Question #370a1

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Answer 1 · 2016-12-08T18:20:11

About 18 hours.

The equation for 1st order decay is:

$sf(N_t=N_0e^(-lambdat))$

$sf(N_0)$ is the initial number of undecayed atoms.

$sf(N_t)$ is the number of undecayed atoms after time t.

$sf(lambda)$ is the decay constant.

We get this using:

$sf(lambda=0.693/t_(1/2)=0.693/6=0.1155color(white)(x)"hr"^(-1))$

Taking natural logs of both sides of the decay equation:

$sf(lnN_t=lnN_0-lambdat)$

$:.$ $sf(lambdat=lnN_0-lnN_t)$

$:.$ $sf(lambdat=ln(0.055)-ln(0.0066)=-2.9-(-5.02)=2.12)$

$:.$ $sf(t=2.12/0.01155=18.3color(white)(x)"hr")$