# Question #2eecb

May 11, 2016

1. $m \left(t\right) = 100 {e}^{- \ln \frac{2}{1590} t}$. Mass in grams and time in years.
2. $m \left(t = 1000\right) = 100 {e}^{- \ln \frac{2}{159} \times 100}$
3. ${t}_{30 g} = - \ln \frac{0.3}{\ln} 2 \times 1590 y e a r s$
4. $m ' \left(t\right) = - \left(\frac{10}{159} \ln 2\right) {e}^{- \ln \frac{2}{1590} t}$

#### Explanation:

Nuclear decay is a $1 s t$ order reaction .
$\frac{\mathrm{dm}}{\mathrm{dt}} = - \lambda m$.
$\lambda$ is the decay constant and $m$ is the amount of matter (mass or number of particles).

Solving the differential equation gives $m = {m}_{o} {e}^{- \lambda t}$.
${m}_{o}$ is the initial amount of reactant.

To find $\lambda$, the given half life-is $1590$ years.

$0.5 {m}_{o} = {m}_{o} {e}^{- \lambda \times 1590}$
Using $\ln$ on both sides of the equation and rearranging it will give $\lambda = \ln \frac{2}{1590}$

Answer $1$ therefore is $m = 100 {e}^{- \ln \frac{2}{1590} t}$.
Question 2; just plug in the value $t = 1000$
Question 3; $m = 30$

Question 4; I assume $m ' \left(t\right)$ is the derivative of $m \left(t\right)$ with respect to time;
$\frac{\mathrm{dm}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} 100 {e}^{- \ln \frac{2}{1590} t}$
$= - \left(\frac{10}{159} \ln 2\right) {e}^{- \ln \frac{2}{1590} t}$, $t = 1000$
$= - \left(\frac{10}{159} \ln 2\right) {e}^{- \ln \frac{2}{159} \times 100}$

This gives the rate of change of your radium in 1000 years time.