Decay law — N = N₀ e^(−λt), half-life, activity: Worked examples

2.3.22 · D3 · Physics › Modern Physics › Decay law — N = N₀ e^(−λt), half-life, activity

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Recall Teen moves aur kab use karein

Multiplicative-halving ::: $N = N_{0} (1/2)^{t / T_{1/2}}$ tab use karo jab $t$ half-life ka clean multiple ho — calculator ki zaroorat nahi. Exponential form ::: $N = N_{0} e^{- λ t}$ (ya $A = A_{0} e^{- λ t}$ ) tab use karo jab $t$ half-lives ka pura number nahi ho — smooth curve kisi bhi $t$ ke liye kaam karti hai. Logarithm ::: $ln$ tab use karo jab unknown exponent mein phansa ho ( $t$ ya $λ$ dhundhna ho). $ln$ ka matlab hai " $e$ kis power pe yeh value deta hai?"

Scenario matrix

Decay ke baare mein har exam question in cells mein se ek (ya mix) hota hai:

#	Case class	Diya → Dhundho	Sahi tool
A	Integer half-lives	$t = n T_{1/2}$ → fraction left	halving $(1/2)^{n}$
B	Non-integer half-lives	arbitrary $t$ → fraction left	exponential
C	Time ke liye backwards	fraction left → $t$	logarithm
D	$λ$ / $T_{1/2}$ ke liye backwards	do readings → $λ$	logarithm
E	Activity numbers	$N_{0}, T_{1/2}$ → $A_{0}$ in Bq	$A = λ N$ + unit care
F	Degenerate / limiting	$t = 0$ aur $t \to \infty$	curve ke ends dekho
G	Real-world word problem	carbon dating style	logarithm
H	Exam twist — do isotopes	mixed sample → baad mein ratio	two exponentials

Neeche ke aath worked examples ek-ek cell cover karte hain. Milake koi cell khaali nahi rehti.

Figure — Decay law — N = N₀ e^(−λt), half-life, activity

Curve dekho: yeh $N_{0}$ se shuru hoti hai (left edge, $t = 0$ ), axis ko kabhi touch nahi karti (yeh cell F hai), aur har case A–H bas "is curve ke kaun se point ke baare mein, ya kaun se slope ke baare mein pooch rahe hain?" hai.

Case A — clean integer half-lives

Case B — non-integer half-lives

Worked example Wahi source,

$T_{1/2} = 8$ days. 20 days baad kitna fraction bacha hai? Forecast: 20 days 2 aur 3 half-lives ke beech hai, toh fraction $1/4$ aur $1/8$ ke beech hoga. Ek number guess karo.

Step 1 — Poore halvings count nahi kar sakte. $20/8 = 2.5$ — integer nahi hai. Toh halving shortcut estimate shuru karta hai lekin exponential finish karta hai. Yeh step kyun? $(1/2)^{2.5}$ ke liye bhi calculator chahiye; hum smooth curve pe jaate hain.

Step 2 — Fractional exponent seedha use karo. $fraction = (\frac{1}{2})^{2.5} = 2^{- 2.5} .$ Yeh step kyun? $N = N_{0} (1/2)^{t / T_{1/2}}$ kisi bhi real $t / T_{1/2}$ ke liye valid hai, integer ho ya nahi.

Step 3 — Evaluate karo. $2^{- 2.5} = 2^{- 2} \cdot 2^{- 0.5} = \frac{1}{4} \cdot \frac{1}{2} \approx 0.25 \times 0.7071 = 0.1768$ .

Verify: Bounds check — $0.125 < 0.1768 < 0.25$ , bilkul 3rd aur 2nd half-life values ke beech, jaisa forecast tha. ✓ Answer: $\approx 0.177 = 17.7%$ .

Case C — time ke liye backwards solve karo

Worked example Ek sample ki activity apni starting value ki

30% reh gayi. Uska decay constant $λ = 0.05 day^{- 1}$ hai. Yeh kitne time mein hua? Forecast: Roughly 1.5 half-lives se 35% milta hai, toh guess: 1.5 half-lives se thoda zyada. $T_{1/2} = ln 2/0.05 \approx 13.9$ days, toh guess ~24 days.

