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

2.3.22 · D4 · Physics › Modern Physics › Decay law — N = N₀ e^(−λt), half-life, activity

Quick symbol reminder (sab parent se):

$N$ = abhi bache hue undecayed nuclei ki sankhya; $N_{0}$ = woh sankhya shuru mein ( $t = 0$ par).
$λ$ = decay constant, har ek nucleus ke decay hone ki constant probability per second. Units $s^{- 1}$ .
$T_{1/2} = \frac{ln 2}{λ}$ = half-life, $N$ ko aadha karne ka samay.
$τ = \frac{1}{λ}$ = mean life.
$A = λ N$ = activity, decays per second (Bq).

Level 1 — Recognition

L1.1 Ek radioactive source ka decay constant $λ = 2.0 \times 1 0^{- 3} s^{- 1}$ hai. Iska mean life $τ$ aur half-life $T_{1/2}$ likho.

Recall Solution L1.1

Hum kya karte hain: sahi one-liners choose karte hain. $τ = 1/ λ = 1/ (2.0 \times 1 0^{- 3}) = 500 s$ . $T_{1/2} = ln 2/ λ = 0.693/ (2.0 \times 1 0^{- 3}) = 346.6 s$ . Sanity check: $τ > T_{1/2}$ hamesha hota hai — long-tail survivors average ko upar kheenchte hain. ✓

L1.2 Exactly 4 half-lives ke baad, original nuclei ka kitna fraction bachta hai?

Recall Solution L1.2

Har half-life $\frac{1}{2}$ se multiply karti hai. $n = 4$ ke baad: fraction $= (\frac{1}{2})^{4} = \frac{1}{16} = 0.0625$ , yaani $6.25%$ . Multiply kyun, add kyun nahi? Baar baar aadha karna repeated multiplication hai — neeche L1 mistake dekho.

L1.3 Ek sample mein $8.0 \times 1 0^{9}$ decays per second hoti hain. Is activity ko becquerel aur curie mein express karo.

Recall Solution L1.3

$1 Bq = 1$ decay/s, isliye activity $= 8.0 \times 1 0^{9} Bq$ . $1 Ci = 3.7 \times 1 0^{10} Bq$ , isliye activity $= \frac{8.0 \times 1 0 ^{9}}{3.7 \times 1 0 ^{10}} = 0.216 Ci$ .

Level 2 — Application

L2.1 Ek sample $N_{0} = 6.0 \times 1 0^{18}$ nuclei se shuru hota hai aur uska $T_{1/2} = 8.0$ hours hai. 24 hours baad kitne nuclei bachenge?

Recall Solution L2.1

Step 1: $n = t / T_{1/2} = 24/8 = 3$ half-lives (pura number hai → halving form use karo). Step 2: $N = N_{0} (1/2)^{3} = 6.0 \times 1 0^{18} \times \frac{1}{8} = 7.5 \times 1 0^{17}$ nuclei.

L2.2 Ek source ke liye jiska $T_{1/2} = 10$ hours aur $N_{0} = 1.0 \times 1 0^{20}$ nuclei hai, initial activity $A_{0}$ Bq mein nikalo.

Recall Solution L2.2

Step 1 (seconds kyun): activity per second hoti hai, isliye $T_{1/2}$ convert karo: $10 \times 3600 = 36000 s$ . $λ = 0.693/36000 = 1.925 \times 1 0^{- 5} s^{- 1}$ . Step 2: $A_{0} = λ N_{0} = (1.925 \times 1 0^{- 5}) (1.0 \times 1 0^{20}) = 1.925 \times 1 0^{15} Bq \approx 1.9 \times 1 0^{15} Bq$ .

L2.3 Agar $λ = 0.05 day^{- 1}$ ho, toh activity ko apni initial value ka $30%$ hone mein kitna time lagega?

Recall Solution L2.3

Exponential kyun, halving kyun nahi: $30%$ koi accha power of $\frac{1}{2}$ nahi hai, isliye $A = A_{0} e^{- λ t}$ use karo. $0.30 = e^{- λ t} \Rightarrow ln 0.30 = - λ t \Rightarrow t = \frac{- ln 0.30}{0.05} = \frac{1.204}{0.05} = 24.1 days$ .

