A metal has half life period of 10 days, $A$ sample originally has a mass of $1000 \mathrm{mg}$, then the…
A metal has half life period of 10 days, $A$ sample originally has a mass of $1000 \mathrm{mg}$,
then the mass remaining after 50 days is
$\frac{225}{8} m g$
$\frac{125}{8} \mathrm{mg}$
$\frac{125}{4} \mathrm{mg}$
$\frac{225}{4} m g$
Solution
A mass has half life period of 10 days. It means every ten days, mass remaining is half of the mass before 10 days.
Initial mass $=1000 \mathrm{mg}$.
$\therefore$ Mass after, 10 days $=\frac{1}{2} \times 1000=500 \mathrm{mg}$
Mass after 20 days $=\frac{1}{2} \times 500=250 \mathrm{mg}$
Mass after 30 days $=\frac{1}{2} \times 250=125 \mathrm{mg}$
Mass after 40 days $=\frac{1}{2} \times 125=\frac{125}{2} \mathrm{mg}$
Mass after 50 days $=\frac{1}{2} \times \frac{125}{2}=\frac{125}{4} \mathrm{mg}$