Question 25.Int.1: On April 26, 1986, an explosion at the nuclear power plant a...
On April 26, 1986, an explosion at the nuclear power plant at Chernobyl, Ukraine, released the greatest quantity of radioactive material ever associated with an industrial accident. (See the discussion and photo on pages 1131-1132.) One of the radioisotopes in this emission was _{}^{131}\text{I}, a \beta^- emitter with a half-life of 8.04 days.
Assume that the total quantity of _{}^{131}\text{I} released was 250 g, and determine the number of curies associated with this _{}^{131}\text{I} 30.0 days after the accident.
Analyze
From the initial mass of _{}^{131}\text{I} and the known half-life of _{}^{131}\text{I} determine how many grams of _{}^{131}\text{I} remain after 30 days. Convert this mass to a number of atoms using the molar mass of _{}^{131}\text{I} and Avogadro’s number. Finally, the product of the number of atoms and the decay constant yields the activity in disintegrations per second, which can be converted to curies.
In equation (25.12), use λ = 0.693/8.04 d and t = 30.0 d. Because the mass of _{}^{131}\text{I} is directly proportional to the number of _{}^{131}\text{I} atoms, we can replace N_0 by {mass}_0 = 250 g, and N_t by {mass}_t, the mass we seek.
\ln \left(\frac{N_t}{N_0}\right)=-\lambda t (25.12)
\ln \frac{\text {mass}_t}{{mass}_0}=\ln \frac{{mass}_t}{250 g } = \frac{-0.693}{8.04 d } \times 30.0 d =-2.59
\text {mass}_t = 250 g \times e ^{-2.59} = 250 g \times 0.0750 = 18.8 g
To determine the number of iodine-131 atoms, use the molar mass, 131 g _{}^{131}\text{I}/mol _{}^{131}\text{I}, and the Avogadro constant.
N_t = 18.8 g \text{ } _{}^{131}\text{I} \times \frac{1 mol \text{ } _{}^{131}\text{I} }{131 g \text{ } _{}^{131}\text{I} } \times \frac{6.022 \times 10^{23} \text{ } _{}^{131}\text{I} \text { atoms }}{1 mol \text{ } _{}^{131}\text{I} }
=8.64 \times 10^{22} \text{ } _{}^{131}\text{I} \text { atoms }
To establish the decay constant in s^{-1} and to obtain λ from t_{1/2}, use the method outlined in Example 25-3a.
\lambda = 0.693 / t_{1 / 2}=\frac{0.693}{8.04 d } \times \frac{1 d }{24 h } \times \frac{1 h }{60 min } \times \frac{1 min }{60 s }
=9.98 \times 10^{-7} s ^{-1}
To determine the decay rate (activity) after 30.0 days, use the two previous results in equation (25.11).
rate of decay ∝ N and rate of decay = A = λN (25.11)
A=\lambda N = 9.98 \times 10^{-7} s ^{-1} \times 8.64 \times 10^{22} \text { atoms }
=8.62 \times 10^{16} \text{ dis } s ^{-1}
Finally, to express the activity in curies use the definition of a curie given in Table 25.4.
A = 8.62 \times 10^{16} \text{ dis } s ^{-1} \times \frac{1 Ci }{3.70 \times 10^{10} \text{ dis } s ^{-1}}
= 2.33 \times 10^6 Ci
Assess
In this calculation we could also have shown that in 30 days (about 4 \times t_{1/2}) approximately 93% of the _{}^{131}\text{I} had disintegrated. In the process, approximately 2 \times 10^6 Ci of radiation was emitted. If we also knew the energy of the \beta^- particles and something about the exposed population, we could also have estimated radiation doses in rads.
| TABLE 25.4 Radiation {Units}^a | |
| Unit | Definition |
| Radioactive decay: | |
| Becquerel, Bq | s^{-1} (disintegrations per second) |
| Curie, Ci | An amount of radioactive material decaying at the same rate as 1 g of radium (3.70 \times 10^{10} \text{ dis } s^{-1}) 1 Ci = 3.70 \times 10^{10} \text{ Bq} |
| Absorbed dose: | |
| Gray, Gy | One gray of radiation deposits one joule of energy per kilogram of matter |
| Rad | 1 rad = 0.01 Gy |
| Equivalent dose: | |
| Sievert, Sv | 1 Sv = 100 rem |
| Rem | 1 rem = 1 rad × Q The quality factor, Q, is about 1 for X rays, γ rays, and \beta^- particles; 3 for slow neutrons; 10 for protons and fast neutrons; and 20 for α particles |
_{}^{a}\text{SI} units are shown in blue. Sources of α radiation are relatively harmless when external to the body and extremely hazardous when taken internally, as in the lungs or stomach. Other forms of radiation (X rays, γ rays), because they are highly penetrating, are hazardous even when external to the body.