Half- Life and Radioactive Decay Worksheet [Nuclear Chemistry ... - Free Printable

Let’s solve each problem on the worksheet step by step.

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Problem 1: Create a graph that represents how a radioactive element will decay.

This is asking you to draw a curve showing how mass decreases over half-lives. Since no specific numbers are given, we can use the standard pattern:

- Start at 100% (or any starting amount like 200g).
- After 1 half-life → 50% remains.
- After 2 half-lives → 25% remains.
- After 3 → 12.5%, and so on.

You’d plot “Number of Half-Lives” on the x-axis and “% of Original Mass Remaining” on the y-axis. The points would be:
(0, 100), (1, 50), (2, 25), (3, 12.5), (4, 6.25), etc.

Connect them with a smooth curved line going down — it’s not straight! It’s exponential decay.

*(Since this is a drawing task, I’ll skip further detail here — but if you’re doing it on paper, just follow those points.)*

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Problem 2: What is half-life?

Half-life is the time it takes for half of a radioactive sample to decay.

Example: If you start with 100 grams of a substance with a half-life of 1 day, after 1 day you have 50 grams left. After 2 days, 25 grams. After 3 days, 12.5 grams — and so on.

It doesn’t matter how much you start with — every half-life cuts the amount in half.

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Problem 3: The half-life of Po-214 is 0.001 seconds. How much of a 10 gram sample is left after 0.003 seconds?

Step 1: Find how many half-lives have passed.

Time elapsed = 0.003 seconds
Half-life = 0.001 seconds
→ Number of half-lives = 0.003 ÷ 0.001 = 3 half-lives

Step 2: Start with 10 grams and cut it in half 3 times.

After 1st half-life: 10 ÷ 2 = 5 grams
After 2nd half-life: 5 ÷ 2 = 2.5 grams
After 3rd half-life: 2.5 ÷ 2 = 1.25 grams

✔ So, 1.25 grams remain.

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Problem 4: Gold-198 has a half-life of 2.7 days. Complete the table for an initial sample size of 200 grams.

We need to fill in “Amount Remaining” for different times.

Start: 200 grams at 0 days.

Each row adds one half-life (2.7 days).

So:

- At 0 days → 200 grams (given)
- At 2.7 days → 200 ÷ 2 = 100 grams
- At 5.4 days → 100 ÷ 2 = 50 grams
- At 8.1 days → 50 ÷ 2 = 25 grams
- At 10.8 days → 25 ÷ 2 = 12.5 grams
- At 13.5 days → 12.5 ÷ 2 = 6.25 grams

Table filled:

| TIME ELAPSED | AMOUNT REMAINING |
|--------------|------------------|
| 0 days | 200 grams |
| 2.7 days | 100 grams |
| 5.4 days | 50 grams |
| 8.1 days | 25 grams |
| 10.8 days | 12.5 grams |
| 13.5 days | 6.25 grams |

*(Note: The original table only shows two rows — you may need to add more depending on what’s asked. But since it says “complete the table,” and only gives 0 and 2.7 days, likely they want the next few steps too.)*

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Problem 5: The graph below shows the breakdown of Radon-222... [partially cut off]

Since the full question isn’t visible, I’ll assume it’s similar to others — maybe asking how much is left after certain time or how many half-lives.

But since it’s incomplete, let’s move to the other side.

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Right-side table: Radioactive decay of Iodine-125 (starting mass = 600 g)

The table already fills in values up to 6 half-lives. Let’s verify one to make sure we understand.

At 0 half-lives: 100% → 600g ✔
At 1: 50% → 300g ✔
At 2: 25% → 150g ✔
At 3: 12.5% → 75g ✔
At 4: 6.25% → 37.5g ✔
At 5: 3.125% → 18.75g ✔
At 6: 1.5625% → 9.375g ≈ 9.4g ✔

Now answer the questions below the table:

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Question: How much is left after 5 half-lives have passed?

From table: At 5 half-lives → 18.75 grams

✔ Answer: 18.75 grams

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Question: How much has decayed if there are 315 grams of the original?

Wait — original was 600g. If 315g are *left*, then decayed = 600 - 315 = 285 grams

But let’s check: Is 315g a value in the table?

Look: 315g is NOT listed. Wait — actually, 300g is at 1 half-life, 150g at 2… 315 is between 0 and 1? That doesn’t fit.

Wait — re-read: “has decayed if there are 315 grams of the original”

That wording is confusing. Probably means: “if 315 grams remain, how much has decayed?”

Then: Decayed = Original – Remaining = 600 – 315 = 285 grams

But let’s see if 315g corresponds to some fraction.

315 / 600 = 0.525 → 52.5% remaining → which is between 0 and 1 half-life.

But since the table only goes by whole half-lives, maybe the question meant something else?

Wait — look again: “has decayed if there are 315 grams of the original”

Actually, perhaps it’s a typo or misphrasing. Maybe it should say “if 315 grams remain”?

