Half Life

[sweetliljenny]

4) The half-life of carbon-14 is eight days.
a) determine the continuous decay rate
b) If a lab acquires 2 g of carbon-14, how much of this sample will remain after 20 days?
c) How long will it be until only 0.01 g remains?

 

[wjm11]

4) The half-life of carbon-14 is eight days.
a) determine the continuous decay rate
b) If a lab acquires 2 g of carbon-14, how much of this sample will remain after 20 days?
c) How long will it be until only 0.01 g remains?

The half-life of C-14 is actually 5730 years. Regardless, what you want to model is accomplished with an exponential function. The can be of the form y = e^(-kt). However, for half-life problems, it is often handy to use A = I(1/2)^n, where I is the initial amount of material, A is the remaining amount, and n is the number of half-lives that have elapsed. “n” can be found by dividing the elapsed time by the half-life period.

n = (elapsed time)/(half-life period) = t/h

A = I(1/2)^n or,

A = I(1/2)^(t/h)

I'll lt you take it from here.

 

[sweetliljenny]

how much percentage it decays by day is how i did it ..

so for part a i got

Q = Q0 e ^ -6.25t


the 0 is a tiny 0

because the equation is Q = Q(tiny)0 e ^ -kt

 

[wjm11]

Q = Q0 e ^ -6.25t

Where did you get -6.25 from? Please show all your work.

 

[sweetliljenny]

from doing 50 (half-life) / 8 (the days of the half-life) = 6.25 = k
so -k = -6.25

 

[wjm11]

from doing 50 (half-life) / 8 (the days of the half-life) = 6.25 = k
so -k = -6.25

Where did the 50 come from? You need to set up your equation with a point that you know plugged, then solve for k.

.5Q0 = Q0(e^(-k(8)))
.5 = e^(-k(8))
ln(.5) = ln (e^(-k(8)))
ln(.5) = -8k

You take over.

 

[sweetliljenny]

gotcha ...diving -8 on both sides
so ln(.5) / -8 = k is approx. .0866

so my continuous decay rate : Q = Q0e ^ -kt = Q = Q0e ^ -.0866t

 

[galactus]

The constant of decay for carbon-14 can be obtained from

\(\displaystyle \L\\k=\frac{-1}{T}ln(2)\)

Where T=half-life. Some texts use 5750 years, but that's neither here nor there.

\(\displaystyle \L\\k=\frac{-1}{5730}ln(2)=\frac{-ln(2)}{5730}\approx{-0.000121}\)

To find the amount after 20 days. Remembering that this is in years

20/365=4/73

\(\displaystyle \L\\y(\frac{4}{73})=2e^{-0.000121}(\frac{4}{73})\)

There is not going to be much decay in carbon-14 in just 20 days.

As wjm said, that 8 day half life you were given is erroneous.

If you want to use 8 days as the half-life, adjust the figures accordingly.