World War II aircraft had instruments with glowing radium-painted dials (see figure ). The activity of one such instrument was \(1.0 \times {10^5}\,{\rm{Bq}}\) when new.

(a) What mass of \(^{{\rm{226}}}{\rm{Ra}}\) was present?

(b) After some years, the phosphors on the dials deteriorated chemically, but the radium did not escape. What is the activity of this instrument \({\rm{57}}{\rm{.0}}\)years after it was made?

For a given number of nuclei, the shorter the half-life, the more decays per unit time. As a result, activity R should be proportional to N, the number of radioactive nuclei, and inversely proportional to t1/2, their half-life. In reality, your instincts are spot on. It can be demonstrated that a source's activity is\(R = \frac{{0.693N}}{{{t_{1/2}}}}\).

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

Where\({t_{1/2}} = \)half-life ,\(R = \)activity and\(N = \)number of atoms.

And

\(\begin{array}{c}{t_{1/2}} = \left( {1.6 \times {{10}^3}\,{\rm{y}}} \right)\\ = 5.05 \times {10^{10}}\,{\rm{s}}\end{array}\)

Now, in the previous equation, enter in the values of\({\rm{R}}\)and\({t_{1/2}}\)and solve for the value of N:

\(\begin{array}{c}N = \frac{{\left( {1.0 \times {{10}^5}\,{\rm{Bq}}} \right)\left( {5.05 \times {{10}^{10}}\,{\rm{s}}} \right)}}{{0.693}}\\ = 7.28 \times {10^{15}}\\N = 7.28 \times {10^{15}}\end{array}\)

Now we must use the following formula to convert the number of atoms into mass:

\(m = N\left( {\frac{{1\;\,{\rm{mol}}}}{{6.02 \times {{10}^{23}}}}} \right)\left( {\frac{M}{{1\;\,{\rm{mol}}}}} \right)\)

Now plugin the value of N and\(226\;\,{\rm{g/mol}}\)for M and solve for the value of m

\(\begin{array}{c}m = \left( {7.28 \times {{10}^{15}}} \right)\left( {\frac{{1\;\,{\rm{mol}}}}{{6.02 \times {{10}^{23}}}}} \right)\left( {\frac{{226\,{\rm{g}}}}{{1\,{\rm{mol}}}}} \right)\\ = 2.73 \times {10^{ - 6}}\,{\rm{g}}\\ = 2.73\,{\rm{\mu g}}\\m = 2.73\,{\rm{\mu g}}\end{array}\)

Therefore, the mass is\(2.73\,{\rm{\mu g}}\).

b) The following is the relationship between beginning activity and activity after time t:

\(R = {R_0}{e^{ - \lambda t}}\)

In the above equation, plug in the values of\(\lambda = \frac{{0.693}}{{{t_{1/2}}}},t = 57\,{\rm{y}}\), and\({t_{1/2}} = 1.6 \times {10^6}\,{\rm{y}}\)and solve to find the value of\({\rm{R}}\).

\(\begin{array}{c}R = \left( {1 \times {{10}^5}} \right)\exp \left( { - \frac{{0.693 \times 57\,{\rm{y}}}}{{1.60 \times {{10}^3}\,{\rm{y}}}}} \right)\\ = 9.76 \times {10^4}\,{\rm{Bq}}\\R = 9.76 \times {10^4}\,{\rm{Bq}}\end{array}\)

Therefore, the activity of this instrument is \(9.76 \times {10^4}\,{\rm{Bq}}\).