# Use the exponential growth model P(t) = P_0e^(kt). 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?

Aug 2, 2016

$t = 7340 \left(\ln \frac{0.2}{\ln} 0.5\right) \cong 3 , 161$ yrs.

#### Explanation:

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

time $t = 0$.

Taking $t = 0$ in, $P \left(t\right) = {P}_{0} {e}^{k t} \ldots \ldots \ldots \left(1\right)$, we get, $P \left(0\right) = {P}_{0}$.

$\therefore$ The Original Amt. of thorium-229 is ${P}_{0}$.

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

$t = 7340 , P \left(t\right) = \frac{1}{2} {P}_{0}$.

Using this data in $\left(1\right)$, we get, $\frac{1}{2} {P}_{0} = {P}_{0} \cdot {e}^{7340 k}$

$\Rightarrow 0.5 = {e}^{7340 k} \Rightarrow \ln 0.5 = 7340 k , \mathmr{and} , k = \ln \frac{0.5}{7340.} \ldots \left(2\right)$.

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

By $\left(1\right) , t h e n , \frac{1}{5} {P}_{0} = {P}_{0} {e}^{k t} \Rightarrow \frac{1}{5} = {e}^{k t} , i . e . , \ln 0.2 = k t$.

But, from $\left(2\right) , k = \ln \frac{0.5}{7340} , s o , \ln 0.2 = \left(\ln \frac{0.5}{7340}\right) t$

Therefore, $t = 7340 \left(\ln \frac{0.2}{\ln} 0.5\right)$

$= 7340 \left\{{\log}_{10} \frac{0.2}{\log} _ 10 e \cdot {\log}_{10} \frac{e}{\log} _ \left(10\right) 0.5\right\}$

$= 7340 \left(\frac{0.3010}{0.6990}\right) \cong 3 , 161$.

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

20% of its original amt.