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

WHY does the decay law exist?

WHY proportional to $N$ ? If every nucleus independently has probability $\lambda\,dt$ of decaying in a tiny time $dt$ , then in a population of $N$ nuclei, the expected number that decay in $dt$ is just $N \times \lambda\, dt$ . More nuclei → more decays. That's it.

HOW to derive $N = N_0 e^{-\lambda t}$ (from scratch)

Step 1 — Write the rate of change. The number that decay in $dt$ is $\lambda N\, dt$ , and each decay removes a nucleus, so: $dN = -\lambda N\, dt$ Why the minus? Because $N$ decreases — $dN$ is negative.

Step 2 — Separate variables. $\frac{dN}{N} = -\lambda\, dt$ Why this step? We collect all $N$ on one side, all $t$ on the other, so each side can be integrated independently.

Step 3 — Integrate both sides from start ( $t=0$ , $N=N_0$ ) to time $t$ : $\int_{N_0}^{N}\frac{dN}{N} = -\lambda\int_0^t dt$ $\ln N - \ln N_0 = -\lambda t \;\Rightarrow\; \ln\frac{N}{N_0} = -\lambda t$

Step 4 — Exponentiate. $\boxed{N = N_0\, e^{-\lambda t}}$ Why exponential? Because the rate is proportional to the amount — the defining property of exponential decay. The fraction lost in equal time intervals is always the same.

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

Half-life $T_{1/2}$ — derive it

Set $N = N_0/2$ : $\frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}} \;\Rightarrow\; \tfrac{1}{2}=e^{-\lambda T_{1/2}}$ Take $\ln$ : $-\ln 2 = -\lambda T_{1/2}$ $\boxed{T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}}$

Mean life $\tau$

Activity $A$ — the thing you actually measure

From $dN/dt = -\lambda N$ : $A = \left|\frac{dN}{dt}\right| = \lambda N$ Substitute $N = N_0 e^{-\lambda t}$ : $\boxed{A = \lambda N_0 e^{-\lambda t} = A_0 e^{-\lambda t}}$ Activity decays with the SAME law and SAME half-life as $N$ . That's why a measurable quantity tells you about the invisible nuclei.

Worked Examples

Common Mistakes (Steel-manned)

Recall Feynman: explain to a 12-year-old

Imagine a huge bag of popcorn kernels in a hot pan. Each kernel pops at a random time — you can't say which pops next. But you can say: "about 1 in 10 of the un-popped ones pops each second." So at the start lots pop (loud!), and as fewer un-popped kernels remain, the popping slows down. Radioactive atoms are exactly like this. Half-life = the time for half the kernels to pop. The "popping noise per second" = activity — loud at first, quieter later, but it never goes totally silent for a long, long time.

Forecast-then-Verify

Connections

Radioactivity — alpha, beta, gamma decay
Binding energy and mass defect
Nuclear reactions and Q-value
Exponential functions and natural log
First-order chemical kinetics (identical math)
Carbon dating (direct application of the decay law)

What is the decay law equation?

N = N_0 e^{-\lambda t}

N_0

= initial number of nuclei,

\lambda

= decay constant.

Why is the decay rate proportional to N?

Each nucleus has a constant probability

\lambda\,dt

of decaying, so total expected decays

= N\lambda\,dt

Derive the differential equation of radioactive decay.

dN = -\lambda N\,dt

(each decay removes one nucleus; minus sign since N falls).

Formula for half-life in terms of λ?

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

How is half-life derived?

Set

N=N_0/2

N=N_0e^{-\lambda t}

; gives

1/2=e^{-\lambda T_{1/2}}

, so

T_{1/2}=\ln2/\lambda

Fraction remaining after n half-lives?

(1/2)^n

What is activity and its formula?

Decays per second;

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

Does activity have the same half-life as N?

Yes — both follow

e^{-\lambda t}

with the same

\lambda

What is mean life τ and its relation to half-life?

\tau = 1/\lambda

;

T_{1/2}=\tau\ln2\approx0.693\tau

, so

\tau>T_{1/2}

Fraction remaining after one mean life?

e^{-1}\approx 0.37

(37%).

Units of activity?

Becquerel (1 decay/s); 1 Ci

=3.7\times10^{10}

Bq.

Why does N never reach exactly zero?

Exponential decay is asymptotic; halving repeatedly always leaves a nonzero amount.

If λ doubles, what happens to half-life?

It halves (inversely proportional).

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, radioactive decay ka core idea bilkul simple hai: ek single nucleus kab decay karega ye purely random hai — predict nahi kar sakte. Lekin jab crores nuclei ho, tab ek pakka rule banta hai — jitne nuclei bache hain, utni hi zyada decay rate hogi. Isi se equation aati hai: $dN = -\lambda N\,dt$ . Yahan $\lambda$ (decay constant) matlab "per second kitni probability hai decay hone ki". Is equation ko integrate karo to milta hai $N = N_0 e^{-\lambda t}$ — yeh hi famous decay law hai.

Ab half-life $T_{1/2}$ matlab woh time jisme nuclei aadhe reh jaate hain. $N=N_0/2$ rakho to $T_{1/2}=0.693/\lambda$ nikalta hai. Important baat: half-lives add nahi hote, multiply hote hain. Do half-life ke baad $1/4$ bacha, na ki zero. $n$ half-life ke baad $(1/2)^n$ fraction bacha. Yaad rakho — decay kabhi exactly zero nahi hota, curve hamesha neeche jhukta rehta hai but touch nahi karta.

Activity woh cheez hai jo Geiger counter actually measure karta hai — per second kitne decay ho rahe hain, $A=\lambda N$ . Kyunki $N$ kam hota jaata hai, isliye $A$ bhi kam hota jaata hai (yaad rakho — $\lambda$ constant hai par activity constant NAHI). Aur sabse common galti: units! Agar $\lambda$ per-second me hai to $t$ bhi seconds me convert karo, warna $\lambda t$ ka dimension gadbad ho jaata hai. Bas itna pakka kar lo aur poora chapter aapka hai.

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

WHY does the decay law exist?

HOW to derive $N = N_0 e^{-\lambda t}$ (from scratch)

Half-life $T_{1/2}$ — derive it

Mean life $\tau$

Activity $A$ — the thing you actually measure

Worked Examples

Common Mistakes (Steel-manned)

Forecast-then-Verify

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Modern Physics

Connections