The half-life of cobalt-60 is 5.27 years. Approximately how much of a 199 g sample will remain after 20 years?

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Answer 1 · 2016-04-18T16:08:36

$14.3"g"$

The expression for radioactive decay is:

$N_t=N_0e^(-lambdat)$

$N_0$ is the initial number of undecayed atoms.

$N_t$ is the number of undecayed atoms remaining at time $t$

$lambda$ is the decay constant

The relationship between $lambda$ and the half - life $t_(1/2)$ is:

$lambda=0.693/t_(1/2)$

$:.lambda=0.693/5.27=0.1315"a"^(-1)$

Taking natural logs of both sides of the decay expression $rArr$

$lnN_t=lnN_0-lambdat$

$:.lnN_t=ln199-(0.1315xx20)$

$lnN_t=5.293-2.63=2.66$

From which:

$N_t=14.29"g"$