Question

There is more than one isotope of natural uranium. If a researcher isolates \(1.00\,{\rm{mg}}\) of the relatively scarce\({}^{{\rm{235}}}{\rm{U}}\) and finds this mass to have an activity of \(80.0\,{\rm{Bq}}\), what is its half-life in years?

Short answer

The half-life is \(7.04 \times {10^8}\,{\rm{y}}\).

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 half life

The activity is performed by,

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

\(R = 80.0\,{\rm{Bq}}\)is the given activity of the\({}^{{\rm{235}}}{\rm{U}}\)sample.

The provided sample has a mass of\(m = 1.00\,{\rm{mg}} = 1.00 \times {10^{ - 3}}\,{\rm{g}}\).

\(M = 235\,{\rm{g}}\)is the molar mass of the element\({}^{{\rm{235}}}{\rm{U}}\). As a result, the total number of atoms in the sample is,

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

Substitute all the value in the above equation

\(\begin{array}{c}N = \frac{{(1.00 \times {{10}^{ - 3}}\,{\rm{g}})}}{{(235\,{\rm{g}})}}\left( {6.02 \times {{10}^{23}}\,{\rm{atoms}}} \right)\\N = 2.56 \times {10^{18}}\,{\rm{atoms}}\end{array}\)

As a result, the half-life is equal to,

\(\begin{array}{c}{t_{1/2}} = \frac{{0.693(2.56 \times {{10}^{18}}\,{\rm{atoms}})}}{{\left( {80\,{\rm{Bq}}} \right)}}\\ = 2.22 \times {10^{16}}\,{\rm{s}}\\ = 7.04 \times {10^8}\,{\rm{y}}\end{array}\)

Therefore, the half-life is \(7.04 \times {10^8}\,{\rm{y}}\).