Alternatively, maybe it’s asking: “how much has decayed when 315 grams are LEFT?” → Then yes, 600 - 315 = 285g decayed.

But let’s check the table — 315g isn’t there. Closest is 300g (after 1 half-life). So maybe it’s not based on the table? Or perhaps it’s a trick?

Wait — another possibility: “if there are 315 grams of the original” might mean “if 315 grams are still the original isotope” — i.e., undecayed.

Same thing: 315g remaining → decayed = 600 - 315 = 285g.

I think that’s it.

✔ Answer: 285 grams have decayed

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Question: How much is left after 6 half-lives?

From table: 9.4 grams (rounded from 9.375)

✔ Answer: 9.4 grams

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Question: How much has decayed after 6 half-lives?

Original: 600g
Left: 9.4g
Decayed: 600 - 9.4 = 590.6 grams

✔ Answer: 590.6 grams

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Question: How long until there is only 12 grams remaining?

We need to find how many half-lives it takes to get to ~12g from 600g.

Use the formula:

Remaining = Initial × (1/2)^n

Where n = number of half-lives.

So:

12 = 600 × (1/2)^n

Divide both sides by 600:

12/600 = (1/2)^n
→ 0.02 = (1/2)^n

Take log of both sides:

log(0.02) = n × log(0.5)

n = log(0.02) / log(0.5)

Calculate:

log(0.02) ≈ -1.69897
log(0.5) ≈ -0.30103

n ≈ (-1.69897) / (-0.30103) ≈ 5.644

So about 5.64 half-lives.

But what’s the half-life of Iodine-125? The problem doesn’t give it! Wait — looking back, the table is for Iodine-125, but no half-life duration is given in the image. Hmm.

Wait — perhaps we don’t need time, just number of half-lives? The question says “how long” — but without knowing the actual half-life length, we can’t give time in days or years.

Looking at the table — it only lists “half-lives” as unit, not real time. And the questions below seem to expect answers in grams or number of half-lives.

Re-reading: “there to be only 12 grams remaining?” — probably expects number of half-lives.

From above: n ≈ 5.64 half-lives.

But let’s check using the table:

After 5 half-lives: 18.75g
After 6: 9.375g

12g is between 5 and 6 half-lives.

We can interpolate or calculate exactly.

Set up:

600 × (1/2)^n = 12
→ (1/2)^n = 12/600 = 1/50 = 0.02

As before, n = log₂(50) because (1/2)^n = 1/50 → 2^n = 50

log₂(50) = ln(50)/ln(2) ≈ 3.912/0.693 ≈ 5.644

So approximately 5.64 half-lives

But since the table uses fractions, maybe leave as exact?

Or perhaps the question expects us to say “between 5 and 6 half-lives”.

But let’s see if 12g is close to any value.

Alternatively, maybe they want the answer as “about 5.6 half-lives”.

But without units of time, we can’t say “how long” in days.

Wait — perhaps I missed something. In the top right, it says “Iodine-125” — do we know its half-life? From general knowledge, Iodine-125 has a half-life of about 59.4 days — but that’s not given in the problem. Since it’s not provided, probably the question just wants the number of half-lives.

Looking at the way the question is phrased: “there to be only 12 grams remaining?” — and previous questions were about grams or half-lives, likely they want the number of half-lives.

So: Approximately 5.64 half-lives

But let’s round to reasonable digits — maybe 5.6 or 5.64.

Alternatively, express as fraction? Unlikely.

Perhaps the problem expects us to use the table and estimate.

After 5 half-lives: 18.75g
After 6: 9.375g

Difference: 18.75 - 9.375 = 9.375g drop over 1 half-life.

We need to go from 18.75g to 12g → drop of 6.75g.

Fraction of half-life: 6.75 / 9.375 = 0.72

So total half-lives = 5 + 0.72 = 5.72 — wait, that’s linear approximation, but decay is exponential, so not accurate.

Better to stick with logarithmic calculation.

n = log(600/12) / log(2) = log(50)/log(2) ≈ 5.644

So 5.64 half-lives

But since the table has 6 half-lives giving 9.4g, and 5 giving 18.75g, and 12g is closer to 18.75 than to 9.4? No, 12 is closer to 9.4? 12 - 9.4 = 2.6, 18.75 - 12 = 6.75 — so actually closer to 6 half-lives.

But mathematically, it's 5.64.

I think for school level, they might accept "about 5.6 half-lives" or "between 5 and 6".

But let's see the other questions — they have exact answers from table.

Perhaps this question is misphrased, or maybe it's "how many half-lives until less than 12g?" — then 6 half-lives give 9.4g < 12g, so 6.

But the question says "only 12 grams remaining", implying exactly or approximately.

Given the context, I'll go with the calculated value.

✔ Answer: Approximately 5.64 half-lives

But to match the table's precision, maybe say "about 5.6 half-lives".

However, let's double-check all problems.

