Half‑Life Practice Worksheet with Answers: A Comprehensive Study Guide
Introduction
Understanding the concept of half‑life is essential for students tackling nuclear physics, radiochemistry, and related fields. A half‑life is the time required for half of a radioactive substance to decay. Mastering this idea enables learners to predict decay rates, calculate remaining activity, and solve real‑world problems involving radioisotopes The details matter here..
This article presents a complete practice worksheet featuring a variety of problems that test comprehension of half‑life calculations, decay chains, and applications. Consider this: each question is followed by a detailed answer, ensuring that students can verify their work and understand the reasoning behind each solution. By working through this worksheet, readers will solidify their grasp of exponential decay and be better prepared for exams and practical applications Surprisingly effective..
Half‑Life Fundamentals
Before diving into the exercises, let’s recap the key formulas and concepts:
| Symbol | Meaning | Formula |
|---|---|---|
| (N(t)) | Number of undecayed nuclei at time (t) | (N(t) = N_0 , e^{-\lambda t}) |
| (N_0) | Initial number of nuclei | — |
| (\lambda) | Decay constant | (\lambda = \dfrac{\ln 2}{t_{1/2}}) |
| (t_{1/2}) | Half‑life | — |
| (A(t)) | Activity at time (t) | (A(t) = A_0 , e^{-\lambda t}) |
| (A_0) | Initial activity | — |
Not the most exciting part, but easily the most useful.
Key points to remember
- Half‑life is independent of the amount of material; it depends only on the isotope’s nuclear properties.
- Exponential decay: the process is continuous and slows over time, never truly reaching zero.
- Decay chains: when a product isotope itself is radioactive, multiple half‑lives interact.
Practice Worksheet
Problem 1 – Basic Half‑Life Calculation
A sample of a radioactive isotope has an initial activity of 200 Bq. After 3 hours, the activity drops to 50 Bq.
(b) How many hours will it take for the activity to reduce to 12.That said, (a) Determine the half‑life of the isotope. 5 Bq?
Problem 2 – Decay Constant from Half‑Life
The half‑life of isotope X is 12.On top of that, 5 days. That's why (a) Calculate the decay constant (\lambda) in day(^{-1}). (b) What fraction of the original quantity remains after 30 days?
Problem 3 – Number of Decays in a Given Time
A 1‑gram sample of a radioactive element contains (5 \times 10^{20}) atoms. Its half‑life is 2 years.
(a) How many atoms decay in the first 6 months?
(b) What is the activity (in Bq) at the start?
Problem 4 – Two‑Step Decay Chain
Isotope A decays to isotope B with a half‑life of 4 days. Isotope B decays to a stable product with a half‑life of 1 day.
A sample initially contains (1 \times 10^{6}) atoms of A and none of B.
(a) After 8 days, how many atoms of B are present?
(b) What is the total activity of the system after 8 days?
Problem 5 – Radiocarbon Dating
A piece of ancient wood has a radiocarbon content of 12.5 % of modern levels.
(a) Estimate its age in years.
(b) If the wood was 2,000 years old, what percentage of its original radiocarbon would remain?
Problem 6 – Half‑Life and Safety
A radioactive waste container holds a nuclide with a half‑life of 50 years.
(a) What fraction of the initial activity remains after 200 years?
(b) If the initial activity was 1000 Bq, what is the activity after 200 years?
Answers and Explanations
Answer 1 – Basic Half‑Life Calculation
(a)
Using (A(t) = A_0 e^{-\lambda t}) and (t_{1/2} = \frac{\ln 2}{\lambda}):
-
Solve for (\lambda): [ \frac{A(t)}{A_0} = e^{-\lambda t} ;\Rightarrow; \frac{50}{200} = e^{-\lambda \times 3} ] [ 0.25 = e^{-3\lambda} ;\Rightarrow; \ln 0.25 = -3\lambda ;\Rightarrow; \lambda = \frac{-\ln 0.25}{3} = \frac{1.3863}{3} \approx 0.4621 \text{ h}^{-1} ]
-
Half‑life: [ t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.6931}{0.4621} \approx 1.5 \text{ h} ]
(b)
We need (t) such that (A(t) = 12.5) Bq:
[
\frac{12.5}{200} = e^{-\lambda t} ;\Rightarrow;
0.0625 = e^{-0.4621 t}
]
[
\ln 0.0625 = -0.4621 t ;\Rightarrow;
t = \frac{-\ln 0.0625}{0.4621} \approx \frac{2.7726}{0.4621} \approx 6.0 \text{ h}
]
Result:
- Half‑life ≈ 1.5 h
- Time to reach 12.5 Bq ≈ 6 h
Answer 2 – Decay Constant from Half‑Life
(a)
[
\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.6931}{12.5 \text{ d}} \approx 0.05545 \text{ d}^{-1}
]
(b)
Fraction remaining after 30 days:
[
N(t) = N_0 e^{-\lambda t} = e^{-0.05545 \times 30} = e^{-1.6635} \approx 0.189
]
So ≈18.9 % of the original quantity remains.
