This worksheet provides practice problems for calculating half-life in chemistry, covering isotopes such as Cesium-137 and Selenium-83.

Let’s solve each problem one by one, step by step. We’ll use the idea of half-life: every half-life period, half of the material decays (disappears). So we keep dividing by 2 for each half-life that passes.

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Problem 1:
Cesium-137 has a half-life of 30 years. Start with 1.0 g. How much remains after 90 years?

→ First, find how many half-lives pass in 90 years:
90 ÷ 30 = 3 half-lives

→ Now, start with 1.0 g and cut it in half 3 times:

After 1st half-life (30 yrs): 1.0 → 0.5 g
After 2nd half-life (60 yrs): 0.5 → 0.25 g
After 3rd half-life (90 yrs): 0.25 → 0.125 g

✔ Final Answer for #1: 0.125 grams

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Problem 2:
Actinium-226 half-life = 29 hours. Start with 100 mg. After 58 hours, how much remains?

→ Number of half-lives: 58 ÷ 29 = 2 half-lives

→ Cut 100 mg in half twice:

After 1st: 100 → 50 mg
After 2nd: 50 → 25 mg

✔ Final Answer for #2: 25 milligrams

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Problem 3:
Sodium-25: took 3.0 minutes to get from reactor to lab. Started with 10.0 µg, ended with 0.84 µg. What is its half-life in seconds?

This one is trickier — we need to figure out how many half-lives passed based on how much decayed.

Start: 10.0 µg
End: 0.84 µg

We can think: how many times do we have to divide 10.0 by 2 to get close to 0.84?

Try:

After 1 half-life: 10.0 → 5.0
After 2: 5.0 → 2.5
After 3: 2.5 → 1.25
After 4: 1.25 → 0.625 ← too low (we have 0.84, which is between 1.25 and 0.625)

Wait — maybe not exact halves? Let’s calculate more precisely.

Actually, let’s use this formula (but explained simply):

Amount left = Initial × (1/2)^n
Where n = number of half-lives

So:
0.84 = 10.0 × (1/2)^n
Divide both sides by 10.0:
0.084 = (1/2)^n

Now take log or guess:

(1/2)^3 = 0.125
(1/2)^4 = 0.0625

0.084 is between them → so n ≈ 3.58? That seems messy.

Wait — perhaps they expect us to notice:

From 10.0 to 0.84 is about 1/12th? Not helpful.

Alternative approach: Maybe count how many half-lives fit into the decay.

But here’s a better way — let’s see what fraction remains:

0.84 / 10.0 = 0.084

What power of 1/2 gives 0.084?

Let me try:

(1/2)^3 = 0.125
(1/2)^3.5 = ? → sqrt(0.125) ≈ 0.353, then 0.125 * 0.353 ≈ no… wait.

Actually, let’s use logs (if allowed), but since this is for students, maybe approximate.

Notice: 10.0 → 5.0 → 2.5 → 1.25 → 0.625

Our value is 0.84, which is between 1.25 and 0.625 → so between 3 and 4 half-lives.

Let’s interpolate:

At 3 half-lives: 1.25 µg
At 4 half-lives: 0.625 µg
Difference: 1.25 - 0.625 = 0.625 µg drop over 1 half-life

We went from 1.25 to 0.84 → drop of 0.41 µg

So fraction of half-life: 0.41 / 0.625 ≈ 0.656

So total half-lives ≈ 3 + 0.656 = 3.656

Time elapsed = 3.0 minutes = 180 seconds

So half-life = total time / number of half-lives = 180 / 3.656 ≈ ?

Calculate: 180 ÷ 3.656 ≈ 49.2 seconds

But that seems odd — maybe I made a mistake.

Wait — let’s check if 0.84 is exactly reachable.

Try: 10.0 × (1/2)^n = 0.84
→ (1/2)^n = 0.084
Take natural log:
ln(0.084) = n × ln(0.5)
n = ln(0.084) / ln(0.5)
ln(0.084) ≈ -2.476
ln(0.5) ≈ -0.693
n ≈ (-2.476)/(-0.693) ≈ 3.573

So n ≈ 3.573 half-lives in 3.0 minutes = 180 seconds

Half-life = 180 / 3.573 ≈ 50.38 seconds

Round to reasonable digits → about 50 seconds

But let’s verify: If half-life is 50 seconds, then in 180 seconds, number of half-lives = 180/50 = 3.6

Then amount left = 10.0 × (1/2)^3.6

(1/2)^3 = 0.125
(1/2)^0.6 ≈ ? → 2^0.6 ≈ 1.5157, so (1/2)^0.6 ≈ 1/1.5157 ≈ 0.6598

So 0.125 × 0.6598 ≈ 0.0825 → times 10.0 = 0.825 µg → very close to 0.84!

Yes! So half-life ≈ 50 seconds.

✔ Final Answer for #3: 50 seconds

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Problem 4:
Isotope X half-life = 2.0 years. Start with 4.0 mg. How long until only 0.50 mg remains?

Start: 4.0 mg
Goal: 0.50 mg

How many half-lives?

4.0 → 2.0 → 1.0 → 0.5 → that’s 3 half-lives

Each half-life = 2.0 years → 3 × 2.0 = 6.0 years

✔ Final Answer for #4: 6.0 years

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Problem 5:
Selenium-83 half-life = 25.0 minutes. Start with 10.0 mg. How long until 1.25 mg remains?

Start: 10.0 mg
Goal: 1.25 mg

Cut in half:

10.0 → 5.0 → 2.5 → 1.25 → that’s 3 half-lives

Each half-life = 25.0 min → 3 × 25.0 = 75.0 minutes

✔ Final Answer for #5: 75.0 minutes

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Problem 6:
Po-218 half-life = 3 minutes. Start with 2.0 g. How much remains after 15 minutes?

Number of half-lives: 15 ÷ 3 = 5 half-lives

Start: 2.0 g

After 1: 1.0
After 2: 0.5
After 3: 0.25
After 4: 0.125
After 5: 0.0625 g

✔ Final Answer for #6: 0.0625 grams

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Final Answers:

1. 0.125 grams
2. 25 milligrams
3. 50 seconds
4. 6.0 years
5. 75.0 minutes
6. 0.0625 grams