Decay kinetics — first-order; half-life, mean life, activity

1. Why first-order? (Derivation from scratch)

WHAT is the starting physical statement? Each nucleus has the same probability $\lambda\,dt$ of decaying in a tiny time $dt$ . The probability does NOT depend on age (a 1-second-old nucleus and a 1000-year-old nucleus are identical). This "memoryless" property is the key.

HOW does this give the rate law? If there are $N$ nuclei, the expected number decaying in $dt$ is $N \times (\lambda\, dt)$ . So the decrease in $N$ :

$-dN = \lambda N\, dt \quad\Rightarrow\quad \frac{dN}{dt} = -\lambda N$

This is first-order because the rate $\propto N^1$ . Notice: the rate law has no concentration units, no collisions, no temperature dependence — it is a pure nuclear property, which is why $\lambda$ is a true constant.

2. Solving the equation → exponential law

WHY separate variables? Because $N$ and $t$ live on opposite sides cleanly.

$\frac{dN}{N} = -\lambda\, dt$

Integrate both sides from $t=0$ (where $N=N_0$ ) to $t$ :

$\int_{N_0}^{N}\frac{dN}{N} = -\lambda\int_0^t dt$ $\ln N - \ln N_0 = -\lambda t$ $\boxed{N = N_0\, e^{-\lambda t}}$

Figure — Decay kinetics — first-order; half-life, mean life, activity

3. Half-life $t_{1/2}$ — derived, not memorized

WHAT is it? The time for the sample to fall to half its current amount: $N = N_0/2$ .

HOW? Plug into the master law:

$\frac{N_0}{2} = N_0 e^{-\lambda t_{1/2}} \;\Rightarrow\; \tfrac12 = e^{-\lambda t_{1/2}} \;\Rightarrow\; \ln 2 = \lambda t_{1/2}$

$\boxed{t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}}$

4. Mean life (average lifetime) $\tau$ — derived

WHAT? The average time a nucleus survives before decaying.

HOW (derivation): The fraction decaying between $t$ and $t+dt$ is $\frac{-dN}{N_0} = \lambda e^{-\lambda t} dt$ . The mean life is the time-weighted average:

$\tau = \int_0^\infty t \,\lambda e^{-\lambda t}\, dt$

Use integration by parts (or recall $\int_0^\infty t e^{-at}dt = 1/a^2$ ):

$\tau = \lambda \cdot \frac{1}{\lambda^2} = \frac{1}{\lambda}$

$\boxed{\tau = \frac{1}{\lambda}, \qquad t_{1/2} = \tau \ln 2 = 0.693\,\tau}$

5. Activity $A$ — what detectors actually measure

Since $-dN/dt = \lambda N$ :

$\boxed{A = \lambda N = \lambda N_0 e^{-\lambda t} = A_0 e^{-\lambda t}}$

WHY does activity decay with the SAME exponential? Because $A = \lambda N$ is just $N$ scaled by a constant $\lambda$ — so it follows the identical curve and the identical half-life.

6. Worked examples

7. Common mistakes (Steel-man + fix)

8. Forecast-then-Verify

Recall Forecast before checking

Q: If a nuclide's activity drops to 6.25% of its initial value in 40 days, what is $t_{1/2}$ ? Forecast: $6.25\% = 1/16 = (1/2)^4$ , so 4 half-lives = 40 days → $t_{1/2} = 10$ days. Verify: $A/A_0 = e^{-\lambda t}$ , $\lambda = \ln(16)/40 = 2.773/40 = 0.0693\ \text{day}^{-1}$ , $t_{1/2}=0.693/0.0693=10$ days ✓

9. Flashcards

Why is radioactive decay first-order?

Each nucleus has a fixed decay probability per unit time, so rate

=\lambda N \propto N^1

Master decay equation?

N=N_0 e^{-\lambda t}

, derived by integrating

dN/dt=-\lambda N

Formula for half-life from

\lambda

t_{1/2}=\ln 2/\lambda = 0.693/\lambda

(set

N=N_0/2

Formula for mean life?

\tau = 1/\lambda

; also

\tau = t_{1/2}/0.693 = 1.44\,t_{1/2}

Why is mean life larger than half-life?

Mean is pulled up by long-lived survivors; half-life is the median,

\tau

is the mean.

Definition of activity?

A=-dN/dt=\lambda N

; it decays as

A=A_0e^{-\lambda t}

At what

N/N_0

is the mean life reached?

N/N_0 = 1/e \approx 0.37

Fraction left after

n

half-lives?

(1/2)^n

1 curie equals how many Bq?

3.7\times10^{10}

disintegrations per second.

Does

t_{1/2}

depend on starting amount or temperature?

No — it depends only on

\lambda

, a nuclear constant.

Recall Feynman: explain to a 12-year-old

Imagine a giant bag of popcorn kernels where each kernel has the same tiny chance of popping every second. You can't predict which kernel pops next, but with millions of kernels you can say "about this many pop each second." The more unpopped kernels left, the more pops you hear — so the popping slows down as the bag empties. The half-life is the time for half the kernels to pop. The next half takes the same time, then the next... so the bag never quite empties. The "popping sound rate" is the activity, and it fades exactly the same way.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, radioactive decay ka core idea bohot simple hai: har ek nucleus ka decay hone ka chance fixed hota hai per second — usko hum decay constant $\lambda$ bolte hain. Tum predict nahi kar sakte ki kaunsa nucleus kab phategaa, lekin agar crores nuclei hain to average rate exactly pata chal jaata hai. Aur rate hamesha bachhe hue nuclei ke proportional hota hai: jitne zyada bache, utni tezi se decay. Isi wajah se equation banti hai $dN/dt = -\lambda N$ , jo first-order hai, aur iska solution exponential nikalta hai: $N = N_0 e^{-\lambda t}$ .

Yahin se half-life aata hai. $N = N_0/2$ rakho to $t_{1/2} = 0.693/\lambda$ . Important baat — half-life kabhi bhi starting amount par depend nahi karta, na temperature par. Har half-life ke baad amount aadha ho jaata hai: 1 → ½ → ¼ → ⅛... isliye sample kabhi pura zero nahi hota, bas zero ke kareeb jaata hai. Mean life $\tau = 1/\lambda$ hota hai, aur yeh half-life se thoda bada hota hai ( $\tau = 1.44\,t_{1/2}$ ) kyunki kuch nuclei bahut lambi umar tak survive karte hain jo average ko upar kheech lete hain.

Activity wo hai jo Geiger counter actually measure karta hai — kitne disintegrations per second: $A = \lambda N$ . Yeh bhi same exponential curve follow karta hai, $A = A_0 e^{-\lambda t}$ , kyunki yeh sirf $N$ ka constant multiple hai. Unit becquerel (Bq) = 1 decay/sec, aur 1 curie = $3.7\times10^{10}$ Bq.

Exam tip / 80-20: bas teen rishte ratto mat — samjho. $A=\lambda N$ , $t_{1/2}=0.693/\lambda$ , $\tau=1/\lambda$ . Carbon dating jaise problems mein, activity ko fraction $(1/2)^n$ mein todo aur $n$ count karo — fatafat answer aa jaata hai bina log lagaye.