Question

An isotope of Sodium, 24 Na, has a half-life of 15 hours. A sample of this isotope has mass 2 g.
(a) Find the amount remaining after 60 hours.
(b) Find the amount remaining after t hours.
(c) Estimate the amount remaining after 4 days (to the nearest thousandth).
(d) Estimate the time (to the nearest tenth) required for the mass to be reduced to .01 g by graphing the function.

Answer 1

we know that, at 0hours, t = 0, the sample is 2g
notice, we're doing b) first, since, without the equation, you can't get a)
b)


thus $\bf \qquad \textit{Amount for Exponential change}\\\\ A=P(1\pm r)^t\qquad \begin{cases} A=\textit{accumulated amount}\to &2\\ P=\textit{starting amount}\\ r=rate\\ t=\textit{elapsed period}\to &0\\ \end{cases} \\\\\\ 2=P(1-r)^0\implies 2=P(1)\implies 2=P\\\\ -------------------------------\\\\ A=2(1-r)^t$

notice, since the isotope is decaying, is a negative rate of change.

now.. hmm let's try to find "r"

well, we know it has a half-life of 15hours, so, 15hours later, the 2g will be come 1g, thus t = 15, A = 1

$\bf A=2(1-r)^t\implies 1=2(1-r)^{15}\implies \cfrac{1}{2}=(1-r)^{15} \\\\\\ \sqrt[15]{\cfrac{1}{2}}=1-r\implies r=1-\sqrt[15]{\cfrac{1}{2}}\implies r\approx 0.045 \\\\\\ A=2(1-0.045)^t\implies \boxed{A=2(0.955)^t}$

a)

well, simply set t = 60 to get A

c)

well, 4days is 96hours, so just set t = 96 to get A

d)

well, that simply requires you to graph it and check where A or A(t) becomes 0.01

you can always get it by using logarithms, but in this case, you're asked to get it from the graph.

Answer 2

Solution:

Half Life formula

$A_{t}=A_{0}\times [\frac{1}{2}]^{\frac{t}{t_{\frac{1}{2}}}$

$A_{t}$=Substance which has not decayed after time t

$A_{0}$=initial amount of Substance

$t_{\frac{1}{2}}$=Half life of Substance

(a)

$A_{60}=2\times[\frac{1}{2}]^{\frac{60}{15}}\\\\ A_{60}=2\times[\frac{1}{2}]^4\\\\ A_{60}=\frac{1}{8}\\\\ A_{60}=0.125$

Amount remaining after 60 hours= 0.125 gm

(b) Amount remaining after t hours.

$A_{t}=2\times[\frac{1}{2}]^{\frac{t}{15}}$

(c)Amount remaining after 4 days that is 96 hours

$A_{96}=2\times[\frac{1}{2}]^{\frac{96}{15}}\\\\ A_{96}=2\times[\frac{1}{2}]^{6.4}\\\\ A_{96}=(0.5)^{6.4}\\\\ A_{96}=0.01184$

Amount remaining after 4 days that is 96 hours=0.012 grams

(d)  Time required for the mass to be reduced to .01 g=

$A_{t}=0.01$ gm

$0.01=2 \times [\frac{1}{2}]^{\frac{t}{15}}\\\\ \frac{1}{200}= [\frac{1}{2}]^{\frac{t}{15}}$

Taking log on both sides

$\frac{t}{15}log(\frac{1}{2})=log(\frac{1}{200})\\\\ t=15 \times \frac{log(200)}{log2}$

Time required for the mass to be reduced to .01 g= 114.658 hours

=4 Days 18 hours 11 minutes