Question 24.4: Calculating the Specific Activity and the Decay Constant of ......

Plan (a) The activity (𝒜) is the number of nuclei (𝒩) emitting a particle in a given time (t), and the specific activity is the activity per gram of sample. We are given 𝒜, which is equal to the number of particles emitted per minute (2.76×10^8  β^− particles), and we know the amount (mol) of sample. Because the sample is an isotope, we use the mass number in g/mol (122 g/mol) to find the mass (g). Converting 𝒜 (in d/min) to curies (in d/s) and dividing by the mass gives the specific activity (in Ci/g). (b) The decay constant (k) relates the activity to the number of nuclei (𝒩). We know 𝒜 (number of nuclei disintegrating/min) and the amount (in mol), so we use Avogadro’s number to find 𝒩 (number of nuclei) and apply Equation 24.2 to get k.

            𝒜  =  \frac{−Δ𝒩}{Δt}  =  k𝒩         (24.2)

Solution a. Finding the mass of the sample:
            \text{Mass (g) = }(2.6×10^{−12}\text{ mol}^{122}Sb)  \left( \frac{122  g  ^{122}Sb}{1\text{ mol }^{122}Sb} \right)
            =  3.2×10^{−10}  g  ^{122}Sb
Using the activity and the mass to find the specific activity:

b. Using the activity and the amount to find the decay constant (Equation 24.2):

           \text{Decay constant (k) = }\frac{𝒜\text{(nuclei/min)}}{𝒩\text{(nuclei)}}  =  \frac{2.76×10^8\text{nuclei/min}}{(2.6×10^{−12}\text{ mol})(6.022×10^{23} \text{ nuclei/mol})}
           =  1.8×10^{−4}  \min^{−1}

Check With so many exponents in the values, rounding is a good way to check the math. In part (b), for example,
           k  =  \frac{3×10^8\text{ nuclei/min}}{(3×10^{−12}\text{ mol})(6×10^{23}\text{ nuclei/mol})}  =  2×10^{−4}  \min^{−1}