Question 1162581
If using model,  {{{y=pe^(-kx)}}} for decay, x for time in years, p for amount at time 0,  then for half life  34.3 years,


{{{1/2=1*e^(-k*34.3)}}}

{{{ln(1/2)=-k*34.3}}}

{{{k=ln(2)/34.3}}}

{{{k=0.0202}}}



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{{{highlight_green(y=1*e^(-0.0202x))}}}


for x=20 years,  amount present then is  {{{e^(-0.0202*20)=highlight(0.68)}}}.