Radiocative carbon14 question (Chemistry)?

Question

Radioactive Carbon-14 has a half-life of 5730 years. Suppose a piece of wood has a decay rate of 15 disintegrations per minute. How many years would it take for the rate to decrease to 4 disintegrations per minute.

Answer 1

The mathematical derivation for the half-life is here: http://en.wikipedia.org/wiki/Half-life

The formula is as follows:
N(t) = N(0)*exp(-Lt) where
N(t) is the final amount (which is 4 in your example),
N(0) is the initial amount (which is 15 in your example),
L is the decay constant,
and t is the number of years you want to calculate.

First, you need to calculate the decay constant.

t(1/2) = half-life = 5730 = ln(2)/L. Solve for L.

L = ln(2)/5730. Plug this in the first equation. So you get ...

4 = 15*exp(-(ln(2)/5730)*t)

4/15 =exp(-(ln(2)/5730)*t)

Take the ln of both sides.

ln (4/15) = -(ln(2)/5730)*t

Solving for t...

t = 10926.48 years.

Answer 2

Let's call it 16 disintegrations per minute for the sake of the calculation. After 5730 years, the rate would be down to 8 per min. After 11,460 years, it would be down to 4 per min. That's what half-life means. The amount of time needed for half of the original to be left.

Answer 3

Reducing the disintegrations from 15 to 4 would take about 2 half-lives, or 11460 years.

Answer 4

Half-life means a radioactive element loses one-half of its radioactivity by the end of one cycle (half-life). Therefore, with a half-life of 5730 years, your example will lose one-half of its radiation level at the end of (one) half-life and the level would be 7.5 dpm. In another 5730 years the level would be 3.75 dpm. So, in approximately 11,030 years your wood (C-14) will have a radiation level of 4 dpm.