Level 3 — Analysis

L3.1 Ek source ki activity $8000 Bq$ se $1000 Bq$ tak $9.0$ minutes mein girती hai. Half-life nikalo.

Recall Solution L3.1

Step 1 (ratio pehchano): $A / A_{0} = 1000/8000 = 1/8 = (1/2)^{3}$ . Toh 3 half-lives guzri hain. Step 2: $3 T_{1/2} = 9.0 min \Rightarrow T_{1/2} = 3.0 min$ . Logs se cross-check: $λ = ln (8) /9 = 2.079/9 = 0.231 min^{- 1}$ , aur $T_{1/2} = 0.693/0.231 = 3.0 min$ . ✓

L3.2 Ek Geiger counter $t = 0$ par $A_{0} = 2400$ counts/s aur $t = 30$ minutes par $A = 300$ counts/s read karta hai. $λ$ ( $min^{- 1}$ mein) aur mean life $τ$ nikalo.

Recall Solution L3.2

Step 1: $A / A_{0} = 300/2400 = 1/8 = (1/2)^{3}$ → 30 min mein 3 half-lives → $T_{1/2} = 10$ min. Step 2: $λ = 0.693/10 = 0.0693 min^{- 1}$ . Step 3: $τ = 1/ λ = 1/0.0693 = 14.43 min$ (sach mein $τ > T_{1/2}$ ). ✓

L3.3 Neeche figure mein decay curve dekho. Graph se padhkar, ek poori half-life kis horizontal gap ke barabar hai, aur pehle half-life interval se doosre mein slope (activity) kaise badalta hai?

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

Recall Solution L3.3

Padhna: count equal horizontal steps par aadha hota jaata hai ( $100 \to 50 \to 25 \to 12.5$ ) — har step ek $T_{1/2}$ hai (yahan 1 unit). Slope: curve shuru mein sabse steep hai aur flat hoti jaati hai. Kyunki $A = λ N$ aur slope ki magnitude hi activity hai, aur $N$ har interval mein aadha hota hai, isliye average steepness bhi har interval mein aadha ho jaati hai. Curve kabhi zero nahi chhuti (asymptotic).

Level 4 — Synthesis

L4.1 Ek source mein do independent isotopes hain:

Isotope X: $N_{0 X} = 1.0 \times 1 0^{12}$ nuclei, $T_{1/2, X} = 2.0$ hours.
Isotope Y: $N_{0 Y} = 4.0 \times 1 0^{12}$ nuclei, $T_{1/2, Y} = 8.0$ hours.

$t = 8.0$ hours par total activity nikalo (decays/hour mein).

Recall Solution L4.1

Activities kyun add karte hain: decays independent events hain, isliye total $A = A_{X} + A_{Y}$ . Decay constants (per hour): $λ_{X} = 0.693/2 = 0.3466 h^{- 1}$ ; $λ_{Y} = 0.693/8 = 0.0866 h^{- 1}$ . $t = 8$ h par bache nuclei:

X: 4 half-lives → $N_{X} = 1.0 \times 1 0^{12} (1/2)^{4} = 6.25 \times 1 0^{10}$ .
Y: 1 half-life → $N_{Y} = 4.0 \times 1 0^{12} (1/2)^{1} = 2.0 \times 1 0^{12}$ . Activities: $A_{X} = λ_{X} N_{X} = 0.3466 \times 6.25 \times 1 0^{10} = 2.166 \times 1 0^{10}$ /h. $A_{Y} = λ_{Y} N_{Y} = 0.0866 \times 2.0 \times 1 0^{12} = 1.733 \times 1 0^{11}$ /h. Total: $A = 2.166 \times 1 0^{10} + 1.733 \times 1 0^{11} = 1.949 \times 1 0^{11}$ decays/hour.

L4.2 Ek parent nuclide $λ = 0.10 day^{- 1}$ se decay karta hai. Aaj uski activity $5.0 \times 1 0^{6} Bq$ hai. Exactly 5 days pehle uski activity kya thi?

