Radioactive Dating Worksheet | Download Free PDF | Radioactive ... - Free Printable

Let’s solve each problem step by step.

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Problem 1:

We’re told that a fossilized bone has only 12.5% of its original Carbon-14 left.
The half-life of Carbon-14 is 5730 years.

Remember: Every half-life, the amount of radioactive material halves.

Start with 100%:

- After 1 half-life → 50% left
- After 2 half-lives → 25% left
- After 3 half-lives → 12.5% left ← This matches!

So, 3 half-lives have passed.

Now multiply:
3 × 5730 = 17,190 years

✔ The bone is 17,190 years old.

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Problem 2:

The igneous intrusion (V) has 25% of its original Uranium-235 left.
Half-life of Uranium-235 = 713 million years

Again, think in half-lives:

- Start: 100%
- After 1 half-life → 50%
- After 2 half-lives → 25% ← This matches!

So, 2 half-lives have passed.

Calculate age:
2 × 713 million = 1,426 million years (or 1.426 billion years)

Now, about relative ages:

In geology, when an igneous intrusion cuts through rock layers, it must be younger than the rocks it cuts through.

Looking at the diagram (even though we don’t describe it), layer V cuts through layers W, X, Y, Z — so those layers are older than V.

Also, fossils found in layers below V are older than V. Fossils above V may be younger or same age depending on context, but since V is an intrusion, anything cut by it is older.

So:

→ Intrusion V is 1,426 million years old.
→ Rock layers and fossils beneath V are older than 1,426 million years.
→ Rock layers and fossils above V could be younger — unless they were also affected, but typically, intrusions are younger than what they cut.

✔ Final answer for Problem 2:
Intrusion is 1,426 million years old.
Rock layers and fossils beneath it are older; those above may be younger.

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Problem 3:

A rock is 3.9 billion years old.
We need to find how much Potassium-40 is left.
Half-life of Potassium-40 = 1.3 billion years

First, figure out how many half-lives have passed:

3.9 ÷ 1.3 = 3 half-lives

Now calculate remaining percentage:

Start: 100%

After 1 half-life → 50%
After 2 → 25%
After 3 → 12.5%

✔ So, 12.5% of Potassium-40 would be left.

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Final Answer:
1.) 17,190 years
2.) The intrusion is 1,426 million years old. Rock layers and fossils beneath it are older; those above are likely younger.
3.) 12.5%