Carbon-14 has a half-life of 5700 years. In a living organism, the ratio of radioactive carbon-14 to

ordinary carbon remains fairly constant during the lifetime of the organism. After the organism’s
death no new carbon is integrated into the organism’s remains, so the proportion of carbon -14 in the
remains decreases according to the law of exponential chance. The variable of interest here is the
percent the original amount of carbon-14 present in the organism’s remains.
a) Find the exact value of k for the decay equation for carbon-14.
b) If a painting attributed to Vermeer (1632-1675) was painted 5 years before his death,
approximately what percentage of carbon-14 should it contain? Round off your answer to the nearest tenth of a percent.
c) The painting was declared a forgery because it contained 99.5% of its original carbon-14. Based
on the law of exponential decay, find the true age of the painting. Round off your answer to the nearest year.

A(t) = Ao * 2^(-t/5700)

If you want that to use e instead of 2, then since 2 = e^ln2,

A(t) = Ao * e^(-ln2/5700 * t)
and k = ln2/5700.

Now just plug in your numbers to answer the other questions.

It cannot be determined, since carbon dating is very inaccurate

such

a) To find the exact value of k for the decay equation for carbon-14, we can use the formula for exponential decay:

A(t) = A₀ * e^(-kt)

where A(t) is the amount of carbon-14 at time t, A₀ is the initial amount of carbon-14, and k is the decay constant.

Since Carbon-14 has a half-life of 5700 years, we know that after 5700 years, the amount of carbon-14 will half. Therefore, we can use this information to find k.

Let's assume the initial amount of carbon-14 is A₀, and after 5700 years, it decreases to A₀/2.

A(5700) = A₀ * e^(-5700k) = A₀/2

Dividing both sides of the equation by A₀, we get:

e^(-5700k) = 1/2

Taking the natural logarithm of both sides, we have:

-5700k = ln(1/2)

Solving for k, we get:

k = ln(1/2) / -5700

Calculating this value, we find:

k ≈ 0.000121

Therefore, the decay equation for carbon-14 is:

A(t) = A₀ * e^(-0.000121t)

b) If the painting was painted 5 years before Vermeer's death, we can use the decay equation to calculate the percentage of carbon-14 it should contain.

Let's assume the original amount of carbon-14 in the painting is A₀. We want to find the percentage of A₀ that remains after 5 years, which is given by:

( A(5) / A₀ ) * 100

Substituting the decay equation for A(t), we have:

( A₀ * e^(-0.000121*5) / A₀ ) * 100

Simplifying, we get:

e^(-0.000605) * 100

Calculating this value, we find:

≈ 99.4%

Therefore, approximately 99.4% of carbon-14 should remain in the painting attributed to Vermeer.

c) If the painting was declared a forgery because it contained 99.5% of its original carbon-14, we can use the decay equation to find the true age of the painting.

Again, let's assume the original amount of carbon-14 in the painting is A₀. We want to find the time 't' when the percentage of A₀ remaining is 99.5%, which is given by:

t = -(1/k) * ln(0.995)

Substituting the value of k we found in part a, we have:

t ≈ -(1/0.000121) * ln(0.995)

Calculating this value, we find:

≈ 2862 years

Therefore, based on the law of exponential decay, the true age of the painting is approximately 2862 years.