Carbon Dating Activity: Its a Puzzle!—ChemTopic™ Lab Activity ... - Free Printable

Let’s work through this step by step.

The problem is about carbon dating — figuring out how old something is by measuring how much carbon-14 is left in it. Carbon-14 decays over time, and we can use math to calculate the age based on how much is gone.

We’re told that after an organism dies, the carbon-14 starts to decay. The half-life of carbon-14 is 5,730 years — meaning every 5,730 years, half of the carbon-14 disappears.

Equation 6 is given as:
> t = -8270 ln(A/A₀) yrs

Where:
- t = age of the sample (in years)
- A = current amount of carbon-14 (or its activity) in the artifact
- A₀ = original amount of carbon-14 (same as in a living organism today)
- ln = natural logarithm (you’ll need a calculator for this)

But wait — let’s check if Equation 6 is correct based on earlier equations.

From Equation 3:
ln(¹⁴C / ¹⁴C₀) = -kt → rearranged to solve for t:
t = -ln(¹⁴C / ¹⁴C₀) / k

And from Equation 4:
k = 0.693 / t₁/₂ = 0.693 / 5730 ≈ 0.000121 per year

So plugging into Equation 5:
t = -ln(¹⁴C / ¹⁴C₀) / 0.000121

Now, 1 / 0.000121 ≈ 8264.46… which rounds to about 8270.

So yes — Equation 6 is just a simplified version using that number:
t = -8270 × ln(A/A₀)

This means: if you know what fraction of carbon-14 is left (A/A₀), you plug it into this formula to get the age.

Example: Suppose a sample has only 25% of its original carbon-14 left. That means A/A₀ = 0.25

Then:
t = -8270 × ln(0.25)
ln(0.25) = ln(1/4) = -ln(4) ≈ -1.3863
So t = -8270 × (-1.3863) ≈ 8270 × 1.3863 ≈ ?

Let me calculate that carefully:

First, 8000 × 1.3863 = 11,090.4
Then 270 × 1.3863 ≈ 270 × 1.386 = let’s do 200×1.386=277.2, 70×1.386≈97.02 → total ≈ 374.22
Add them: 11,090.4 + 374.22 = 11,464.62

So approximately 11,465 years.

That makes sense — because 25% left means two half-lives have passed (50% → 25%), so 2 × 5730 = 11,460 years. Our answer matches!

So the formula works.

Now — since no specific numbers were given in your question (like “a sample has X% carbon-14 left”), I think you might be asking for help understanding how to use the formula or verify it.

But looking back at the document — it doesn’t actually give a numerical problem to solve. It explains the theory and gives formulas.

Wait — maybe you want to know how to use Equation 6? Or perhaps there’s a hidden question?

Actually, re-reading your request: “Solve the problem accurately.” But the image is informational — it teaches carbon dating, shows equations, but doesn’t pose a specific calculation problem like “Calculate the age if 30% remains.”

Hmm.

Perhaps the “problem” is to understand and apply Equation 6 correctly — so let’s assume a typical textbook-style question that would go with this material.

Common question:
“If a wooden artifact has 60% of the carbon-14 found in living trees, how old is it?”

Let’s solve that as an example — since it’s not specified, but this is standard.

Given: A/A₀ = 0.60

Use Equation 6:
t = -8270 × ln(0.60)

Calculate ln(0.60):
Using calculator: ln(0.6) ≈ -0.51082562376

So:
t = -8270 × (-0.51082562376) = 8270 × 0.51082562376

Compute:

First, 8000 × 0.51082562376 = 4086.60499008
Then 270 × 0.51082562376 ≈ 270 × 0.5108 = let’s compute:

200 × 0.5108 = 102.16
70 × 0.5108 = 35.756
Total = 102.16 + 35.756 = 137.916

Add to previous: 4086.605 + 137.916 ≈ 4224.521

So t ≈ 4225 years

Check with half-life logic:
After one half-life (5730 yrs), 50% remains. We have 60%, which is more than 50%, so less than one half-life — 4225 is less than 5730 → makes sense.

Alternatively, using exact k value:

k = 0.693 / 5730 ≈ 0.000120942

t = -ln(0.6)/k = -(-0.5108256)/0.000120942 ≈ 0.5108256 / 0.000120942 ≈ ?

0.5108256 ÷ 0.000120942 ≈ 4223.7 → same as above.

So our answer is consistent.

Therefore, if the problem was to find the age when 60% carbon-14 remains, the answer is about 4225 years.

But since no specific values were given in the original prompt, perhaps the intent was to explain how to use the formula — but you asked to “solve the problem,” so I’ll assume a standard problem is implied.

Alternatively — maybe the “puzzle” refers to deriving or verifying Equation 6? Let’s see.

From Equation 3: ln(C/C₀) = -kt → t = -ln(C/C₀)/k

k = 0.693 / t₁/₂ = 0.693 / 5730

So t = -ln(C/C₀) / (0.693/5730) = -ln(C/C₀) * (5730 / 0.693)

Calculate 5730 / 0.693:

5730 ÷ 0.693

First, 0.693 × 8268 = ? Let’s divide directly.

5730 ÷ 0.693 = 5730000 ÷ 693 (multiply numerator and denominator by 1000)

Divide 5730000 ÷ 693:

693 × 8268 = let’s not — better to compute:

693 × 8000 = 5,544,000
Subtract from 5,730,000: 5,730,000 - 5,544,000 = 186,000

693 × 268 = 693×200=138,600; 693×68=693×70 - 693×2=48,510 - 1,386=47,124 → total 138,600+47,124=185,724

Close to 186,000 — difference 276

So 8000 + 268 = 8268, and remainder 276 → so ≈ 8268 + 276/693 ≈ 8268.4

But earlier we had 8270 — probably rounded.

Actually, 0.693 is approximate — actual ln(2) = 0.693147...

So k = ln(2)/t₁/₂ = 0.693147 / 5730 ≈ 0.000120968

Then 1/k = 5730 / 0.693147 ≈ 8267.0

But in the text they used 0.693 exactly and got 8270 — likely rounding up.

In any case, for educational purposes, we use the given Equation 6: t = -8270 ln(A/A₀)

So to summarize — without a specific input, I’ve demonstrated how to use the formula with an example.

Final Answer:
If a sample has 60% of its original carbon-14 remaining, its age is approximately 4225 years.