Half-life Problems Worksheet featuring ten questions on radioactive decay calculations for various isotopes.
A worksheet titled "Half-life Problems" with ten questions related to radioactive decay, including calculations involving cesium-137, carbon-14, germanium-66, strontium-90, cobalt-60, phosphorus-32, radon-222, sodium-25, polonium-214, and iodine-131, requiring students to show work on separate paper.
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Step-by-step solution for: Half Life Worksheet Answers
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Show Answer Key & Explanations
Step-by-step solution for: Half Life Worksheet Answers
Here are the step-by-step solutions for the half-life problems on your worksheet.
The general formula used for these problems is:
$$ \text{Remaining Amount} = \text{Initial Amount} \times (0.5)^{\frac{\text{Time Elapsed}}{\text{Half-Life}}} $$
Or, if you need to find the time or initial amount, you can rearrange it using logarithms or algebra.
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1. Cesium-137
* Given: Half-life ($t_{1/2}$) = 30.2 years, Initial Mass ($N_0$) = 1.00 kg, Time ($t$) = 151 years.
* Step 1: Find how many half-lives have passed.
$$ n = \frac{151}{30.2} = 5 \text{ half-lives} $$
* Step 2: Calculate remaining mass.
$$ N_t = 1.00 \times (0.5)^5 $$
$$ N_t = 1.00 \times 0.03125 $$
$$ N_t = 0.03125 \text{ kg} $$
2. Carbon-14
* Given: Half-life = 5730 years, Initial Mass ($N_0$) = 24 g, Remaining Mass ($N_t$) = 1.5 g.
* Step 1: Set up the equation to solve for time ($t$).
$$ 1.5 = 24 \times (0.5)^{\frac{t}{5730}} $$
* Step 2: Divide by 24.
$$ \frac{1.5}{24} = 0.0625 $$
$$ 0.0625 = (0.5)^{\frac{t}{5730}} $$
* Step 3: Recognize that $0.0625$ is $(0.5)^4$. So, 4 half-lives have passed.
$$ 4 = \frac{t}{5730} $$
$$ t = 4 \times 5730 = 22,920 \text{ years} $$
3. Germanium-66
* Given: Initial Mass ($N_0$) = 64 g, Remaining Mass ($N_t$) = 2.0 g, Time ($t$) = 12.5 hours.
* Step 1: Determine how many half-lives occurred.
$$ 2.0 = 64 \times (0.5)^n $$
$$ \frac{2.0}{64} = \frac{1}{32} = 0.03125 $$
Since $(0.5)^5 = 0.03125$, then $n = 5$ half-lives.
* Step 2: Calculate the length of one half-life.
$$ 5 \times t_{1/2} = 12.5 \text{ hours} $$
$$ t_{1/2} = \frac{12.5}{5} = 2.5 \text{ hours} $$
4. Strontium-90
* Given: Half-life = 28.8 years, Initial Mass ($N_0$) = 1 g, Remaining Mass ($N_t$) = 125 mg.
* Step 1: Convert units so they match.
$$ 1 \text{ g} = 1000 \text{ mg} $$
* Step 2: Solve for number of half-lives ($n$).
$$ 125 = 1000 \times (0.5)^n $$
$$ \frac{125}{1000} = 0.125 $$
Since $(0.5)^3 = 0.125$, then $n = 3$ half-lives.
* Step 3: Calculate total time.
$$ t = 3 \times 28.8 = 86.4 \text{ years} $$
5. Cobalt-60
* Given: Half-life = 5.3 years, Time ($t$) = 26.5 years, Remaining Mass ($N_t$) = 14.5 g.
* Step 1: Find number of half-lives.
$$ n = \frac{26.5}{5.3} = 5 \text{ half-lives} $$
* Step 2: Solve for Initial Mass ($N_0$).
$$ 14.5 = N_0 \times (0.5)^5 $$
$$ 14.5 = N_0 \times 0.03125 $$
$$ N_0 = \frac{14.5}{0.03125} = 464 \text{ g} $$
6. Phosphorus-32
* Given: Initial Mass ($N_0$) = 1.000 kg, Half-life = 14.3 days, Time ($t$) = 100.1 days.
* Step 1: Find number of half-lives.
$$ n = \frac{100.1}{14.3} = 7 \text{ half-lives} $$
* Step 2: Calculate remaining mass.
$$ N_t = 1.000 \times (0.5)^7 $$
$$ N_t = 1.000 \times 0.0078125 $$
$$ N_t \approx 0.00781 \text{ kg} $$ (or 7.81 g)
7. Radon-222
* Given: Half-life = 3.8 days, Time ($t$) = 7.6 days, Remaining Mass ($N_t$) = 6.5 $\mu$g.
