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

Every symbol is built before use. Let me list the cast of characters we will meet, one at a time:

• $N$ — the number of nuclei that have not yet decayed, right now.
• $t$ — the time elapsed, measured from the moment we start watching.
• $\lambda$ (Greek letter "lambda") — the decay constant: the chance, per second, that any one nucleus decays.
• $dN$, $dt$ — a tiny change in $N$ and a tiny slice of time. The little "$d$" means "a sliver of."

Related maths lives in Exponential functions and natural log; the parent is the decay-law topic note.

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Step 1 — Picture the pile

WHAT. Start with a heap of undecayed nuclei. Call the count $N$. At the very beginning ($t=0$) we name this starting count $N_0$ — the little "0" just means "the value at time zero."

WHY. Before any equation, we must agree on what we are counting. Everything downstream is a statement about how this pile shrinks.

PICTURE. Look at the figure: each dot is one undecayed nucleus. There are many of them — that "many" is what makes the randomness smooth out.

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Step 2 — One nucleus, one coin-flip

WHAT. Zoom in on a single nucleus. In a tiny time slice $dt$, it either decays or it doesn't. The chance it decays is $\lambda dt$.

WHY THIS FORM. Why is the probability written as $\lambda \times dt$ and not just $\lambda$? Because a chance must be a pure number between 0 and 1, but $\lambda$ has units of "per second." Multiplying by the time slice $dt$ (in seconds) cancels the "per second" and leaves a bare probability. Longer slice → bigger chance to decay. That is the whole meaning of $\lambda$: the chance per second.

PICTURE. The figure shows one nucleus and a little clock ticking through $dt$. Two arrows leave it: a plum arrow "stays" (probability $1-\lambda dt$) and a burnt-orange arrow "decays" (probability $\lambda dt$).

$\underset{\text{a pure number}}{\underbrace{P_{\text{decay}}}}=\underset{\text{chance per second}}{\underbrace{\lambda }}\cdot \underset{\text{tiny seconds}}{\underbrace{dt}}$

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Step 3 — From one nucleus to the whole pile

WHAT. Now count decays across the whole pile in the slice $dt$. If each of the $N$ nuclei independently has chance $\lambda dt$ to go, the expected number that decay is $N\times \lambda dt$.

WHY. "Expected number" just means: chance-per-item times number-of-items. Ten dice, each with a $1/6$ chance of a six → expect $10\times 1/6$ sixes. Same logic, no new idea.

PICTURE. The figure highlights, in burnt orange, the handful of nuclei that popped in this one slice — their count is proportional to how full the pile currently is.

$\underset{\text{orange dots}}{\underbrace{(\text{decays in }dt)}}=\underset{\text{how many are left}}{\underbrace{N}}\cdot \underset{\text{each one’s chance}}{\underbrace{\lambda dt}}$

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Step 4 — Write the rate of change

WHAT. Each decay removes one nucleus from the pile, so the change in $N$ during $dt$ is the negative of the number that decayed: $dN=-\lambda Ndt.$

WHY THE MINUS SIGN. $dN$ is "the change in $N$." The pile only ever loses members, never gains, so $dN$ must come out negative. The minus sign is not decoration — it is the physics saying "shrinking."

WHY WE NEED CALCULUS HERE. We could try to track the pile in fixed jumps, but decay happens continuously — nuclei go at every instant, not on a metronome. The tools $dN$ and $dt$ ("slivers") let us talk about instantaneous rate. This is the moment calculus earns its place: it is the only language for "rate that changes as you go."

PICTURE. The figure plots $N$ dropping a little over one slice $dt$; the little step down is $dN$, and its steepness is set by how tall the pile currently is.

$\underset{\text{tiny change in }N}{\underbrace{dN}}=\underset{\text{it falls}}{\underbrace{-}}\underset{\text{current decay rate}}{\underbrace{\lambda N}}\underset{\text{tiny time}}{\underbrace{dt}}$

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Step 5 — Separate the two variables

WHAT. We rearrange so that everything about $N$ sits on the left and everything about $t$ sits on the right: $\frac{dN}{N}=-\lambda dt.$

WHY. We have two changing quantities, $N$ and $t$, tangled together. To add up their slivers (integrate) we must untangle them — one variable per side. This trick is called separation of variables, and it works precisely because our equation is a clean product $N\cdot dt$.

PICTURE. The figure is a simple balance: on the left pan, the fractional shrink $dN/N$; on the right pan, the fixed drip $-\lambda dt$. Equal pans mean: each equal fractional loss takes an equal chunk of time. That single sentence is the seed of the whole exponential.

$\underset{\text{fractional loss}}{\underbrace{\frac{dN}{N}}}=\underset{\text{a fixed drip}}{\underbrace{-\lambda dt}}$

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Step 6 — Add up all the slivers (integrate)

WHAT. We sum every tiny sliver from the start ($t=0$, $N=N_0$) to the present ($t$, $N$): ${\int }_{N_0}^N\frac{dN}{N}=-\lambda {\int }_0^{t}dt.$ The left sum is the natural logarithm; the right sum is just $\lambda t$: $\ln N-\ln N_0=-\lambda t⟹\ln \frac{N}{N_0}=-\lambda t.$

WHY THE LOGARITHM SHOWS UP. Adding up slivers of $\frac{1}{N}$ is a special sum whose answer is $\ln N$. That is the definition of the natural log — it is the running total of $1/N$. So the log is not pulled from a hat; it is exactly the quantity that accumulates a "fractional loss." See Exponential functions and natural log.

