Answer key for half-life word problems involving carbon-14, iodine-131, titanium-51, fluorine-18, chromium-46, and radon-222.

It looks like you've uploaded an image of a worksheet titled "Half-Life Word Problems" with answers provided. The task appears to be solving half-life problems involving radioactive decay, and the answers are already filled in. I'll go through each problem step by step, explain the reasoning behind the solution, and verify the answers.

---

🔬 Understanding Half-Life

The half-life of a substance is the time it takes for half of the radioactive material to decay. This concept is modeled using exponential decay:

$$
N(t) = N_0 \times \left(\frac{1}{2}\right)^{t/T}
$$

Where:
- $ N(t) $ = amount remaining after time $ t $
- $ N_0 $ = initial amount
- $ T $ = half-life
- $ t $ = elapsed time

---

✔ Problem 1:

> The half-life of Carbon-11 is 20 minutes. A sample of Carbon-11 starts with 120 mg. How much is left after 1 hour?

Step-by-step:

- Half-life $ T = 20 $ minutes
- Time $ t = 1 $ hour = 60 minutes
- Number of half-lives: $ \frac{60}{20} = 3 $
- So, we halve the amount 3 times:
- After 20 min: $ 120 \div 2 = 60 $ mg
- After 40 min: $ 60 \div 2 = 30 $ mg
- After 60 min: $ 30 \div 2 = 15 $ mg

✔ Answer: 15 mg — ✔️ Correct!

---

✔ Problem 2:

> Iodine-131 has a half-life of 8 days. How much of a 545 g sample would remain at the end of 24 days?

- Half-life $ T = 8 $ days
- $ t = 24 $ days → $ \frac{24}{8} = 3 $ half-lives
- $ 545 \to 272.5 \to 136.25 \to 68.125 $

So, after 3 half-lives:
$$
545 \times \left(\frac{1}{2}\right)^3 = 545 \times \frac{1}{8} = 68.125 \text{ g}
$$

✔ Answer: 68.125 g — ✔️ Correct!

---

✔ Problem 3:

> The half-life of Titanium-51 is 6 minutes. If we start with a 21 mg sample, how much would be left after one hour?

- $ T = 6 $ min, $ t = 60 $ min → $ \frac{60}{6} = 10 $ half-lives
- $ 21 \times \left(\frac{1}{2}\right)^{10} = 21 \times \frac{1}{1024} \approx 0.0205 $ mg

But the answer given is 0.02 mg, which is rounded appropriately.

✔ Answer: 0.02 mg — ✔️ Correct (rounded)

---

✔ Problem 4:

> Carbon-14 has a radioactive half-life of 5,730 years. How much is left after one minute if we start with a 150 mg sample?

This is a trick question. One minute is negligible compared to 5,730 years.

- In such a short time, almost no decay occurs.
- So, the amount remains nearly unchanged.

✔ Answer: about 150 mg — ✔️ Correct

---

✔ Problem 5:

> Chromium-46 decays with a half-life of about 1 minute. After 6 half-lives, what percentage of a 30 g sample remains?

- After each half-life, amount halves.
- After 6 half-lives: $ \left(\frac{1}{2}\right)^6 = \frac{1}{64} $
- $ \frac{1}{64} \approx 0.015625 = 1.5625\% $

Rounded to 1.56% or 1.6% depending on precision.

✔ Answer: 1.56% — ✔️ Correct

---

✔ Problem 6:

> Carbon-14 has a half-life of 5,730 years. If a sample contains 19 mg originally, how much is left after 17,190 years?

- $ T = 5,730 $ years
- $ t = 17,190 $ years → $ \frac{17,190}{5,730} = 3 $ half-lives
- $ 19 \times \left(\frac{1}{2}\right)^3 = 19 \times \frac{1}{8} = 2.375 $ mg

But the answer says 11.25 mg, which is incorrect.

Wait — let’s double-check.

Wait! 17,190 ÷ 5,730 = 3 → exactly 3 half-lives.

$ 19 \div 2 = 9.5 $
$ 9.5 \div 2 = 4.75 $
$ 4.75 \div 2 = 2.375 $ mg

So correct answer should be 2.375 mg, not 11.25 mg.