* Step 1: Find number of half-lives.
$$ n = \frac{7.6}{3.8} = 2 \text{ half-lives} $$
* Step 2: Solve for Initial Mass ($N_0$).
$$ 6.5 = N_0 \times (0.5)^2 $$
$$ 6.5 = N_0 \times 0.25 $$
$$ N_0 = \frac{6.5}{0.25} = 26 \text{ \mu g} $$
8. Sodium-25
* Given: Half-life = 1.0 minute, Initial Mass ($N_0$) = 1.0 kg, Time ($t$) = 30 minutes (half an hour).
* Step 1: Find number of half-lives.
$$ n = \frac{30}{1.0} = 30 \text{ half-lives} $$
* Step 2: Calculate remaining mass.
$$ N_t = 1.0 \times (0.5)^{30} $$
$$ N_t \approx 9.31 \times 10^{-10} \text{ kg} $$
9. Polonium-214
* Given: Initial Mass ($N_0$) = 1.0 g, Remaining Mass ($N_t$) = 31.25 mg, Time ($t$) = 820 seconds.
* Step 1: Convert units.
$$ 1.0 \text{ g} = 1000 \text{ mg} $$
* Step 2: Find number of half-lives ($n$).
$$ 31.25 = 1000 \times (0.5)^n $$
$$ \frac{31.25}{1000} = 0.03125 $$
Since $(0.5)^5 = 0.03125$, then $n = 5$ half-lives.
* Step 3: Calculate half-life duration.
$$ 5 \times t_{1/2} = 820 \text{ seconds} $$
$$ t_{1/2} = \frac{820}{5} = 164 \text{ seconds} $$
10. Iodine-131
* Given: Half-life = 8.0 days, Initial Mass ($N_0$) = 225 g, Time ($t$) = 40 days.
* Step 1: Find number of half-lives.
$$ n = \frac{40}{8.0} = 5 \text{ half-lives} $$
* Step 2: Calculate remaining mass.
$$ N_t = 225 \times (0.5)^5 $$
$$ N_t = 225 \times 0.03125 $$
$$ N_t = 7.03125 \text{ g} $$
Final Answer:
1. 0.03125 kg
2. 22,920 years
3. 2.5 hours
4. 86.4 years
5. 464 g
6. 0.00781 kg (or 7.81 g)
7. 26 µg
8. $9.31 \times 10^{-10}$ kg
9. 164 seconds
10. 7.03 g
The general formula used for these problems is:
$$ \text{Remaining Amount} = \text{Initial Amount} \times (0.5)^{\frac{\text{Time Elapsed}}{\text{Half-Life}}} $$
Or, if you need to find the time or initial amount, you can rearrange it using logarithms or algebra.
---
1. Cesium-137
* Given: Half-life ($t_{1/2}$) = 30.2 years, Initial Mass ($N_0$) = 1.00 kg, Time ($t$) = 151 years.
* Step 1: Find how many half-lives have passed.
$$ n = \frac{151}{30.2} = 5 \text{ half-lives} $$
* Step 2: Calculate remaining mass.
$$ N_t = 1.00 \times (0.5)^5 $$
$$ N_t = 1.00 \times 0.03125 $$
$$ N_t = 0.03125 \text{ kg} $$
2. Carbon-14
* Given: Half-life = 5730 years, Initial Mass ($N_0$) = 24 g, Remaining Mass ($N_t$) = 1.5 g.
* Step 1: Set up the equation to solve for time ($t$).
$$ 1.5 = 24 \times (0.5)^{\frac{t}{5730}} $$
* Step 2: Divide by 24.
$$ \frac{1.5}{24} = 0.0625 $$
$$ 0.0625 = (0.5)^{\frac{t}{5730}} $$
* Step 3: Recognize that $0.0625$ is $(0.5)^4$. So, 4 half-lives have passed.
$$ 4 = \frac{t}{5730} $$
$$ t = 4 \times 5730 = 22,920 \text{ years} $$
3. Germanium-66
* Given: Initial Mass ($N_0$) = 64 g, Remaining Mass ($N_t$) = 2.0 g, Time ($t$) = 12.5 hours.
* Step 1: Determine how many half-lives occurred.
$$ 2.0 = 64 \times (0.5)^n $$
$$ \frac{2.0}{64} = \frac{1}{32} = 0.03125 $$
Since $(0.5)^5 = 0.03125$, then $n = 5$ half-lives.
