Question

A radio-isotope ${}_{15}^{32}P$ has half life of $15$ days. Calculate the time in which the radioactivity of $1mg$ quantity will fall to $10$% of the initial value.

Answer 1

$t1/2=15$ days

$t=?$

$A_0=1mg$

$A=\frac{10}{100}\times 1=0.1mg$

Radioactive reactions are first order

$t=\frac{2.303}{K}log\frac{1}{0.1}$

$t=\frac{2.303\times 15}{0.693}\times log10$