $Z$ dono unchanged rehte hain, ek photon nikalta hai.

Recall Solution 1.1

Yeh mnemonic use karo: "Alpha 2-and-2, Beta plus-one to Z, Gamma changes nothing but the energy."

(a) $A - 4, Z - 2$ → ek $_{2}^{4} He$ chunk nikla → alpha decay.
(b) $A$ fixed, $Z + 1$ → ek neutron proton ban gaya → beta-minus ( $β^{-}$ ).
(c) Balance scale par kuch nahi hilta, sirf energy light ke roop mein nikali → gamma decay.

Problem 1.2 (L1)

$α$ , $β^{-}$ , $β^{+}$ , aur $γ$ ke liye emitted particle, aur uska charge batao.

Recall Solution 1.2

Decay	Emitted	Charge
$α$	$_{2}^{4} He$ nucleus	$+ 2 e$
$β^{-}$	electron $e^{-}$ + antineutrino $\overset{ν}{ˉ}_{e}$	$- e$ (neutrino: $0$ )
$β^{+}$	positron $e^{+}$ + neutrino $ν_{e}$	$+ e$ (neutrino: $0$ )
$γ$	photon	$0$

Level 2 — Application

Problem 2.1 (L2)

Alpha decay complete karo aur daughter ke name-numbers batao: $_{88}^{226} Ra \to_{?}^{?} Y + α$

Recall Solution 2.1

WHAT: upar di gayi figure mein scale use karke $A$ aur $Z$ balance karo. WHY: ek alpha $A = 4, Z = 2$ le jaata hai, toh daughter mein baaki sab hona chahiye. $A_{Y} = 226 - 4 = 222, Z_{Y} = 88 - 2 = 86.$ $Z = 86$ radon hai. Toh daughter $_{86}^{222} Rn$ hai. $_{88}^{226} Ra \to_{86}^{222} Rn +_{2}^{4} α .$

Problem 2.2 (L2)

$_{90}^{234} Th$ $β^{-}$ decay se guzarta hai. Neutrino samet poori equation likho, aur daughter ka $Z$ identify karo.

Recall Solution 2.2

WHAT: $β^{-}$ matlab nucleus ke andar $n \to p + e^{-} + \overset{ν}{ˉ}_{e}$ . WHY: ek neutron proton ban gaya, toh $Z$ 1 se badhta hai jabki $A$ unchanged rehta hai; antineutrino lepton number balance karta hai. $_{90}^{234} Th \to_{91}^{234} Pa + e^{-} + \overset{ν}{ˉ}_{e} .$ $Z : 90 \to 91$ (protactinium), $A$ $234$ rehta hai.

Problem 2.3 (L2)

$_{92}^{238} U \to_{90}^{234} Th + α$ ke liye $Q = 4.27 MeV$ ke saath, alpha ki kinetic energy nikalao.

Recall Solution 2.3

WHY this formula: parent rest par hai, toh momentum conservation se alpha aur daughter equal aur opposite momenta ke saath alag hote hain. Isliye halka alpha zyada bada share lета hai (neeche split figure dekho). Use karo: $K E_{α} = Q \cdot \frac{m _{Y}}{m _{Y} + m _{α}} \approx Q \cdot \frac{A - 4}{A} .$ $K E_{α} = 4.27 \times \frac{234}{238} = 4.20 MeV .$ Daughter chhota sa remainder rakhti hai, $Q - K E_{α} \approx 0.07 MeV$ .

Level 3 — Analysis

Problem 3.1 (L3)

$_{6}^{14} C$ , $Q \approx 0.156 MeV$ ke saath $β^{-}$ decay karta hai. Ek student $0.100 MeV$ kinetic energy wala emitted electron measure karta hai. (a) Kya yeh allowed hai? (b) Antineutrino ne kitni energy carry ki (nitrogen ke tiny recoil ko ignore karo)? (c) Do alag $^{14}$ C nuclei alag energies ke electrons kyun de sakte hain?

Recall Solution 3.1

(a) Endpoint (maximum electron KE) $\approx Q = 0.156 MeV$ . Kyunki $0.100 < 0.156$ , yeh allowed hai. (b) Energy teen products mein share hoti hai; negligible nitrogen recoil ignore karke, $E_{\overset{ν}{ˉ}} = Q - K E_{e} = 0.156 - 0.100 = 0.056 MeV .$ (c) Beta decay ek three-body split hai (daughter + $e^{-}$ + $\overset{ν}{ˉ}_{e}$ ). Ek fixed $Q$ share karne ke teen tarike hone se, electron $0$ se lekar $Q$ tak kuch bhi le sakta hai — ek continuous spectrum, single line nahi. Yahi parent note ka neutrino argument predict karta hai.

Problem 3.2 (L3)

Ek excited daughter $_{28}^{60} Ni^{*}$ ground se $E_{i} = 2.506 MeV$ upar hai; yeh $1.332 MeV$ par ek level se hote hue do steps mein de-excite hota hai. Do gamma photon energies nikalao. (Recoil ignore karo.)

Recall Solution 3.2

WHY: ek nuclear energy level gap bilkul $E_{γ} = E_{i} - E_{f}$ ka photon emit karta hai, nuclear analog ki tarah jaise atom light emit karta hai.

Step 1: $2.506 \to 1.332$ : $E_{γ 1} = 2.506 - 1.332 = 1.174 MeV$ .
Step 2: $1.332 \to 0$ : $E_{γ 2} = 1.332 - 0 = 1.332 MeV$ . Do discrete lines — isliye gamma spectra level diagram map karte hain.