Step 1 — Activity ratio likho. $\frac{A}{A _{0}} = 0.30 = e^{- λ t}$ . Yeh step kyun? Activity same law $A = A_{0} e^{- λ t}$ follow karti hai, isliye uska fractional drop nuclei ke fractional drop ke barabar hai.

Step 2 — Unknown $t$ exponent mein phansa hai — $ln$ lo. $ln (0.30) = - λ t .$ $ln$ kyun? Yeh $e^{(\cdot)}$ ka inverse hai; yeh exponent ko neeche kheench leta hai taaki hum $t$ isolate kar sakein.

Step 3 — Solve karo. $t = \frac{- ln ( 0.30 )}{λ} = \frac{1.2040}{0.05} = 24.1$ days. Minus sign kyun bacha rehta hai? $ln (0.30)$ negative hai (1 se chhoti number); do minuses milke $t$ positive banate hain, jaisa time hona chahiye.

Verify: Waapis substitute karo: $e^{- 0.05 \times 24.1} = e^{- 1.204} = 0.300$ . ✓ Units: $day / day^{- 1}$ ... actually $[ln (\cdot)] / [day^{- 1}] = day$ . ✓ Answer: $\approx 24.1 days$ .

Case D — $λ$ aur $T_{1/2}$ ke liye backwards solve karo

Worked example

$t = 0$ pe ek counter $A_{0} = 8000$ Bq read karta hai. Do ghante baad yeh $A = 2000$ Bq read karta hai. $λ$ aur $T_{1/2}$ dhundho. Forecast: Activity 2 h mein factor 4 se drop hui. Factor 4 = do halvings. Toh guess $T_{1/2}$ = 1 hour.

Step 1 — Do readings ka ratio lo. $\frac{A}{A _{0}} = \frac{2000}{8000} = \frac{1}{4} = e^{- λ t}$ jahan $t = 2$ h. Ratio kyun? Unknown $A_{0}$ cancel ho jaata hai — hume sirf reading kitni badi thi, absolute counts nahi chahiye.

Step 2 — $λ$ free karne ke liye $ln$ lo. $ln \frac{1}{4} = - λ (2) \Rightarrow λ = \frac{- ln ( 1/4 )}{2} = \frac{ln 4}{2} = \frac{1.3863}{2} = 0.6931 h^{- 1} .$ Yeh kyun kaam karta hai: $ln$ multiplicative drop ko ek additive wale mein convert karta hai jisko $t$ linearly scale karta hai.

Step 3 — $λ$ ko $T_{1/2}$ mein convert karo. $T_{1/2} = \frac{ln 2}{λ} = \frac{0.6931}{0.6931} = 1.00$ h.

Verify: Ek half-life = 1 h, aur 2 h = 2 half-lives → factor $(1/2)^{2} = 1/4$ , jo $2000/8000$ se match karta hai. Forecast confirm. ✓ Answer: $λ = 0.693 h^{- 1}, T_{1/2} = 1.0 h$ .

Case E — activity in becquerels (unit care)

Case F — degenerate aur limiting inputs

Worked example

$N = N_{0} e^{- λ t}$ ke liye: (i) $t = 0$ pe $N$ kya hai? (ii) $t \to \infty$ hone par $N$ kya approach karta hai? (iii) Kya $N$ kabhi exactly zero ho sakta hai? Forecast: Teeno padhne se pehle predict karo. Khaaskar (iii) — haan ya nahi?

Step 1 — $t = 0$ daalo. $N = N_{0} e^{0} = N_{0} \times 1 = N_{0}$ . Kyun? $e^{0} = 1$ ; start mein kuch bhi decay nahi hua. Yeh figure ka left edge hai.

Step 2 — $t \to \infty$ jaane do. Jaise $t$ badhta hai, $- λ t \to - \infty$ aur $e^{- λ t} \to 0$ , toh $N \to 0$ . Curve axis ke paas kyun rehti hai: exponential shrink hoti hai lekin hamesha positive rehti hai.

Step 3 — Poocho kya $N = 0$ kabhi reach hota hai. Har finite $t$ ke liye $e^{- λ t} > 0$ . Toh $N > 0$ hamesha; zero sirf limit hai, kabhi actual value nahi. Yeh kyun matter karta hai: yeh "2 half-lives ke baad sab khatam" wale myth ko khatam karta hai — curve asymptotic hai.

