13.17 Radioactive isotopes decay according to first-order kinetics. For one particular isotope, 1.00 nmole registers $$ 1.2 imes 10^{6} $$ decays in 1.00 min. (a) How many decays will occur in 1.00 min if 5.00 nmoles of this isotope are present? (b) What fraction of the isotope decays per minute in each case? (c) Explain the relationship between your answers to (a) and (b).

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13.17 Radioactive isotopes decay according to first-order kinetics. For one particular isotope, 1.00 nmole registers $$ 1.2 	imes 10^{6} $$ decays in 1.00 min. (a) How many decays will occur in 1.00 min if 5.00 nmoles of this isotope are present? (b) What fraction of the isotope decays per minute in each case? (c) Explain the relationship between your answers to (a) and (b).

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Let the number of radioactive atoms be $$N$$. For first‐order kinetics the rate of decay is

$$
\frac{dN}{dt} = -kN,
$$

so the number of decays in a short time is proportional to the number of atoms present. In particular, if the number of decays during one minute is small compared to the total, then

$$
\text{Number of decays in 1 min} \approx kN.
$$

Because 1.00 nmol of the isotope gives $$1.2\times10^6$$ decays, we have

$$
kN_{\text{nmol}=1} \approx 1.2\times10^6.
$$

The total number of atoms in 1.00 nmol is

$$
N_{\text{nmol}=1} = (1.00 \times 10^{-9}\, \text{mol}) \times (6.02\times10^{23}\, \text{atoms/mol})
= 6.02\times10^{14}\, \text{atoms}.
$$

Thus, the rate constant in units of per minute is approximately

$$
k \approx \frac{1.2\times10^6}{6.02\times10^{14}}
\approx 2.0\times10^{-9}\, \text{min}^{-1}.
$$

Notice that the fraction of atoms that decay in one minute is

$$
f = k \quad (\text{if } k \ll 1).
$$

Thus, for 1.00 nmol, the fraction decaying per minute is

$$
f \approx 2.0\times10^{-9}.
$$

Now we address each part:

**(a)** If 5.00 nmoles are present, the total number of atoms is 5 times that in 1.00 nmol, and since the decay process is first order, the number of decays in one minute is simply 5 times as many. Thus,

$$
\text{Decays in 1 min} \approx 5.00 \times 1.2\times10^6 = 6.0\times10^6.
$$

**(b)** The fraction of the isotope that decays per minute is given by

$$
f = 1 - e^{-k} \approx k \quad \text{(since } k=2.0\times10^{-9} \ll 1\text{)}.
$$

Thus, in both cases (whether 1.00 or 5.00 nmoles) the fraction decaying per minute is approximately

$$
f \approx 2.0\times10^{-9}.
$$

**(c)** The relationship is that while the absolute number of decays in one minute scales with the amount of sample present, the fraction of the sample that decays per minute is determined solely by the rate constant $$k$$ and is independent of the number of atoms present. This is a hallmark of first‐order kinetics: the decay process is memoryless so that every atom has the same probability of decaying per unit time regardless of how many atoms are initially present.
**(a)** If there are 5.00 nmoles, approximately $$6.0 \times 10^6$$ decays occur in 1 minute. **(b)** About $$2.0 \times 10^{-9}$$ of the isotope decays per minute in both cases. **(c)** The number of decays increases with more isotope, but the fraction decaying per minute remains the same because the decay rate depends only on the rate constant, not on the total amount of isotope.

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