Question

Carbon–14 is a radioactive isotope that decays exponentially at a rate of 0.0124 percent a year. How many years will it take for carbon–14 to decay to 10 percent of its original amount? The equation for exponential decay is At = A0e–rt.

Answer 1

Answer:

18569.234 years

Step-by-step explanation:

Given : Carbon–14 is a radioactive isotope that decays exponentially at a rate of 0.0124 percent a year.

To Find: How many years will it take for carbon–14 to decay to 10 percent of its original amount?

Solution:

The equation for exponential decay is$A(t) = A0e^{-rt}$

$A_0$ = initial amount

A(t) = Amount after t time

Now we are supposed to find after how many years will it take for carbon–14 to decay to 10 percent of its original amount.

So,$A(t)=10\% A_0$

$A(t)=0.1 A_0$

r = 0.0124 % = 0.000124

Substitute the values in the equation:

$0.1 A_0 = A0e^{- 0.000124t}$

$- 0.000124t = ln [ \frac{0.1 A_0}{ A_0}]$

$t = ln [ \frac{0.1 A_0}{ A_0}] \times \frac{1}{- 0.000124}$

$t = ln [0.1] \times \frac{1}{-0.000124}$

$t = 18569.234$

Hence it will take 18569.234 years to decay to 10 percent of its original amount.

Answer 2

The decay rate should have units, it should be negative and it should be 100 times smaller than what you posted.
k = -.000124 / years

k = -.000124 / years
Half-Life = ln (.5) / k
Half-Life = -.693147 / -.000124
Half-Life = 5589.8951612903
Half-Life = 5,590 (rounded)

elapsed time = half-life * log(bgng amt / end amt) / log(2)
elapsed time = 5,590 * log(10) / 0.3010299957
elapsed time = 5,590 * 1 / 0.3010299957
elapsed time = 18,569 years

Source:
http://www.1728.org/halflife.htm