Chemistry · Ch 13 — Nuclear Chemistry and Radioactivity

Half-Life of a Radioelement

13.5.4

Half-Life of a Radioelement

The half-life of a radioelement, $t_{1/2}$ , is defined as the time needed for a given number of its nuclei to decay to exactly half of their initial value. Every radioisotope has its own characteristic half-life, which can range from a tiny fraction of a second to billions of years, and is expressed in whatever time unit is convenient -- seconds, minutes, hours, days or years.

The relationship between half-life and the decay constant $\lambda$ follows directly from substituting the defining condition of half-life -- at $t = t_{1/2}$ , $N = N_0/2$ -- into $\lambda = \frac{2.303}{t}\log_{10}\frac{N_0}{N}$ (section 13.5.3): $\lambda = \frac{2.303}{t_{1/2}}\log_{10}\frac{N_0}{N_0/2} = \frac{2.303}{t_{1/2}}\log_{10}2 = \frac{2.303}{t_{1/2}} \times 0.3010 = \frac{0.693}{t_{1/2}}$ . …

Misc Problem 13.2Half-life of 41Ar from two activity readings

Worked out. Worked example. 41Ar decays initially at a rate of 575 Bq (dps); after 75 minutes the rate has fallen to 358 dps. Using λ = (2.303/t) log10(A0/A) = (2.303/75 min) x log10(575/358) = (2.303/75) x log10(1.6061) = (2.303/75) x 0.2058 = 6.32 x 10^-3 min^-1. Then t(1/2) = 0.693/λ = 0.693/(6.32 x 10^-3 min^-1) ≈ 109.7 minutes. …

Misc Problem 13.3Percentage of 32P remaining after 40 days

Worked out. Worked example. Half-life of 32P is 14.26 d, so λ = 0.693/14.26 d ≈ 0.0486 d^-1. Taking N0 = 100 (as a percentage basis), log10(N0/N) = λt/2.303 = (0.0486 x 40)/2.303 ≈ 0.8441; taking the antilog gives N0/N ≈ 6.984, so N ≈ 100/6.984 ≈ 14.32. Hence about 14.32% of the original 32P sample remains after 40 days. …

Misc Problem 13.4Time for 99.9% of 34Cl to decay

Worked out. Worked example. Half-life of 34Cl is 1.53 s, so λ = 0.693/1.53 s ≈ 0.453 s^-1. If 99.9% has decayed, N0 = 100 and N = 0.1 remain, so N0/N = 1000. Using λ = (2.303/t) log10(N0/N): t = (2.303/λ) x log10(1000) = (2.303/0.453 s^-1) x 3 ≈ 15.25 seconds. …

Misc Problem 13.5Amount of 209Po decaying in 62 years

Worked out. Worked example. Half-life of 209Po is 102 y, so λ = 0.693/102 y ≈ 6.794 x 10^-3 y^-1. log10(N0/N) = λt/2.303 = (6.794 x 10^-3 x 62)/2.303 ≈ 0.1829; antilog gives N0/N ≈ 1.524. Starting from a 1 mg sample, the amount remaining N = 1 mg / 1.524 ≈ 0.656 mg, so the amount that has decayed in 62 years = 1 mg - 0.656 mg = 0.34 …

Misc Problem 13.8Alpha particles emitted per second by 2 mg of 209Po

Worked out. Worked example, with a genuine printing/OCR inconsistency in the source flagged below. Half-life of 209Po is 102 y, so λ = 0.693/(102 x 365 x 24 x 3600 s) = 0.693/(3.2167 x 10^9 s) ≈ 2.154 x 10^-10 s^-1. Number of atoms in a 2 mg sample: N = (2 x 10^-3 g / 209 g mol^-1) x 6.022 x 10^23 mol^-1 ≈ 5.763 x 10^18 atoms -- this figure for N is printed correctly in the source. Activity = λN ≈ (2.154 x 10^-10 s^-1) x (5.763 x 10^18) ≈ 1.24 x 10^9 alpha particles per second. FLAGGED FIDELITY ISSUE: the source's own final line of working substitutes N as '5.763 x 10^12' (dropping six orders of magnitude from the correctly-derived 5.763 x 10^18 two lines above it) and prints a final answer of '1241 particles s^-1'. Recomputing independently and consistently from the source's own correct atom count (5.763 x 10^18) gives an answer six orders of magnitude larger, about 1.24 x 10^9 particles per second, which is also the physically sensible order of magnitude for 2 mg of a 102-year-half-life alpha emitter. This is treated as a probable exponent-extraction/printing error in the source rather than silently repr …

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