Level 4 — Synthesis

Problem 4.1 (L4)

Atomic masses diye hain: $m (^{226} Ra) = 226.025410 u$ , $m (^{222} Rn) = 222.017578 u$ , $m (^{4} He) = 4.002603 u$ , aur $1 u = 931.494 MeV / c^{2}$ : alpha decay ke liye $Q$ nikalao, phir $K E_{α}$ .

Recall Solution 4.1

Step 1 — mass defect. Decay chalane ke liye parent products se heavier hona chahiye (mass defect energy ke roop mein release hoti hai): $Δ m = 226.025410 - (222.017578 + 4.002603) = 0.005229 u .$ Step 2 — energy mein convert karo ( $Q = Δ m c^{2}$ ): $Q = 0.005229 \times 931.494 = 4.871 MeV .$ Step 3 — alpha ka share (mass-ratio split): $K E_{α} = Q \cdot \frac{222}{226} = 4.871 \times 0.98230 = 4.785 MeV (\approx 4.79 MeV) .$ Positive $Q$ ⇒ decay spontaneous hai, aur alpha iska almost sab kuch carry karta hai.

Problem 4.2 (L4)

Parent note kehta hai zyada energetic alphas ki half-lives bahut zyada chhoti hoti hain (Geiger–Nuttall). Explain karo, tunnelling through Coulomb barrier use karke, kyun thodi zyada $K E_{α}$ dramatically chhoti half-life produce karti hai. (Qualitative.)

Recall Solution 4.2

WHAT: alpha ek repulsive Coulomb wall ke peeche trapped hai; classically yeh escape nahi kar sakta, lekin quantum mechanics ise ek chhoti si tunnelling probability $P$ deta hai. WHY sensitivity: zyada energy wala alpha barrier par upar se shuru hota hai, toh cross karne wala barrier ka slice thinner aur lower hota hai. Tunnelling probability us barrier width/height par exponentially depend karti hai. Barrier thickness mein thodi si kami ⇒ $P$ mein bada jump ⇒ per second bahut zyada escapes. WHAT it looks like: neeche ki figure mein, alpha ki energy line uthane se shaded barrier chhoti ho jaati hai jise isse paarna hota hai. Kyunki $P$ ek exponent ke andar hai, half-life $\sim 1/ P$ moderate energy rise ke liye orders of magnitude gir sakti hai — exactly wahi steep Geiger–Nuttall trend.

Level 5 — Mastery

Problem 5.1 (L5)

$_{83}^{212} Bi$ do branches se decay kar sakta hai: (i) $α$ se $_{81}^{208} Tl$ tak, aur (ii) $β^{-}$ se $_{84}^{212} Po$ tak. Dono daughters phir stable $_{82}^{208} Pb$ tak continue karti hain. (a) Har branch ke liye, agla decay ( $^{208} Tl \to Pb$ aur $^{212} Po \to Pb$ ) $α$ ya $β^{-}$ ke roop mein likho, $A, Z$ check karte hue. (b) Confirm karo ki dono branches same final nucleus tak pahunchen.

Recall Solution 5.1

Branch (i): $_{83}^{212} Bi α_{81}^{208} Tl$ . Ab $_{81}^{208} Tl \to_{82}^{208} Pb$ : $A$ fixed, $Z : 81 \to 82$ ⇒ $β^{-}$ . ✓ Branch (ii): $_{83}^{212} Bi β^{-}_{84}^{212} Po$ . Ab $_{84}^{212} Po \to_{82}^{208} Pb$ : $A : 212 \to 208$ ( $- 4$ ), $Z : 84 \to 82$ ( $- 2$ ) ⇒ $α$ . ✓ (b) Dono routes $_{82}^{208} Pb$ (stable lead) par khatam hote hain. Bi se net change kisi bhi taraf: $Δ A = - 4, Δ Z = - 1$ (kisi bhi order mein ek $α$ + ek $β^{-}$ ). $α$ aur $β^{-}$ ka order alag hai, destination nahi — decay chain mein yeh ek classic branching hai.

Problem 5.2 (L5)

Ek $^{60}$ Co source $β^{-}$ phir do gammas (Problem 3.2 se) se guzarta hai. Decay law $N = N_{0} e^{- λ t}$ use karke, half-life $T_{1/2} = 5.27 yr$ ke saath, $10.54 yr$ baad source ka kitna fraction bachta hai? Kaun sa emission ( $β$ ya $γ$ s) yeh clock set karta hai?

Recall Solution 5.2

WHY the exponential: har nucleus ek fixed probability per second $λ$ se decay karta hai, age se independent, toh population $e^{- λ t}$ ki tarah girti hai. Step 1 — half-lives count karo: $t = 10.54 yr = 2 \times 5.27 yr$ , toh exactly 2 half-lives. Step 2 — fraction left: $\frac{N}{N _{0}} = (\frac{1}{2})^{2} = \frac{1}{4} = 0.25.$ $λ = ln 2/ T_{1/2}$ se check karo: $e^{- λ t} = e^{- (l n 2/5.27) (10.54)} = e^{- 2 l n 2} = 0.25$ . ✓ Kaun sa clock: slow, rate-limiting step $β^{-}$ decay hai (half-life years). Gammas essentially instantly follow karte hain (sub-nanosecond de-excitation) jaise hi beta ho jaata hai — woh timescale set nahi karte.

Recall One-line self-check (sabhi problems ke baad hi kholo)

Alpha split ::: $K E_{α} = Q m_{Y} / (m_{Y} + m_{α})$ , daughter recoils with the rest. Beta spectrum ::: continuous 0→ $Q$ because three bodies share the energy. Gamma energy ::: $E_{γ} = E_{i} - E_{f}$ , discrete lines. Q from masses ::: $Q = (m_{parent} - m_{products}) c^{2}$ , use atomic masses. Chain destination ::: fixed by total $Δ A, Δ Z$ , not by the order of decays.