Question #0522c

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Answer 1 · 2017-04-23T14:36:16

$Q (60.50 yrs) = 3.50 \times 10^{- 4} g$ $Q(60.50" yrs") = 3.50xx10^-4" g"$

The general equation for exponential decay is:

$Q (t) = Q (0) e^{λ t}$ $Q(t) = Q(0)e^(lambdat)$

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

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

$\frac{1}{2} Q (0) = Q (0) e^{λ (5.27 yrs)}$ $1/2Q(0) = Q(0)e^(lambda(5.27" yrs"))$

Divide both sides by Q(0):

$\frac{1}{2} = e^{λ (5.27 yrs)}$ $1/2 = e^(lambda(5.27" yrs"))$

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

$ln (\frac{1}{2}) = λ (5.27 yrs)$ $ln(1/2) = lambda(5.27" yrs")$

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

$- ln (2) = λ (5.27 yrs)$ $-ln(2) = lambda(5.27" yrs")$

$λ = - \frac{ln (2)}{5.27 yrs}$ $lambda = -ln(2)/(5.27" yrs")$

Now we can evaluate at the equation at $t = 60.50 yrs$ $t= 60.50" yrs"$ and $Q (0) = 1 g$ $Q(0) = 1" g"$

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

$Q (60.50 yrs) = 3.50 \times 10^{- 4} g$ $Q(60.50" yrs") = 3.50xx10^-4" g"$