A \({}^{{\rm{60}}}{\rm{Co}}\) source is labelled \(4.00\,{\rm{mCi}}\), but its present activity is found to be \(1.85 \times {10^7}\,{\rm{Bq}}\). (a) What is the present activity in \({\rm{mCi}}\)? (b) How long ago did it actually have a \({\rm{4}}{\rm{.00mCi}}\) activity?

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

(a) As \(1\,{\rm{mCi}}\) equals \(3.70 \times {10^7}\,{\rm{Bq}}\), the source's current activity is,

\(\begin{array}{c}{R_f} = 1.85 \times {10^7}\,{\rm{Bq}}\\ = \frac{{(1.85 \times {{10}^7}\,{\rm{Bq}})}}{{(3.70 \times {{10}^7}\,{\rm{Bq/mCi}})}}\\ = 0.500\,{\rm{mCi}}\end{array}\)

Therefore, the source's current activity is\(0.500\,{\rm{mCi}}\).

(b) A radioactive substance's activity is determined by,

\(R = \frac{{In(2)N}}{{{t_{1/2}}}}\)

As a result, the first activity is,

\(\begin{array}{c}{R_i} & = \frac{{In(2){N_i}}}{{{t_{1/2}}}}\\ & = 4.0\,{\rm{mCi}}\end{array}\)

the last activity is,

\(\begin{array}{c}{R_f} & = \frac{{In(2){N_f}}}{{{t_{1/2}}}}\\ & = 0.500\,{\rm{mCi}}\end{array}\)

To get the answer, divide equations,

\(\begin{array}{c}\frac{{{R_f}}}{{{R_i}}} & = \frac{{{N_f}}}{{{N_i}}}\\ & = \frac{{0.500\,{\rm{mCi}}}}{{4.0\,{\rm{mCi}}}}\\ & = \frac{1}{8}\\ & = {\left( {\frac{1}{2}} \right)^3}\end{array}\)

As a result, the time spent is,

\(t = 3{t_{1/2}}\)

Now that\({}^{{\rm{60}}}{\rm{Co}}\)has a half-life of\({t_{1/2}} = 5.27\,{\rm{y}}\), we may say that time is,

\(\begin{array}{c}t &= 3\left( {5.27\,{\rm{y}}} \right)\\ & = 15.8\,{\rm{y}}\end{array}\)

Therefore, the activity was \(4.00\,{\rm{mCi}}\),\(15.8\,{\rm{y}}\) ago.