Question

Show that the activity of the \({}^{{\rm{14}}}{\rm{C}}\) in \(1.00\,{\rm{g}}\) of \({}^{{\rm{12}}}{\rm{C}}\) found in living tissue is \(0.250\,{\rm{Bq}}\).

Short answer

The sample's activity is \(0.25\,{\rm{Bq}}\).

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.

Explanation

Carbon has a molar mass of\(12\,{\rm{g}}\). As a result, with\(1.00\,{\rm{g}}\)of carbon, the number of carbon atoms equals,

\(\begin{array}{c}{N_1} = \frac{{\left( {1.00\,{\rm{g}}} \right)}}{{\left( {12\,{\rm{g}}} \right)}}\left( {6.02 \times {{10}^{23}}} \right)\\ = 5.01 \times {10^{22}}\end{array}\)

The\({}^{{\rm{14}}}{\rm{C}}\)isotope occurs in nature at a rate of\({\rm{1}}{\rm{.3}}\)atoms per\({\rm{1}}{{\rm{0}}^{{\rm{12}}}}\)atoms. As a result, the total number of\({}^{{\rm{14}}}{\rm{C}}\)atoms in the sample are,

\(\begin{array}{c}N = (5.01 \times {10^{22}})(1.3 \times {10^{ - 12}})\\ = 6.52 \times {10^{10}}\end{array}\)

\({t_{1/2}} = 5730\,{\rm{y}} = 1.81 \times {10^{11}}\,{\rm{s}}\)is the half-life of the\({}^{{\rm{14}}}{\rm{C}}\)isotope. As a result, the activity is,

\(\begin{array}{c}R = \frac{{0.693N}}{{{t_{1/2}}}}\\ = \frac{{0.693(6.52 \times {{10}^{10}})}}{{(1.81 \times {{10}^{11}}\,{\rm{s}})}}\\ = 0.250\,{\rm{decay/s}}\end{array}\)

Since\({\rm{1Bq}}\)equals\({\rm{1}}\)decay per second, the sample's activity is\(0.25\,{\rm{Bq}}\).