Q.A radioactive sample has decay constant $\lambda = 0.1\ \text{per day}$. Find the half-life of the sample and the time taken for its activity to fall to one-quarter of its initial value.

Substituting $t=T_{1/2}/2$ into the exponential decay law gives $N=N_0/\sqrt2$.

Working

The radioactive decay law is

$N = N_0 e^{-\lambda t} = N_0\left(\dfrac12\right)^{t/T_{1/2}}$

(using $\lambda=\ln2/T_{1/2}$).

At $t = \dfrac{T_{1/2}}{2}$:

$N = N_0\left(\dfrac12\right)^{1/2} = \dfrac{N_0}{\sqrt{2}}$

Half-life of a radioactive substance is T½ = 0.6931/λ → option (C).

Radioactive decay follows N = N₀e^(−λt). The half-life is the time for the number of nuclei to fall to half, so putting N = N₀/2:

$\frac{1}{2} = e^{-\lambda T_{1/2}} \;\Rightarrow\; T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.6931}{\lambda}$

Thus the half-life is inversely proportional to the decay constant λ.

20 days = 4 half-lives ⇒ fraction left = (½)⁴ = 1/16.

Number of half-lives elapsed:

$n = \frac{t}{T_{1/2}} = \frac{20}{5} = 4.$

Fraction of nuclei remaining undecayed:

$\frac{N}{N_0} = \left(\frac{1}{2}\right)^{n} = \left(\frac{1}{2}\right)^{4} = \frac{1}{16}.$

Mean life and half life of a radioactive sample are related by a factor of ln 2.

$T_{1/2}=\tau \ln 2 = 100\times 0.693 = 69.3\ \text{s}$

Half-life T_1/2 = 0.693/λ = ln 2 / λ.

Radioactive decay follows N = N₀e^(−λt). At the half-life, N = N₀/2:

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

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

Half-life is inversely proportional to the decay constant λ.

$\tau = T_{1/2}/0.693 = 1.44\,T_{1/2}$.

Definitions in terms of decay constant $\lambda$:

$\text{mean life }\tau=\frac{1}{\lambda},\qquad \text{half-life }T_{1/2}=\frac{\ln 2}{\lambda}=\frac{0.693}{\lambda}.$

Relation. Dividing,

$\frac{\tau}{T_{1/2}}=\frac{1/\lambda}{0.693/\lambda}=\frac{1}{0.693}=1.44.$

Converting each undecayed fraction into a number of half-lives elapsed in the same 8-hour interval gives half-life periods of 2 h for A and 1 h for B, a ratio of 2:1.

For a radioactive sample, the undecayed fraction remaining after $n$ half-lives is $\left(\dfrac{1}{2}\right)^n$.

For substance A, the undecayed fraction after 8 hours is $\dfrac{1}{16}$. Since $\dfrac{1}{16} = \left(\dfrac{1}{2}\right)^4$, exactly $n_A = 4$ half-lives have elapsed in 8 hours. So the half-life of A is $T_{1/2,A} = \dfrac{8\text{ h}}{4} = 2$ h.

For substance B, the undecayed fraction after the same 8 hours is $\dfrac{1}{256}$. Since $\dfrac{1}{256} = \left(\dfrac{1}{2}\right)^8$, exactly $n_B = 8$ half-lives have elapsed in 8 hours. So the half-life of B is $T_{1/2,B} = \dfrac{8\text{ h}}{8} = 1$ h.

The ratio of the half-life periods is $T_{1/2,A} : T_{1/2,B} = 2\text{ h} : 1\text{ h} = 2:1$.

This is physically sensible: B decays much faster (needing 8 half-lives to reach $1/256$ in the same real time that A needs only 4 half-lives to reach $1/16$), so B must have the shorter half-life.

Mean life $\tau$ and half-life $T_{1/2}$ are related by $T_{1/2} = 0.693\,\tau$, i.e. $\tau = 1.44\,T_{1/2}$.

Concept. Both quantities are tied to the decay constant $\lambda$: $T_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{\lambda}$ and $\tau = \dfrac{1}{\lambda}$.

Relation. Dividing,

$T_{1/2} = 0.693\,\tau \qquad\text{or}\qquad \tau = \frac{T_{1/2}}{0.693} = 1.44\,T_{1/2}.$

The time to fall to half the initial amount is the half-life.

By definition, the half-life T₁/₂ is the time in which the number of undecayed nuclei of a radioactive substance reduces to one-half of its initial value. It relates to the decay constant λ by

$T_{1/2} = \frac{0.693}{\lambda}.$

(Average/mean life τ = 1/λ is a different, longer time.)

The undecayed mass is $1/8$ of the original, corresponding to 3 half-lives, i.e. 9 minutes.

Step 1: Mass remaining undecayed $= 1\text{ g} - 0.875\text{ g} = 0.125\text{ g}$.

Step 2: Fraction remaining $= \dfrac{0.125}{1} = \dfrac{1}{8} = \left(\dfrac{1}{2}\right)^3$.

Step 3: Since the fraction remaining after $n$ half-lives is $(1/2)^n$, here $n=3$.

Step 4: Time elapsed $= n\times T_{1/2} = 3\times3\text{ min} = 9$ minutes.

After 3 half-lives ($15\,\text{yr}/5\,\text{yr}=3$), the substance left is $N_0(1/2)^3 = N_0/8$.

Radioactive decay law.

$N = N_0 \left(\dfrac12\right)^{t/T}$

where $T$ is the half-life and $t$ is elapsed time.

Step 1 -- number of half-lives.

$n = \dfrac{t}{T} = \dfrac{15\,\text{years}}{5\,\text{years}} = 3$

Step 2 -- apply the formula.

$N = N_0\left(\dfrac12\right)^{3} = \dfrac{N_0}{8}$

So after 5 years, $N_0/2$ remains; after 10 years, $N_0/4$; after 15 years, $N_0/8$.