Question

The activity of a radioactive element reduces to $\left(1/16\right)th$ of its original value in $30$ years. Find its half life?

• A. $7.5$ yrs
• B. $75$ yrs
• C. $7.5$ days
• D. $75$ month

Answer 1

The correct option is A $7.5$ yrs

$N=N_0{e}^{-\lambda t}$

$\frac{N_0}{16}=N_0{e}^{-\lambda \ast 30}$

$\frac{1}{16}=e^{-30\lambda }$

$ln16=30\lambda$

$\lambda =\frac{ln16}{30}$

$\lambda =\frac{4ln2}{30}$

Half life $t_{0.5}=\frac{ln2}{\lambda }{t}_{0.5}=\frac{ln2}{\frac{4ln2}{30}}{t}_{0.5}=\frac{30}{4}$

$t_{0.5}=7.5$ yrs

Hence option 'A' is correct.