✘ Given Answer: 11.25 mg — INCORRECT

Possibility: Maybe the problem meant after 1/3 of a half-life? But no — 17,190 is exactly 3 × 5,730.

Alternatively, maybe there's a typo in the problem or answer key.

Let me check:
Is 17,190 really 3 × 5,730?

$ 5,730 × 3 = 17,190 $ → Yes.

So:
$$
19 \times \left(\frac{1}{2}\right)^3 = 19 \times 0.125 = 2.375 \text{ mg}
$$

✔ Correct Answer: 2.375 mg — Given answer 11.25 mg is wrong

---

✔ Problem 7:

> If we start with a sample of 100 g of Radon-221, and its half-life is 3 days, how much is left after 21 days?

- Half-life = 3 days
- $ t = 21 $ days → $ \frac{21}{3} = 7 $ half-lives
- $ 100 \times \left(\frac{1}{2}\right)^7 = 100 \times \frac{1}{128} \approx 0.78125 $ g

But the answer says 3.75 g — that’s way off.

Wait — let’s see: $ 100 \div 2^7 = 100 / 128 ≈ 0.78 $ g

But 3.75 g is close to $ 100 / 26.66... $ — doesn't match.

Wait — maybe the half-life is not 3 days? Or perhaps it's 100 g after 3 days?

No — the problem says: “half-life is 3 days” and “after 21 days”

21 ÷ 3 = 7 half-lives

After 7 half-lives:
- $ 100 \to 50 \to 25 \to 12.5 \to 6.25 \to 3.125 \to 1.5625 \to 0.78125 $

So final amount = ~0.78 g

But the answer says 3.75 g — this is incorrect.

Wait — could it be after 6 days? Then 2 half-lives: $ 100 \to 50 \to 25 $ → still not 3.75

Or maybe after 12 days: 4 half-lives → 6.25 g

Still not 3.75.

Wait — $ 100 \times (1/2)^n = 3.75 $

Then $ (1/2)^n = 0.0375 $

Take log: $ n \log(0.5) = \log(0.0375) $

$ n ≈ \log(0.0375)/\log(0.5) ≈ (-1.443)/(-0.301) ≈ 4.8 $

So about 4.8 half-lives → 4.8 × 3 = 14.4 days — not 21.

So 3.75 g is not correct.

But wait — maybe the problem was misread?

Let me recheck:
“After 21 days” → 7 half-lives → $ 100 / 128 = 0.78125 $ g

✔ Correct Answer: ~0.78 g

✘ Given answer: 3.75 g — WRONG

---

✘ Summary of Errors:

| Problem | Given Answer | Correct Answer | Notes |
|--------|--------------|----------------|-------|
| 1 | 15 mg | ✔ 15 mg | Correct |
| 2 | 68.125 g | ✔ 68.125 g | Correct |
| 3 | 0.02 mg | ✔ ~0.0205 mg | Rounded correctly |
| 4 | 150 mg | ✔ ~150 mg | Correct |
| 5 | 1.56% | ✔ 1.56% | Correct |
| 6 | 11.25 mg | ✘ 2.375 mg | Incorrect — should be 2.375 mg |
| 7 | 3.75 g | ✘ 0.78125 g | Incorrect — should be ~0.78 g |

---

📌 Final Correction:

Problem 6:
Carbon-14, 19 mg, half-life 5,730 years, after 17,190 years (3 half-lives):
→ $ 19 \div 8 = 2.375 $ mg

Problem 7:
Radon-221, 100 g, half-life 3 days, after 21 days (7 half-lives):
→ $ 100 \div 128 = 0.78125 $ g ≈ 0.78 g

---

✔ Corrected Answers:

1. 15 mg
2. 68.125 g
3. 0.02 mg
4. 150 mg
5. 1.56%
6. 2.375 mg (not 11.25 mg)
7. 0.78 g (not 3.75 g)

---

💡 Conclusion:

The worksheet has two errors in the answer key:
- Problem 6: Should be 2.375 mg
- Problem 7: Should be 0.78 g

All others are correct.

Let me know if you'd like a printable version or explanation in simpler terms!