Phosphorus-32 is radioactive and has a half life of 14.3 days. How long would it take a sample to decay from 2.20 mg to 600 $mu$ g?

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Answer 1 · 2016-11-01T20:05:27

26.8 days.

The equation for 1st order decay is:

$sf(N_t=N_(0)e^(-lambdat))$

We can get the value of the decay constant $sf(lambda)$ from the 1/2 life:

$sf(lambda=0.693/t_(1/2)=0.693/14.3=0.0484color(white)(x)d^(-1))$

Taking natural logs of both sides:

$sf(lnN_t=lnN_0-lambdat)$

Putting in the numbers and converting to milligrams:

$sf(ln0.6=ln2.2-0.0484t)$

$:.$ $sf(-0.5108=0.7884-0.0484t)$

$:.$ $sf(t=1.30/0.0484=26.8color(white)(x)"days")$

Phosphorus-32 is radioactive and has a half life of 14.3 days. How long would it take a sample to decay from 2.20 mg to 600 mumug?