Answer to Question #266691 in General Chemistry for Mikay

Question

Natural potassium contains K-40, which has a half-life of 1.277 x 109 years.

a. What mass of K-40 in a person would have a decay rate of 4140 Bq?

b. What is the fraction of K-40 in natural potassium, given that the person has 140 g in his body? (These numbers are typical for a 70-kg adult.)

Answer 1

$t_{1/2} = 1.277 \times 10^9 \\ = 1.277 \times 10^9 \times 365 \times 24 \times 60 \times 60 \\ = 4.0271 \times 10^{16} \\ R = 4140 \;Bq$

a. The expression to find the activity for a radioactive is,

$R = \frac{0.693N}{t_{1/2}} \\ N = \frac{R(T_{1/2})}{0.693} \\ N = \frac{4140 \times 4.0271 \times 10^{16}}{0.693} = 2.41 \times 10^{20}$

The conversion from mass to number of atoms is

$N = \frac{m}{M} \times 6.02 \times 10^{23} \\ M(K) = 40 \; g/mol \\ m = \frac{N \times M}{6.02 \times 10^{23}} \\ = \frac{2.41 \times {20} \times 40}{6.02 \times 10^{23}} \\ = 0.0160 \; g \\ = 0.16 \; mg$

b. Now, the expression to find the fraction of natural K-40 in a human body, if a person has 140 g.

$\%fraction = \frac{0.0160 \; g}{140 \; g} \times 100 \; \% = 1.14 \times 10^{-2} \\ = 0.0114 \; \%$