PICTURE. The figure shades the area under the $1/N$ curve between $N_0$ and $N$ — that shaded area is $\ln (N/N_0)$, and it equals the height $\lambda$ times width $t$ on the right.

$\underset{\text{accumulated fractional loss}}{\underbrace{\ln \frac{N}{N_0}}}=\underset{\text{steady drip}\times \text{time}}{\underbrace{-\lambda t}}$

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Step 7 — Undo the logarithm

WHAT. To free $N$ from inside the log, apply the log's opposite operation: the exponential $e^{(\cdot )}$. Since $e^{\ln x}=x$: $\frac{N}{N_0}=e^{-\lambda t}⟹\boxed{N=N_0{e}^{-\lambda t}}.$

WHY $e$ AND NOT SOME OTHER BASE. The log that appeared in Step 6 was the natural log (base $e$), because it is the one whose running total is exactly $1/N$. Its perfect inverse is $e^{(\cdot )}$. Using any other base would drag along an extra clutter factor. The natural base keeps the equation clean.

PICTURE. The figure shows the finished curve: starts at $N_0$, curves down, flattens toward — but never touches — zero.

$\underset{\text{left now}}{\underbrace{N}}=\underset{\text{started with}}{\underbrace{N_0}}\underset{\text{the shrinking factor}}{\underbrace{e^{-\lambda t}}}$

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Step 8 — Where the half-life lives on the curve

WHAT. Ask: how long until the pile is half its start? Set $N=N_0/2$: $\frac{1}{2}=e^{-\lambda T_{1/2}}\Rightarrow -\ln 2=-\lambda T_{1/2}\Rightarrow \boxed{T_{1/2}=\frac{\ln 2}{\lambda }=\frac{0.693}{\lambda }}.$

WHY $\ln 2$ APPEARS. "Halving" means the shrinking factor $e^{-\lambda t}$ hits $\frac{1}{2}$. Undoing the exponential with a log turns "one half" into $\ln 2\approx 0.693$. So $0.693$ is nothing mysterious — it is just $\ln 2$, the log of "one half."

PICTURE. The figure marks equal ledges: the curve drops from $N_0$ to $N_0/2$ over the same horizontal width $T_{1/2}$ that it later takes to fall from $N_0/2$ to $N_0/4$, and again to $N_0/8$. Equal widths, halving heights — the staircase of decay.

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Step 9 — The edge cases (never skip these)

WHAT & PICTURE. The figure overlays three regimes so you never meet an unshown scenario:

• $t=0$ (the start). $e^{-\lambda \cdot 0}=e^0=1$, so $N=N_0$. The curve begins exactly at the top. Sanity check: nothing has decayed yet.
• $t\to \infty$ (the far future). $e^{-\lambda t}$ shrinks toward 0 but is always positive, so $N$ approaches — yet never reaches — zero. The curve is asymptotic: mathematically the pile never fully empties. (In reality the last few nuclei do go, because $N$ is a whole number, but the smooth law says "vanishingly small.")
• $\lambda =0$ (a stable nucleus). Zero chance per second → $e^0=1$ for all $t$ → $N=N_0$ forever. A flat line: nothing ever decays. This is the degenerate case, and the formula handles it automatically.
• Large vs small $\lambda$. Bigger $\lambda$ = steeper drop = shorter half-life; smaller $\lambda$ = gentle slope = long half-life. Same shape, different stretch.

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The one-picture summary

This final figure compresses all nine steps into a single storyboard: the pile (Step 1) → one nucleus flipping its coin (Step 2) → the rate law $dN=-\lambda Ndt$ (Steps 3–4) → separate and integrate into $\ln$ (Steps 5–6) → exponentiate to $N=N_0{e}^{-\lambda t}$ (Step 7) → read the half-life ledges off the curve (Step 8), with the edge cases pinned in place (Step 9).

Recall Feynman: tell the whole walk in plain words

Picture a huge jar of jumping beans, and each bean has the same small chance of hopping out of the jar every second — that chance is $\lambda$. Because they all share the same chance, the number that hop out right now just depends on how many are still inside: a fuller jar loses beans faster. Write that as "the tiny change equals minus lambda times how many are left, times a tiny time." To find how the jar empties over a long stretch, we add up all the tiny changes — that adding-up (calculus) naturally produces the natural logarithm, and undoing the log gives $e^{-\lambda t}$. The jar's count starts full ($N_0$), curves down, and slides toward empty without ever quite touching bottom. The time to lose half the beans is always the same length, because losing a half is losing a half no matter where you start — and that repeated equal step is the half-life, $T_{1/2}=0.693/\lambda$, where $0.693$ is simply the log of one-half.

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Connections

• Parent topic
• Exponential functions and natural log (why $\ln$ and $e$ appear)
• Radioactivity — alpha, beta, gamma decay
• First-order chemical kinetics (identical mathematics)
• Carbon dating (this derivation, applied)