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

Every claim below rests on four facts you already earned:

Decay constant $λ$ = a nucleus's constant probability of decaying per unit time (units $s^{- 1}$ ).
The law: $N = N_{0} e^{- λ t}$ , where $N$ is the number of undecayed nuclei left after time $t$ .
Activity $A = λ N$ = the number of decays counted per second (what a detector reads).
Mean life $τ = 1/ λ$ = the average lifetime of a nucleus (how long a nucleus lasts on average). This is not the half-life — we compare them below.

Figure 1 — the master decay curve (with half-life and mean life marked):

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

Recall Full step-by-step: how

$N = N_{0} e^{- λ t}$ comes out of $d N / d t = - λ N$ Some traps below say "the exponential falls out of the maths." Here is that maths, spelled out so nobody has to take it on faith. Step 1 — the rule of change. In a tiny time $d t$ , each of the $N$ nuclei has probability $λ d t$ of decaying, so $λ N d t$ vanish. Since $N$ drops, the change $d N$ is negative: $\frac{d N}{d t} = - λ N$ Step 2 — gather like with like. Divide by $N$ and multiply by $d t$ so every $N$ sits left, every $t$ right — now each side can be summed on its own: $\frac{d N}{N} = - λ d t$ Step 3 — add up all the tiny pieces (integrate) from the start ( $t = 0, N = N_{0}$ ) to now ( $t, N$ ): $\int_{N_{0}}^{N} \frac{d N}{N} = - λ \int_{0}^{t} d t \Rightarrow ln N - ln N_{0} = - λ t$ Why does $ln$ show up here? We are summing the quantity $1/ N$ as $N$ shrinks. Picture stacking thin rectangles under the curve $y = 1/ N$ (Figure 2): each rectangle has height $1/ N$ and width $d N$ , so its area is $d N / N$ . The total shaded area from $N_{0}$ down to $N$ is exactly what $ln$ measures — $ln$ is defined as the running area under $1/ N$ . That is the only reason a logarithm appears; it is the natural "area-adder" for a $1/ N$ shape. Step 4 — undo the log with $e$ . Rewrite as $ln (N / N_{0}) = - λ t$ , then apply $e$ to both sides ( $e$ cancels $ln$ ): $\frac{N}{N _{0}} = e^{- λ t} \Rightarrow N = N_{0} e^{- λ t}$ That is the whole derivation — see it plotted in Figure 1. For the deeper "why $ln$ and $e$ at all," see Exponential functions and natural log.

Figure 2 — why $ln$ appears: the area under $y = 1/ N$ is exactly $ln N$ :

True or false — justify

A nucleus is "safer" the longer it has already survived, since it's overdue.

False. Decay is memoryless — the probability

λ d t

of decaying in the next instant is the same whether the nucleus is one second or one million years old. Nuclei do not age or wear out.

Doubling the number of nuclei doubles the half-life.

False. Half-life

T_{1/2} = ln 2/ λ \approx 0.693/ λ

depends only on

λ

, a property of the isotope. Twice as many nuclei just means twice the activity, but each still halves in the same time.

If the activity is now

A

, then after one half-life the activity is

A /2

True. Since

A = λ N

and

λ

is constant, activity is directly proportional to

N

, so it follows the exact same

e^{- λ t}

curve and halves in one

T_{1/2}

After 10 half-lives the sample is completely decayed.

False. After 10 half-lives the fraction left is

(1/2)^{10} \approx 1/1024

— tiny, but nonzero. Exponential decay is asymptotic and never mathematically reaches zero (see Figure 3's flattening tail).

Two half-lives leave one quarter, so three leave zero.

False. Half-lives multiply:

(1/2)^{3} = 1/8

remains after three. Each half-life halves whatever survived the previous one; they never subtract to zero (Figure 3 shows the ladder

1 \to \frac{1}{2} \to \frac{1}{4} \to \frac{1}{8}

Mean life

τ

equals the half-life because both describe "how long nuclei live."

False.

τ = 1/ λ

(the mean lifetime) while

T_{1/2} = ln 2/ λ \approx 0.693/ λ

(the median lifetime), so

τ > T_{1/2}

. The rare long-lived survivors drag the average lifetime above the median — see the two marked times in Figure 1.

A sample with a larger

λ

is more radioactive per nucleus.

True. Larger

λ

means each nucleus decays with higher probability per second, so for the same

N

the activity

A = λ N

is higher — and the half-life is shorter.

Heating or compressing a radioactive sample speeds up its decay.

False.

λ

is set by nuclear forces inside the nucleus, essentially untouchable by ordinary temperature, pressure, or chemistry. Decay rate is fixed for a given isotope.

Spot the error

Each item below is also drawn as "wrong path vs. right path" in Figure 4 — match the letter to the panel.

"(A)

N = N_{0} /2

after time

t

, so

λ = 2/ t

Wrong. From

1/2 = e^{- λ t}

you must take a logarithm to pull

λ

out of the exponent:

λ = ln 2/ t = 0.693/ t

, not

2/ t

. You cannot cancel an exponent by ordinary algebra — only

ln

undoes

e

"(B)

λ t

is in seconds because

t

is in seconds."

