Question 25.3: Using the Half-Life Concept and the Radioactive Decay Law to......
Using the Half-Life Concept and the Radioactive Decay Law to Describe the Rate of Radioactive Decay
The phosphorus isotope ^{32}P is used in biochemical studies to determine the pathways of phosphorus atoms in living organisms. Its presence is detected through its emission of β^{−} particles. (a) What is the decay constant for ^{32}P expressed in the unit s^{−1}? (b) What is the activity of a 1.00 mg sample of ^{32}P (that is, how many atoms disintegrate per second)? (c) Approximately what mass of ^{32}P will remain in the original 1.00 mg sample after 57 days? (See Table 25.1.) (d) What will be the rate of radioactive decay after 57 days?
| TABLE 25.1 Some Representative Half-Lives | |||||
| Nuclide | Half-Life^{a} | Nuclide | Half-Life^{a} | Nuclide | Half-Life^{a} |
| ^{3}_{1}H | 12.33 a | ^{40}_{19}K | 1.26 × 10^{9} a | ^{214}_{84}Po | 1.64 × 10^{−4} s |
| ^{14}_{6}C | 5715 a | ^{80}_{35}Br | 17.6 min | ^{222}_{86}Rn | 3.823 d |
| ^{13}_{8}O | 8.7 × 10^{−3} s | ^{90}_{38}Sr | 29.1 a | ^{226}_{88}Ra | 1.599 × 10³ a |
| ^{28}_{12}Mg | 21 h | ^{131}_{53}I | 8.021 d | ^{234}_{99}Th | 24.10 d |
| ^{32}_{15}P | 14.3 d | ^{137}_{55}Cs | 30.2 a | ^{238}_{92}U | 4.47 × 10^{9} a |
| ^{35}_{16}S | 87.9 d | ||||
^{a}s, second; min, minute; h, hour; d, day; a, year.
Source: CRC Handbook of Chemistry and Physics, 92 ed., by W. M. Haynes – Editor in Chief.
Analyze
To solve these types of problems, we need to determine the decay constant, λ, of the radioactive species, which is related to the concept of half-life through equation (25.13). After determining the decay constant from the half-life of the sample, we can then use it to determine the activity for part (b).
t_{1/2} = \frac{\ln (2)}{λ} (25.13)
Solve
(a) We can determine λ from t_{1/2} with equation (25.13). The first result we get has the unit d^{−1}. We must convert this unit to h^{−1}, min^{−1}, \text{and} s^{−1}.
λ = \frac{0.693}{14.3 d} × \frac{1 d}{24 h} × \frac{1 h}{60 min} × \frac{1 min}{60 s} = 5.61 × 10^{−7} s^{−1}
(b) First, let us find the number of atoms, N, in 1.00 mg of ^{32}P.
N (^{32}P atoms) = 0.00100 g × \frac{1 mol ^{32}P}{32.0 g} × \frac{6.022 × 10^{23} ^{32}P atoms}{1 mol ^{32}P}
= 1.88 × 10^{19} ^{32}P atoms
Then, we can multiply this number by the decay constant to get the activity or decay rate.
activity = λN = 5.61 × 10^{−7} s^{−1} × 1.88 × 10^{19} atoms
= 1.05 × 10^{13} atoms s^{−1}
(c) A period of 57 days is 57/14.3 = 4.0 half-lives. As shown in Figure 25-4, the quantity of radioactive material decreases by one-half for every half-life. The quantity remaining is (\frac{1}{2})^{4} of the original quantity.
? mg ^{32}P = 1.00 mg × (\frac{1}{2})^{4} = 1.00 mg × \frac{1}{16} = 0.063 mg ^{32}P
(d) The activity is directly proportional to the number of radioactive atoms remaining (activity = λN), and the number of atoms is directly proportional to the mass of ^{32}P. When the mass of ^{32}P has dropped to one-sixteenth its original mass, the number of ^{32}P atoms also falls to one sixteenth the original number, and the rate of decay is one-sixteenth the original activity.
rate of decay = \frac{1}{16} × 1.05 × 10^{13} atoms s^{−1} = 6.56 × 10^{11} atoms s^{−1}
Assess
Phosphorus-32 isotope is an ideal radionuclide to use because of its short half-life. This time is long enough to carry out experiments, yet short enough that disposal is not a major problem. We see in this example that after 57 days, approximately 6% of the original mass of phosphorus remains.
