Question 0522c

Apr 23, 2017

Q(60.50" yrs") = 3.50xx10^-4" g"

Explanation:

The general equation for exponential decay is:

$Q \left(t\right) = Q \left(0\right) {e}^{\lambda t}$

Where Q(t) is the quantity at a given elapsed time, Q(0) is the initial quantity, and $\lambda$ is a decay coefficient

To find $\lambda$ given the half-life, you set $Q \left(t\right) = \frac{1}{2} Q \left(0\right)$, t = the given time and then solve for $\lambda$

$\frac{1}{2} Q \left(0\right) = Q \left(0\right) {e}^{\lambda \left(5.27 \text{ yrs}\right)}$

Divide both sides by Q(0):

$\frac{1}{2} = {e}^{\lambda \left(5.27 \text{ yrs}\right)}$

To make the exponential function disappear, we use the natural logarithm:

$\ln \left(\frac{1}{2}\right) = \lambda \left(5.27 \text{ yrs}\right)$

Replace $\ln \left(\frac{1}{2}\right)$ with -ln(2)

$- \ln \left(2\right) = \lambda \left(5.27 \text{ yrs}\right)$

$\lambda = - \ln \frac{2}{5.27 \text{ yrs}}$

Now we can evaluate at the equation at $t = 60.50 \text{ yrs}$ and $Q \left(0\right) = 1 \text{ g}$

$Q \left(60.50 \text{ yrs") = (1" g")e^(-ln(2)/(5.27" yrs")60.50" yrs}\right)$

$Q \left(60.50 \text{ yrs") = (1" g")e^(-ln(2)/(5.27" yrs")60.50" yrs}\right)$

Q(60.50" yrs") = 3.50xx10^-4" g"#