IB Physics HL E 3 12 Half Life Changes QuestionsSolve HL half-life questions by calculating remaining parent and daughter fractions, elapsed time and activity changes after repeated half-lives.SyllabusFirst assessment 2025CoursePhysics HLLevelHL
Exam pointscalculate the remaining parent fraction and daughter fraction after a stated number of half-livesdetermine elapsed time, remaining mass or activity from a half-life and an initial valueinterpret fractional activity changes such as a seven-eighths decrease using repeated half-life intervals
IB Physics HL E 3 12 Half Life Changes Questions question 1[Maximum number: 3]The radioactive nuclide beryllium-10 (Be-10) undergoes beta minus ( β\betaβ-) decay to form a stable boron (B) nuclide.The initial number of nuclei in a pure sample of beryllium-10 is N0\mathrm{N}_{0}N0. The graph shows how the number of remaining beryllium nuclei in the sample varies with time.After 4.3×1064.3 \times 10^{6}4.3×106 years, number of produced boron nuclei number of remaining beryllium nuclei =7.\frac{\text { number of produced boron nuclei }}{\text { number of remaining beryllium nuclei }}=7 . number of remaining beryllium nuclei number of produced boron nuclei =7.Show that the half-life of beryllium-10 is 1.4×1061.4 \times 10^{6}1.4×106 years.Show AnswerALTERNATIVE 1fraction of Be=18,12.5%\mathrm{Be}=\frac{1}{8}, 12.5\%Be=81,12.5%, or 0.125therefore 3 half-lives have elapsedt12=4.3×1063=1.43×106 y≈1.4×106 yt_{\frac{1}{2}}=\frac{4.3\times10^{6}}{3}=1.43\times10^{6}\ \mathrm{y}\approx1.4\times10^{6}\ \mathrm{y}t21=34.3×106=1.43×106 y≈1.4×106 yALTERNATIVE 2fraction of Be=18,12.5%\mathrm{Be}=\frac{1}{8}, 12.5\%Be=81,12.5%, or 0.12518=e−λ(4.3×106), leading to λ=4.836×10−7 y−1\frac{1}{8}=e^{-\lambda(4.3\times10^{6})}\text{, leading to }\lambda=4.836\times10^{-7}\ \mathrm{y}^{-1}81=e−λ(4.3×106), leading to λ=4.836×10−7 y−1ln2λ=1.43×106 y\frac{\ln 2}{\lambda}=1.43\times10^{6}\ \mathrm{y}λln2=1.43×106 yAdd to Test