Question 688702
 Suppose that 5g of a radioactive substance decrease to 4g in 30 seconds .
find the half-life of the this substance
A = Ao*2^(-t/h), where
A = amt after t time
Ao = initial amt
t = time of decay
h = half-life of substance
:
5*2^(-30/h) = 4
2^(-30/h) = 4/5
2^(-30/h) = .8
{{{-30/h}}}ln(2) = ln(.8)
{{{-30/h}}} = {{{ln(.8)/ln(2)}}}
{{{-30/h}}} = -.322
h = {{{(-30)/(-.322))}}}
t ~ 93 seconds is the half life of the substance
:
 How long does it take for 3/5 of the substance to decay?
3/5 decay of 5g leaves 2g remaining
5*2^(-t/93) = 2
2^(-t/93) = 2/5
2^(-t/93) = .4
{{{-t/93}}} = {{{ln(.4)/ln(2)}}}
{{{-t/93}}} = -1.322
t = -93 * -1.322
t ~ 123 seconds

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