How long would it take for fifteen-sixteenths of a sample of radon-222 to decay?

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Answer 1 · 2016-07-03T03:53:17

$sf(15.3" ""days")$

The radioactive decay of an atom is a random event that obeys the laws of chance.

The more atoms you have, the greater chance there is of one decaying.

In mathematical terms we can say that the rate of decay is proportional to the number of undecayed atoms N.

$sf(-(dN)/dt=lambdaN)$

Where $sf(lambda)$ is the decay constant.

By integration we get the expression:

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

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

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

Suppose we start with 100 atoms of Rn - 222. So $sf(N_0=100)$ .

We are told that 15/16 of this decay.

So the number of atoms which have decayed = 100 x 15/16 = 93.75.

This means the number remaining must be 100 - 93.75 = 6.25

$:.$ $sf(N_t=6.25)$

Taking natural logs of our decay equation gives:

$sf(lnN_t=lnN_0-lambdat)$

$:.$ $sf(lambdat=ln[(N_0)/(N_t)]=ln[(100)/(6.25)]=ln(16)=2.772)$

We can get the value of $sf(lambda)$ from the half - life:

$sf(lambda=0.693/t_(1/2)=0.693/3.82=0.181" ""d"^(-1)$

$:.$ $sf(t=2.772/lambda=2.772/0.181=15.3" ""days")$