Answer 3 – Number of Decays in a Given Time
(a)
Number of atoms after 6 months (0.5 yr):
[
N(t) = N_0 e^{-\lambda t}
]
First, find (\lambda):
[
t_{1/2} = 2 \text{ yr} ;\Rightarrow; \lambda = \frac{\ln 2}{2} \approx 0.3466 \text{ yr}^{-1}
]
Then:
[
N(0.5) = 5 \times 10^{20} , e^{-0.3466 \times 0.5} = 5 \times 10^{20} , e^{-0.1733} \approx 5 \times 10^{20} \times 0.841 \approx 4.205 \times 10^{20}
]
Decays in 6 months:
[
\Delta N = N_0 - N(0.5) = 5 \times 10^{20} - 4.205 \times 10^{20} \approx 0.795 \times 10^{20} = 7.95 \times 10^{19}
]
(b)
Activity (A = \lambda N_0):
[
A = 0.3466 \text{ yr}^{-1} \times 5 \times 10^{20} \text{ atoms}
]
Convert yr(^{-1}) to s(^{-1}):
[
0.3466 \text{ yr}^{-1} = \frac{0.3466}{3.1536 \times 10^{7}} \text{ s}^{-1} \approx 1.099 \times 10^{-8} \text{ s}^{-1}
]
Thus:
[
A \approx 1.099 \times 10^{-8} \times 5 \times 10^{20} \approx 5.495 \times 10^{12} \text{ Bq}
]
Approximately 5.5 × 10¹² Bq.
Answer 4 – Two‑Step Decay Chain
Let (N_A) and (N_B) be the numbers of atoms of A and B, respectively.
For a two‑step chain, the solution after time (t) is:
[
N_B(t) = N_{A0} \frac{\lambda_A}{\lambda_B - \lambda_A}
\left(e^{-\lambda_A t} - e^{-\lambda_B t}\right)
]
where (\lambda_A = \frac{\ln 2}{4}) day(^{-1}) and (\lambda_B = \frac{\ln 2}{1}) day(^{-1}) Took long enough..
Compute: [ \lambda_A = 0.Think about it: 250 ] [ e^{-\lambda_B t} = e^{-0. On the flip side, 0039 ] Thus: [ N_B(8) = 1 \times 10^{6} \times 0. 6931 \times 8} = e^{-5.Now, 5448} \approx 0. 333 ] Now evaluate the exponentials for (t = 8) days: [ e^{-\lambda_A t} = e^{-0.5198} \approx 0.333 \times 0.2461 \approx 81,900 ] (a) Approximately 8.1733}{0.Still, 1733 \times 8} = e^{-1. 0039) \approx 1 \times 10^{6} \times 0.1733 \text{ d}^{-1}, \quad \lambda_B = 0.That's why 3864} \approx 0. 6931 \text{ d}^{-1} ] [ \frac{\lambda_A}{\lambda_B - \lambda_A} = \frac{0.250 - 0.So 333 \times (0. 2 × 10⁴ atoms of B.
(b) Activity: [ A_A = \lambda_A N_A, \quad A_B = \lambda_B N_B ] First find (N_A(8)): [ N_A(8) = N_{A0} e^{-\lambda_A t} = 1 \times 10^{6} \times 0.250 = 250,000 ] So: [ A_A = 0.1733 \times 250,000 \approx 43,325 \text{ Bq} ] [ A_B = 0.6931 \times 81,900 \approx 56,780 \text{ Bq} ] Total activity: [ A_{\text{total}} = 43,325 + 56,780 \approx 100,105 \text{ Bq} ] ≈1.0 × 10⁵ Bq.
Answer 5 – Radiocarbon Dating
Radiocarbon half‑life (modern convention) ≈ 5,730 years.
(a)
Fraction remaining (f = 0.125).
Using (f = 2^{-t/t_{1/2}}):
[
0.125 = 2^{-t/5730}
]
Take log base 2:
[
-3 = -\frac{t}{5730} ;\Rightarrow; t = 3 \times 5730 = 17,190 \text{ yr}
]
So the wood is ≈17,200 years old.
(b)
Fraction remaining after 2,000 years:
[
f = 2^{-2000/5730} \approx 2^{-0.349} \approx 0.78
]
Thus about 78 % of the original radiocarbon remains Which is the point..
Answer 6 – Half‑Life and Safety
(a)
Fraction remaining after 200 years:
[
f = 2^{-t/t_{1/2}} = 2^{-200/50} = 2^{-4} = \frac{1}{16} = 0.0625
]
So 6.25 % remains.
(b)
Initial activity (A_0 = 1000) Bq.
After 200 years:
[
A = A_0 \times f = 1000 \times 0.0625 = 62.5 \text{ Bq}
]
≈63 Bq.
Frequently Asked Questions (FAQ)
| Question | Answer |
|---|---|
| **What if the activity is given instead of the number of atoms?Day to day, ** | Use (A = \lambda N). Also, knowing either activity or atom count lets you compute the other via the decay constant. |
| **Do half‑lives change with temperature or pressure?Day to day, ** | No. Half‑life is a quantum mechanical property of the nucleus and remains constant under normal environmental conditions. |
| **Can I use the simple “half every X time” rule for decay chains?So ** | Only for the first step. Still, subsequent products may have different half‑lives, so exponential decay equations are required. |
| How does half‑life affect radiological safety? | Short half‑lives mean rapid decay and low long‑term hazard; long half‑lives require strict containment and monitoring. |
| What is the significance of the decay constant? | It quantifies the probability per unit time that a nucleus will decay. It’s the reciprocal of the mean lifetime. |
Conclusion
Mastering half‑life calculations equips students with a powerful tool to analyze radioactive decay, design experiments, and assess safety protocols. The worksheet above offers a spectrum of problems—from basic half‑life determination to complex decay chains—providing ample practice for diverse learning needs. By consistently applying the exponential decay equations, students will gain confidence and precision, ensuring they can tackle both academic questions and real‑world scenarios with competence.
