Question

(a) Calculate the activity \({\rm{R}}\) in curies of \(1.00\,{\rm{g}}\) of \({}^{{\rm{226}}}{\rm{Ra}}\). (b) Discuss why your answer is not exactly \(1.00\,{\rm{Ci}}\), given that the curie was originally supposed to be exactly the activity of a gram of radium.

Short answer

(a) The activity \({\rm{R}}\) is \(0.989\,{\rm{Ci}}\).

(b) We now know more about the half-life than we did when the unit was initially formed.

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 activity R

(a) The number of nuclei is computed using the formula \(N = \frac{m}{M}{N_A}\)

Where\({\rm{m}}\)is the mass of\({}^{{\rm{226}}}{\rm{Ra}}\),\({\rm{M}}\)is the molar mass of\({}^{{\rm{226}}}{\rm{Ra}}\), and\({{\rm{N}}_{\rm{A}}}\)is the Avogadro number, the number of atoms may be computed as follows,

\(\begin{array}{c}N = (1.00\,{\rm{g}})\left( {\frac{{{\rm{mol}}}}{{226\,{\rm{g}}}}} \right)\frac{{6.022 \times {{10}^{23}}\,{\rm{atoms}}}}{{{\rm{mol}}}}\\ = 2.6646 \times {10^{21}}\,{\rm{atoms}}\end{array}\)

By the equation\(R = \frac{{0.693}}{{{t_{1/2}}}}\), where we know that\({}^{{\rm{226}}}{\rm{Ra}}\)has a half-life of\(1.6 \times {10^3}\,{\rm{y}}\).

\(\begin{array}{c}R = \frac{{\left( {0.693} \right)\left( {2.6646 \times {{10}^{21}}} \right)}}{{1.6 \times {{10}^3}\,{\rm{y}}}} \times \frac{{1\,{\rm{y}}}}{{3.156 \times {{10}^7}\,{\rm{s}}}}\\ = 3.66 \times {10^{10}}\,{\rm{Bq}}\left( {\frac{{1\,{\rm{Ci}}}}{{3.70 \times {{10}^{10}}\,{\rm{Bq}}}}} \right)\\ = 0.989\,{\rm{Ci}}\end{array}\)

Therefore, the activity \({\rm{R}}\) is \(0.989\,{\rm{Ci}}\).

Explanation

(b) We now have a better understanding of the half-life of\({}^{{\rm{226}}}{\rm{Ra}}\) than we had when the\({\rm{Ci}}\) unit was first established.