Verify: Figure se, red curve horizontal axis ko kabhi cross nahi karti; woh sirf uski taraf flatten hoti hai. Numerically, $t = 20 T_{1/2}$ pe ek fraction $2^{- 20} \approx 9.5 \times 1 0^{- 7}$ abhi bhi survive karta hai — chhota lekin nonzero. ✓ Answer: $N (0) = N_{0}, N (\infty) \to 0, N = 0 kabhi exactly reach nahi hota$ .

Case G — real-world word problem (dating)

Worked example Carbon-14 ka

$T_{1/2} = 5730$ years hai. Ek wooden artefact mein fresh wood ki C-14 activity ka sirf 25% dikhta hai. Yeh kitna purana hai? (Dekho Carbon dating.) Forecast: 25% exactly do halvings hai, toh guess $2 \times 5730 = 11460$ years — calculator ki zaroorat nahi.

Step 1 — Activity fraction likho. $\frac{A}{A _{0}} = 0.25 = (\frac{1}{2})^{2}$ . Activity kyun, nuclei nahi? Lab mein hum decays/second measure karte hain; dono same law share karte hain.

Step 2 — Clean power pehchano. $0.25 = (1/2)^{2}$ , toh $t / T_{1/2} = 2$ half-lives. Yahan shortcut kyun? 25% exactly $1/2$ ki power hai, toh Case A logic apply hota hai — $ln$ ki zaroorat nahi.

Step 3 — Multiply karo. $t = 2 \times 5730 = 11460$ years.

Verify (log route se, consistency prove karne ke liye): $t = \frac{- ln ( 0.25 )}{λ}$ , $λ = ln 2/5730$ . Phir $t = \frac{ln 4}{ln 2} \times 5730 = 2 \times 5730 = 11460$ . ✓ Answer: $11460 years old$ .

Case H — exam twist: do-isotope mixture

Worked example Ek sample

equal numbers $N_{0}$ do isotopes ke saath shuru hota hai: X jiska $T_{1/2} = 1$ h hai aur Y jiska $T_{1/2} = 2$ h hai. 2 hours baad (nuclei of X) : (nuclei of Y) ka ratio kya hai? Forecast: X zyada tezi se halve hota hai, toh 2 h baad X kam hona chahiye Y se — 1:2 jaisa ratio guess karo jo 1 se neeche ho.

Step 1 — Har isotope ko alag track karo. Yeh independently decay karte hain, har ek ka apna $λ$ hai. $N_{X} = N_{0} (\frac{1}{2})^{2/1} = N_{0} (\frac{1}{2})^{2} = \frac{N _{0}}{4} .$ $N_{Y} = N_{0} (\frac{1}{2})^{2/2} = N_{0} (\frac{1}{2})^{1} = \frac{N _{0}}{2} .$ Alag kyun? Ek nucleus ka decay doosre type ki presence se affect nahi hota — koi interaction nahi.

Step 2 — Ratio banao. $\frac{N _{X}}{N _{Y}} = \frac{N _{0} /4}{N _{0} /2} = \frac{1}{2}$ . $N_{0}$ cancel kyun hota hai? Dono equal shuru hue, toh common factor drop ho jaata hai.

Verify: X ne 2 half-lives run kiye → 1/4 bacha; Y ne 1 half-life run ki → 1/2 bacha. Ratio $(1/4) / (1/2) = 1/2$ , forecast se match karta hai ki X zyada kam hai. ✓ Answer: $N_{X} : N_{Y} = 1 : 2$ .

Ek picture jo saare cases unify karti hai

Mint curve (slow, $T_{1/2} = 2$ h) hamesha coral curve (fast, $T_{1/2} = 1$ h) ke upar rehti hai — yeh Case H ek nazar mein. Dashed markers dikhate hain jahan Cases A, C, aur G axis se height read karte hain.

Connections

Parent: Decay law
Exponential functions and natural log — Cases C, D, G mein $ln$ move
First-order chemical kinetics — identical maths
Carbon dating — Case G real world mein
Radioactivity — alpha, beta, gamma decay