Question

Initial amount of the radioactive element with half life $10$ days is $16$ $g$. What amount in $gm$ of this element will remain after $40$ days?

Answer 1

$t1/2=10$ days

$Ao=16$ g

$t=\frac{2.303\times 10}{0.693}\times log\left(\frac{16}{A}\right)$

$40=\frac{2.303\times 10}{0.693}\times log\left(\frac{16}{A}\right)$

$A=1$ g