Question

a) Natural potassium contains \({}^{{\rm{40}}}{\rm{K}}\), which has a half-life of \(1.277 \times {10^9}\,{\rm{y}}\). What mass of \({}^{{\rm{40}}}{\rm{K}}\) in a person would have a decay rate of \(4140\,{\rm{Bq}}\)? (b) What is the fraction of \({}^{{\rm{40}}}{\rm{K}}\) in natural potassium, given that the person has \({\rm{140g}}\) in his body? (These numbers are typical for a \({\rm{70}}\)-kg adult.)

Short answer

(a) The mass is\(16.0\,{\rm{mg}}\).

(b) The fraction is \(1.14 \times {10^{ - 4}}\).

Step-by-step solution

Define radioactivity

Radioactivity is a phenomenon in which a few substances spontaneously release energy and subatomic particles. The nuclear instability of an atom causes radioactivity.

Evaluating the mass

(a) The \({}^{{\rm{40}}}{\rm{K}}\) has a half-life of \({t_{1/2}} = 1.277 \times {10^9}\,{\rm{y}} = 4.03 \times {10^{16}}\,{\rm{s}}\).

The activity is performed by,

\(\begin{array}{c}R = \frac{{0.693N}}{{{t_{1/2}}}}\\N = \frac{{R{t_{1/2}}}}{{0.693}}\end{array}\)

Substitute all the value in the above equation

\(\begin{array}{c}N = \frac{{\left( {4410\,{\rm{Bq}}} \right)\left( {4.027 \times {{10}^{16}}\,{\rm{s}}} \right)}}{{0.693}}\\N = 2.405828 \times {10^{20}}\,{\rm{atoms}}\end{array}\)

\(M = 40\,{\rm{g}}\)is the molar mass of\({}^{{\rm{40}}}{\rm{K}}\). As a result, the mass that will produce the activity is,

\(m = \frac{N}{{{N_A}}}M\)

Substitute all the value in the above equation

\(\begin{array}{c}m = \frac{{2.405828 \times {{10}^{20}}\,{\rm{atoms}}}}{{6.02 \times {{10}^{23}}\,{\rm{atoms}}}}\left( {40\,{\rm{g}}} \right)\\ = 1.598 \times {10^{ - 2}}\,{\rm{g}}\\m \approx 16.0\,{\rm{mg}}\end{array}\)

Therefore, the mass is \(16.0\,{\rm{mg}}\).

Explanation

(b) As a result, the natural potassium fraction is, \(\begin{array}{c}f = \frac{{1.5985 \times {{10}^{ - 2}}\,{\rm{g}}}}{{140\,{\rm{g}}}}\\ = 1.14 \times {10^{ - 4}}\end{array}\)

Therefore, the fraction is \(1.14 \times {10^{ - 4}}\).