Wrong.

λ t

must be dimensionless — you exponentiate it. If

λ

is in

s^{- 1}

then

t

must be in seconds so the units cancel; the product itself carries no unit.

"(C) Activity is constant because the decay constant

λ

is constant."

Wrong.

A = λ N

;

λ

is fixed but

N

keeps falling, so activity falls too. A constant probability applied to a shrinking population gives a shrinking count.

"(D) The fraction remaining after

n

half-lives is

1 - n /2

Wrong. It is

(1/2)^{n}

. The linear formula hits

0

n = 2

and goes negative beyond, which is impossible; Figure 4 overlays the straight wrong line on the true curve. Halving is multiplicative, not a fixed subtraction each time.

"(E)

T_{1/2} = λ / ln 2

Wrong — it's upside down.

T_{1/2} = ln 2/ λ \approx 0.693/ λ

. Check the physics: a large

λ

(fast decay) must give a small half-life, so

λ

belongs in the denominator.

"(F) After one mean life

τ

, half the nuclei are gone."

Wrong. After one mean life the fraction remaining is

e^{- 1} \approx 0.37

, so about 63% have decayed. Half is gone only after one half-life, which is shorter than

τ

"(G) Activity of 1 curie means 1 decay per second."

Wrong. 1 becquerel (Bq) is 1 decay/s; 1 curie

= 3.7 \times 1 0^{10}

Bq. The curie is a huge unit tied historically to 1 gram of radium.

Why questions

Why is the decay rate proportional to

N

and not to

N^{2}

or a constant?

Because each nucleus decays independently with probability

λ d t

; the expected number decaying is simply the count

N

times that per-nucleus probability, giving

λ N

. There is no interaction between nuclei to produce

N^{2}

Why does an exponential function fall out of the derivation, rather than a straight line?

Because the rate of loss is proportional to the amount present (

d N / d t = - λ N

). A function whose rate of change is proportional to itself is precisely the exponential — that's its defining property (see the boxed step-by-step above).

Why can we measure activity but almost never count

N

directly?

A detector registers the events (decays per second) that reach it, which is

A = λ N

; it cannot see the invisible undecayed nuclei sitting in the sample. But since

A = λ N

, measuring

A

tells us

N

Why does the same

λ

govern both

N

and

A

Because

A = λ N

is just

N

scaled by the constant

λ

; scaling a curve by a constant doesn't change its shape or its half-life, so both decay as

e^{- λ t}

Why is the mean life longer than the half-life?

The half-life is the median time (count halves), but a small population of nuclei survives for very long times. These long-lived stragglers pull the mean lifetime above the median time.

Why must we convert

t

and

T_{1/2}

to the same units before using

λ = ln 2/ T_{1/2}

Because

λ

inherits its units from

T_{1/2}

, and then

λ t

must be dimensionless. Mixing hours and seconds silently rescales the exponent and gives a wildly wrong answer.

Edge cases

What is

N

at exactly

t = 0

N = N_{0} e^{0} = N_{0}

— the full starting population, as it must be. The formula correctly anchors to the initial condition used in the integration.

What happens to

N

t \to \infty

e^{- λ t} \to 0

, so

N \to 0

but never actually equals zero. Decay approaches the axis asymptotically; there is always a mathematically tiny remnant (Figure 3's tail).

What if

λ = 0

(a perfectly stable, non-radioactive nuclide)?

Then

N = N_{0} e^{0} = N_{0}

for all time — nothing decays. The half-life

ln 2/ λ

becomes infinite, which correctly means "never halves."

What if the half-life is much shorter than your measuring time step?

Almost everything decays before you finish the first measurement, so you'd record near-zero activity and might wrongly think the sample was inactive. Very short-lived isotopes need very fast detectors.

Is it meaningful to ask when a single nucleus will decay?

No — for one nucleus you can only give a probability, never a definite time. The smooth exponential law is a statistical statement that only emerges for large numbers of nuclei.

Can activity ever exceed its initial value

A_{0}

during ordinary decay?

No.

A = A_{0} e^{- λ t}

only decreases with time for a single decaying isotope. (An apparent rise happens only in decay chains where a daughter is being fed — a different scenario.)

After the number of nuclei drops to a handful, does the smooth exponential curve still describe them exactly?

No — with only a few nuclei left, randomness dominates and the actual count jumps in integer steps around the curve. The exponential is an average that becomes exact only for enormous

N

Connections

Radioactivity — alpha, beta, gamma decay
Exponential functions and natural log
First-order chemical kinetics
Carbon dating

Recall One-line summary of every trap

Randomness is memoryless, half-lives multiply, decay is asymptotic (never zero), $λ$ is immutable by heat/pressure, $τ$ (mean) $> T_{1/2}$ (median), and activity falls even though $λ$ is constant because $N$ falls.