Question

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

• A. $10\text{min}$
• B. $15\text{min}$
• C. $5\text{min}$
• D. $7\frac{1}{2}\text{min}$

Answer 1

The correct option is C $5\text{min}$

Radioactivity decay law gives the number of undecayed nuclei out of initial $N_0$ nuclei at time t to be,
$N=N_0\left(e^{-\lambda t}\right)$
Now, decayed part
$\Rightarrow \frac{N_0-N}{N_0}=\frac{7}{8}$

$\Rightarrow 1-\frac{7}{8}=\frac{N}{N_0}$
$\therefore \frac{1}{8}=e^{-\lambda t}\Rightarrow 8=e^{\lambda t}$
taking logarithm on both sides
$\text{ln}8=\lambda t$
$3ln2=\lambda t\lambda =\frac{3\times 0.693}{15}$
Half - life period,
$t_{\frac{1}{2}}=\frac{ln2}{\lambda }=\frac{0.693}{3\times 0.693}\times 15=5\text{min}$