* Step 2: Calculate the length of one half-life.
$$ 5 \times t_{1/2} = 12.5 \text{ hours} $$
$$ t_{1/2} = \frac{12.5}{5} = 2.5 \text{ hours} $$
4. Strontium-90
* Given: Half-life = 28.8 years, Initial Mass ($N_0$) = 1 g, Remaining Mass ($N_t$) = 125 mg.
* Step 1: Convert units so they match.
$$ 1 \text{ g} = 1000 \text{ mg} $$
* Step 2: Solve for number of half-lives ($n$).
$$ 125 = 1000 \times (0.5)^n $$
$$ \frac{125}{1000} = 0.125 $$
Since $(0.5)^3 = 0.125$, then $n = 3$ half-lives.
* Step 3: Calculate total time.
$$ t = 3 \times 28.8 = 86.4 \text{ years} $$
5. Cobalt-60
* Given: Half-life = 5.3 years, Time ($t$) = 26.5 years, Remaining Mass ($N_t$) = 14.5 g.
* Step 1: Find number of half-lives.
$$ n = \frac{26.5}{5.3} = 5 \text{ half-lives} $$
* Step 2: Solve for Initial Mass ($N_0$).
$$ 14.5 = N_0 \times (0.5)^5 $$
$$ 14.5 = N_0 \times 0.03125 $$
$$ N_0 = \frac{14.5}{0.03125} = 464 \text{ g} $$
6. Phosphorus-32
* Given: Initial Mass ($N_0$) = 1.000 kg, Half-life = 14.3 days, Time ($t$) = 100.1 days.
* Step 1: Find number of half-lives.
$$ n = \frac{100.1}{14.3} = 7 \text{ half-lives} $$
* Step 2: Calculate remaining mass.
$$ N_t = 1.000 \times (0.5)^7 $$
$$ N_t = 1.000 \times 0.0078125 $$
$$ N_t \approx 0.00781 \text{ kg} $$ (or 7.81 g)
7. Radon-222
* Given: Half-life = 3.8 days, Time ($t$) = 7.6 days, Remaining Mass ($N_t$) = 6.5 $\mu$g.
* Step 1: Find number of half-lives.
$$ n = \frac{7.6}{3.8} = 2 \text{ half-lives} $$
* Step 2: Solve for Initial Mass ($N_0$).
$$ 6.5 = N_0 \times (0.5)^2 $$
$$ 6.5 = N_0 \times 0.25 $$
$$ N_0 = \frac{6.5}{0.25} = 26 \text{ \mu g} $$
8. Sodium-25
* Given: Half-life = 1.0 minute, Initial Mass ($N_0$) = 1.0 kg, Time ($t$) = 30 minutes (half an hour).
* Step 1: Find number of half-lives.
$$ n = \frac{30}{1.0} = 30 \text{ half-lives} $$
* Step 2: Calculate remaining mass.
$$ N_t = 1.0 \times (0.5)^{30} $$
$$ N_t \approx 9.31 \times 10^{-10} \text{ kg} $$
9. Polonium-214
* Given: Initial Mass ($N_0$) = 1.0 g, Remaining Mass ($N_t$) = 31.25 mg, Time ($t$) = 820 seconds.
* Step 1: Convert units.
$$ 1.0 \text{ g} = 1000 \text{ mg} $$
* Step 2: Find number of half-lives ($n$).
$$ 31.25 = 1000 \times (0.5)^n $$
$$ \frac{31.25}{1000} = 0.03125 $$
Since $(0.5)^5 = 0.03125$, then $n = 5$ half-lives.
* Step 3: Calculate half-life duration.
$$ 5 \times t_{1/2} = 820 \text{ seconds} $$
$$ t_{1/2} = \frac{820}{5} = 164 \text{ seconds} $$
10. Iodine-131
* Given: Half-life = 8.0 days, Initial Mass ($N_0$) = 225 g, Time ($t$) = 40 days.
* Step 1: Find number of half-lives.
$$ n = \frac{40}{8.0} = 5 \text{ half-lives} $$
* Step 2: Calculate remaining mass.
$$ N_t = 225 \times (0.5)^5 $$
$$ N_t = 225 \times 0.03125 $$
$$ N_t = 7.03125 \text{ g} $$
Final Answer:
1. 0.03125 kg
2. 22,920 years
3. 2.5 hours
4. 86.4 years
5. 464 g
6. 0.00781 kg (or 7.81 g)
7. 26 µg
8. $9.31 \times 10^{-10}$ kg
9. 164 seconds
10. 7.03 g
Parent Tip: Review the logic above to help your child master the concept of half life worksheet extra practice.
