general equation
amount = starting (1/2)^(t/k) , where t is the time and k is the half-life period
a)
.945 = 1(.5)^1/k
ln .945 = (1/k)ln.5
1/k = ln.945/ln.5
k = ln.5/ln.945 = 12.25 years
b)
.65 = (.5)^t/12.25
t/12.25 = ln.65/ln.5 = .621488..
t = 7.61 years
A sample of a radioactive substance decayed to 94.5% of its original amount after a year. (Round your answers to two decimal places.)
(a) What is the half-life of the substance?
(b) How long would it take the sample to decay to 65% of its original amount?
2 answers
A certain radioactive isotope decays at a rate of 2% per 100 years. If t represents time in years and y
represents the amount of the isotope left then the equation for the situation is y= y0e-0.0002t. In how many years will there be 93% of the isotope left?
represents the amount of the isotope left then the equation for the situation is y= y0e-0.0002t. In how many years will there be 93% of the isotope left?
2 answers