91Ó°ÊÓ

The following is the relationship between activity, half-life, and the number of atoms:

\(R = \frac{{0.693N}}{{{t_{1/2}}}}\)

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

The relationship\({{\rm{M}}_{\rm{0}}}\)between initial and subsequent power.

After time t, power is supplied by:

\(\begin{aligned}P = {P_0}{e^{ - \lambda t}}\\{P_0} = P{e^{\lambda t}}\end{aligned}\)

\({\rm{Where }}{{\rm{P}}_{\rm{0}}}{\rm{ = initial powerP = final power \lambda = }}\frac{{{\rm{0}}{\rm{.693}}}}{{{{\rm{t}}_{{\rm{1/2}}}}}}{\rm{ and t = time Now plug in the required value and solve for }}{{\rm{P}}_{\rm{0}}}\)

\(\begin{aligned}{P_0} = \left( {12.48 \times {{10}^3}\;\,{\rm{W}}} \right){e^{\frac{{0.693(12\,{\rm{y}})}}{{87.7\,{\rm{y}}}}}}\\ = 13.72\,\;{\rm{kW}}\\{P_0} = 13.72\;\,{\rm{kW}}\end{aligned}\)

The initial power and initial activity are related using the following formula:

\({P_0} = \frac{{\Delta E}}{{\Delta t}} \Rightarrow {P_0} = (\Delta E)(R) \Rightarrow R = \frac{{{P_0}}}{{\Delta E}}\)

Fill in the missing value and solve for R.

\(\begin{aligned}R = \frac{{13.72\;\,{\rm{kW}}\left( {\frac{{{{10}^3}\,{\rm{J/s}}}}{{1\;\,{\rm{kW}}}}} \right)}}{{(5.59\,{\rm{MeV}})\left( {\frac{{1.6 \times {{10}^{ - 13}}\,{\rm{J}}}}{{1\,{\rm{MeV}}}}} \right)}}\\ = \frac{{\left( {13.72 \times {{10}^3}\,{\rm{J/s}}} \right)}}{{8.944 \times {{10}^{ - 13}}\,{\rm{J}}}}\\ = 1.534 \times {10^{16}}\,{\rm{counts /s}}\end{aligned}\)

As a result, \({\rm{N}}\)may be determined using the formula \(N = \frac{{R{t_{\frac{1}{2}}}}}{{0.693}}\).

So, here's what you'll need to do:

\(\begin{aligned}N = \frac{{\left( {1.534 \times {{10}^{16}}\,{\rm{Bq}}} \right)(87.7\,{\rm{y}})\left( {3.156 \times {{10}^7}\,{\rm{s}}} \right)}}{{0.693}}\\ = 6.13 \times {10^{25}}\,{\rm{atoms }}\\N = 6.13 \times {10^{25}}\,{\rm{atoms }}\end{aligned}\)

The formula for calculating the number of atoms is:

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

After that, plug in the appropriate value and solvefor \({\rm{m}}\).

\(\begin{aligned}m = \frac{{(238\;\,{\rm{g}})\left( {6.13 \times {{10}^{25}}\,{\rm{atoms }}} \right)}}{{6.022 \times {{10}^{23}}\,{\rm{atoms }}}}\\ = 24.23\;\,{\rm{kg}}\\ = 24230\,{\rm{g}}\\m = 24.23\,{\rm{kg}}\end{aligned}\)

Therefore, the initial mass source is \(m = 24.23\,{\rm{kg}}\).