Question

An old bone contains 80% of its original carbon-14. Use the half-life model to find the age of the bone. Find an equation equivalent. How old is the bone

Answer 1

Answer:

Part 1 = C. (the LAST one)  /// Part 2 = 0.8 ////// Part 3 = B. (about 1,845 years)

Step-by-step explanation:

theyre all correct.

Answer 2

Answer:

An old bone contains 80% of its original carbon-14 in 1844.6479 years

Step-by-step explanation:

We know that

half life time of C-14 is 5730 years

so, h=5730

now, we can use formula

$P(t)=P_0(\frac{1}{2})^{\frac{t}{h} }$

we can plug back h

and we get

$P(t)=P_0(\frac{1}{2})^{\frac{t}{5730} }$

An old bone contains 80% of its original carbon-14

so,

P(t)=0.80Po

we can plug it and then we solve for t

$0.80P_0=P_0(\frac{1}{2})^{\frac{t}{5730} }$

$0.80=(\frac{1}{2})^{\frac{t}{5730} }$

$\ln \left(0.8\right)=\ln \left(\left(\frac{1}{2}\right)^{\frac{t}{5730}}\right)$

$t=-\frac{5730\ln \left(0.8\right)}{\ln \left(2\right)}$

$t=1844.6479$

So,

An old bone contains 80% of its original carbon-14 in 1844.6479 years