Recall Solution L4.2

Key idea (clock ulta chalao): $A (t) = A_{0} e^{- λ t}$ . 5 days peeche jaane ka matlab hai pehle ki activity zyada thi, $e^{+ λ \cdot 5}$ se. $A_{past} = A_{now} e^{λ (5)} = 5.0 \times 1 0^{6} \times e^{0.10 \times 5} = 5.0 \times 1 0^{6} \times e^{0.5}$ . $e^{0.5} = 1.6487$ , isliye $A_{past} = 8.24 \times 1 0^{6} Bq$ .

Level 5 — Mastery

L5.1 Ek sample mein $t = 0$ par activity $A_{0}$ hai. Jo bhi decays kabhi bhi honge unka kitna fraction pehle mean life $τ$ mein hoga?

Recall Solution L5.1

Kabhi bhi hone wali total decays: $N_{0}$ nuclei mein se har ek eventually decay karta hai → total $= N_{0}$ . $[0, τ]$ mein decays: woh hai $N_{0} - N (τ)$ . Kyunki $N (τ) = N_{0} e^{- λ τ} = N_{0} e^{- 1}$ (kyunki $λ τ = 1$ ), ab tak ki decays $= N_{0} (1 - e^{- 1})$ . Fraction $= 1 - e^{- 1} = 1 - 0.3679 = 0.6321$ , yaani $\approx 63.2%$ . Matlab: lagbhag do-tihaai decays ek mean life se pehle ho jaati hain — process apni activity ko shuru mein front-load karti hai.

L5.2 Ek sample ki activity teen times measure ki jaati hai: $t = 0$ par $A_{0}$ , phir $t = T$ par $A_{0} /2$ hai. $t = 3 T$ par activity predict karo, aur dikhao ki general rule $A (k T) = A_{0} (1/2)^{k}$ purely constant-fraction property se aata hai, bina $λ$ compute kiye.

Recall Solution L5.2

Diya gaya: $T$ samay mein activity aadhi hoti hai — toh $T$ yahan half-life hi hai. Constant-fraction property: har equal interval mein same fraction bachti hai. Ek $T$ mein, factor $\frac{1}{2}$ . $k$ intervals of length $T$ mein: factor $(\frac{1}{2})^{k}$ . Isliye $A (3 T) = A_{0} (1/2)^{3} = A_{0} /8$ . $λ$ kyun nahi chahiye: exponential $A_{0} e^{- λ t}$ ko $A_{0} (1/2)^{t / T}$ likhaa ja sakta hai; $t = k T$ par exponent exactly $k$ hota hai. Physics ("equal times → equal fractions") seedha answer deta hai.

L5.3 Ek radioactive nuclide constant rate $R = 1.0 \times 1 0^{9}$ nuclei/s par produce ho raha hai jabki woh $λ = 1.0 \times 1 0^{- 3} s^{- 1}$ se decay bhi kar raha hai. Steady-state nuclei ki sankhya $N_{eq}$ nikalo (woh count jis par production aur decay balance ho jaati hain).

Recall Solution L5.3

"Steady state" ka matlab: sankhya change hona band ho jaati hai, isliye $d N / d t = 0$ . Balance equation: $\frac{d N}{d t} = R - λ N$ . Zero set karo: $R = λ N_{eq}$ . $N_{eq} = \frac{R}{λ} = \frac{1.0 \times 1 0 ^{9}}{1.0 \times 1 0 ^{- 3}} = 1.0 \times 1 0^{12}$ nuclei. Equilibrium par activity production rate ke barabar hoti hai: $A = λ N_{eq} = R = 1.0 \times 1 0^{9} Bq$ — har bana hua nucleus (on average) ek decay hone wale nucleus se match hota hai. Yahi "secular equilibrium" idea hai jo reactor aur medical-isotope production mein use hota hai.

Connections

Parent: Decay Law — $N = N_0 e^{-\lambda t}$, Half-life, Activity
Exponential functions and natural log — L2.3, L4.2 mein log step.
First-order chemical kinetics — identical math (rate ∝ amount).
Carbon dating — L2.3-style " $t$ ke liye solve karo" problems.
Radioactivity — alpha, beta, gamma decay

Recall One-line self-test

Do-isotope sample (L4.1) ki koi single half-life kyun nahi hoti? ::: Uska total count do exponentials ka sum hai jinka $λ$ alag-alag hai; sirf ek single exponential ki ek single half-life hoti hai.