# The half-life of arsenic-81 is 33 seconds. Suppose you have 100 g - how much will remain undecayed after 12.5 days?

Dec 18, 2014

The Rutherford-Soddy law of radioactive decay is :
N(t)=N_0 e^(-\lambdat); \qquad \lambda = \ln(2)/t_{1/2}.

Given : t_{1/2} = 33 sec; => \lambda = 2.1004\times10^{-2} sec^{-1}

t=12.5 days=1.08\times10^6 sec;
$\frac{N \left(t\right)}{N} _ o = {e}^{- \setminus \lambda t} = 0.3679 \setminus \times {10}^{- 5}$

${N}_{o} -$ Number of As-81 nuclei at the beginning ($t = 0$),
$N \left(t\right) -$ Number of As-81 nuclei after time $t$,
${M}_{o} -$ Mass of As-81 at the beginning ($t = 0$)
$M \left(t\right) -$ Mass of As-81 nuclei after time $t$,

(M(t))/M_o = (N(t))/N_o = 0.3679\times10^{-5};
$M \left(t\right) = 0.3679 \setminus \times {10}^{- 5} \setminus \times {M}_{o} = 0.3679$ milli-grams