Question

A certain radioactive element has half-life of $4$ days. The fraction of material that decays in $2$ days is

• A. $1/2$
• B. $1/\sqrt{2}$
• C. $\sqrt{2}$
• D. $(\sqrt{2}-1)/\sqrt{2}$

Answer 1

Answer: (D)

$(\sqrt{2}-1)/\sqrt{2}$

$N=N_o{e}^{-\lambda t}$
$t_{1/2}=\frac{ln2}{\lambda }$
If $t=2$, $\to N=N_o{e}^{\left(\frac{ln2}{4}\right)\times 2}$
$N=\frac{1}{\sqrt{2}}{N}_o$
Fraction decayed $=\frac{N-N_o}{N_o}$$=(\sqrt{2}-1)/\sqrt{2}$