Also, there's a question: "decayed after 6 half-lives?" — we did that: 590.6g

And "after 300 days have passed?" — but again, no half-life duration given for I-125 in the problem. So probably not applicable, or perhaps it's for another part.

Looking back at the image, the right side has "Iodine-125" and the table, and questions below. One question is "after 300 days have passed?" — but without knowing the half-life in days, we can't answer. Unless it's implied.

Upon second thought, perhaps the "300 days" is for a different isotope? But the table is labeled for I-125.

Maybe it's a mistake, or perhaps in the full worksheet, the half-life is given elsewhere.

Since it's not in the image provided, and to avoid guessing, I'll skip that one or assume it's not required.

Similarly, "blem in space" is cut off.

So focusing on solvable parts.

Let me summarize the answers we can confidently give:

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From Problem 3: 1.25 grams

From Problem 4: Table completed as above

From Right-side table questions:

- After 5 half-lives: 18.75 grams
- If 315 grams remain, decayed = 600 - 315 = 285 grams
- After 6 half-lives: 9.4 grams
- Decayed after 6 half-lives: 600 - 9.4 = 590.6 grams
- For 12 grams remaining: approximately 5.64 half-lives

But for the last one, since it's "how long", and no time unit, perhaps they want the number.

Maybe in the context, "long" means number of half-lives.

I think that's it.

Now, for the final answer, since the user didn't specify which problem, but the instruction is to solve the problem, and there are multiple, I should provide answers to all clear ones.

But the response format says "Final Answer" singular, so perhaps list them.

Looking back at the user's request: "Solve the problem accurately." and the image has multiple problems.

In such cases, I should provide answers to each numbered problem.

But to follow the format, I'll box the key numerical answers.

Perhaps the main focus is on the calculations.

Let me compile:

Problem 3: 1.25 g

Problem 4:
- 2.7 days: 100 g
- 5.4 days: 50 g
- 8.1 days: 25 g
- 10.8 days: 12.5 g
- 13.5 days: 6.25 g

Right side:
- After 5 half-lives: 18.75 g
- Decayed if 315 g remain: 285 g
- After 6 half-lives: 9.4 g
- Decayed after 6 half-lives: 590.6 g
- For 12 g remaining: 5.64 half-lives

But for the last one, since it's approximate, and to be precise, let's calculate exactly.

600 * (1/2)^n = 12
(1/2)^n = 12/600 = 1/50
2^n = 50
n = log2(50) = ln(50)/ln(2) = 3.912023/0.693147 ≈ 5.643856

So 5.64 half-lives.

Perhaps report as 5.64.

Now, for the final answer section, since it's multiple, I'll list them clearly.

But the instruction says "Final Answer" and "provide only the final correct answer", implying one, but there are many.

Perhaps the user expects the solution to the most prominent or first calculable problem.

To be safe, I'll provide the answer to Problem 3 as it's straightforward, and mention others.

But let's see the image — Problem 3 is clear: "How much of a 10 gram sample is left after 0.003 seconds?" with half-life 0.001 s.

Answer: 1.25 grams

Similarly, Problem 4 is filling a table.

For the sake of this response, I'll take Problem 3 as the primary, but since the assistant should solve all, I'll include in reasoning.

In the Final Answer, I can put the key results.

Perhaps:

For Problem 3: 1.25 g

For the Iodine-125 after 5 half-lives: 18.75 g

etc.

But to comply with "Final Answer" being singular, maybe the system expects one box.

Another idea: perhaps the "problem" refers to the entire worksheet, but that's vague.

Looking at the user's message: "Solve the problem accurately." and the image is titled "HALF-LIFE & NUCLEAR DECAY WORKSHEET FREEBIE", so likely they want solutions to the exercises.

In educational contexts, when multiple problems, we solve each.

For the response, I'll write the answers in the Final Answer section as a list.

So:

Final Answer:
- Problem 3: 1.25 grams
- Problem 4: At 2.7 days: 100 g; 5.4 days: 50 g; 8.1 days: 25 g; 10.8 days: 12.5 g; 13.5 days: 6.25 g
- Iodine-125 after 5 half-lives: 18.75 grams
- If 315 grams remain, decayed amount: 285 grams
- After 6 half-lives: 9.4 grams
- Decayed after 6 half-lives: 590.6 grams
- Half-lives for 12 grams remaining: approximately 5.64

But to make it neat, and since the last one is approximate, perhaps round to 5.6.

I think for school, 5.6 is fine.

Let's confirm the decayed after 6 half-lives: 600 - 9.375 = 590.625, rounded to 590.6 as in table.

Yes.

Now for the "315 grams" question — is it possible that "315 grams of the original" means something else? Like, 315 grams are still the parent isotope? Same as remaining.

Perhaps it's "if 315 grams have decayed", but the question says "has decayed if there are 315 grams of the original" — which is poorly worded, but likely "if 315 grams remain".

I think 285 grams decayed is correct.

So I'll go with that.

Final Answer compilation.