Use the exponential growth model P(t) = P_0e^(kt)P(t)=P0ekt. The half-life of thorium-229 is 7,340 years. How long will it take for a sample of this substance to decay to 20% of its original amount?

1 Answer
Aug 2, 2016

t=7340(ln0.2/ln0.5)~=3,161t=7340(ln0.2ln0.5)3,161 yrs.

Explanation:

We first find the original amt. of thorium-229, i.e., its amt. P(0)P(0)at

time t=0t=0.

Taking t=0t=0 in, P(t)=P_0e^(kt).........(1), we get, P(0)=P_0.

:. The Original Amt. of thorium-229 is P_0.

We are given that, its, Half-life is 7,340 yrs, which means that, for

t=7340, P(t)=1/2P_0.

Using this data in (1), we get, 1/2P_0=P_0*e^(7340k)

rArr 0.5=e^(7340k) rArr ln0.5=7340k, or, k=ln0.5/7340....(2).

What we require, now, is time t=? for P(t)=20%of P_0=1/5P_0

By (1), then, 1/5P_0=P_0e^(kt) rArr 1/5=e^(kt), i.e., ln0.2=kt.

But, from (2), k=ln0.5/7340, so, ln0.2=(ln0.5/7340)t

Therefore, t=7340(ln0.2/ln0.5)

=7340{log_(10)0.2/log_10e*log_10e/log_(10)0.5}

=7340(0.3010/0.6990)~=3,161.

Thus, it will take (approx.) 3,161 yrs. for thorium-229 to decay to

20% of its original amt.