Question

Starting with a sample of pure ${}^{66}\mathrm{Cu}$, $\frac{7}{8}$ of its decays into $\mathrm{Zn}$ in $15\min$. The corresponding half-life is

Answer 1

Step 1: Analyse the formula for radioactive decay.

If $\mathrm{N}$ is the number of disintegrated nucleus in time $\mathrm{t}$. then

$\mathrm{N}={\mathrm{N}}_0\left({\mathrm{e}}^{-\lambda t}\right)$

Where, ${\mathrm{N}}_0=$ the initial quantity of the substance.

$\mathrm{N}$ is the quantity still remained and not yet decayed.

$\mathrm{t}$ is the half-life of the decaying quantity.

$\mathrm{e}$ is the Euler's number equal to $2.71828$.

Step 2: Analyze the decayed part.

According to the question, a sample of pure ${}^{66}\mathrm{Cu},\frac{7}{8}$of its decays into $\mathrm{Zn}$ in $15\min$

So, decayed part: $\frac{N_0-N}{N_0}=\frac{7}{8}$
$\Rightarrow 1-\frac{7}{8}=\frac{\mathrm{N}}{{\mathrm{N}}_0}$

Now, from the radioactive decay equation:
$$\begin{gathered} \Rightarrow \frac{1}{8}={\mathrm{e}}^{-\mathrm{\lambda t}} \\ \Rightarrow 8={\mathrm{e}}^{\mathrm{\lambda }} \end{gathered}$$

Taking ln both sides:
$$\begin{gathered} \ln 8=\mathrm{\lambda t} \\ \Rightarrow \ln 2^3=\mathrm{\lambda t} \\ \Rightarrow 3\ln 2=\mathrm{\lambda t} \end{gathered}$$

So, $\mathrm{\lambda }=\frac{3\times 0.693}{15}$
Step 3: Find the half-life period

Half-life period can be found out as:

${\mathrm{t}}_{\frac{1}{2}}=\frac{0.693}{\mathrm{\lambda }}$

Putting the values, we get:

${\mathrm{t}}_{\frac{1}{2}}=\frac{0.693\times 15}{3\times 0.693}$

${\mathrm{t}}_{\frac{1}{2}}=5\min$

So, the corresponding half-life is $5\min$.