The \({{\rm{\beta }}^{\rm{ - }}}\) particles emitted in the decay of \(^{\rm{3}}{\rm{H}}\) (tritium) interact with matter to create light in a glow-in-the-dark exit sign. At the time of manufacture, such a sign contains \(15.0\,{\rm{Ci}}\) of \(^{\rm{3}}{\rm{H}}\).

a) What is the mass of the tritium?

b) What is its activity \(5.00\,{\rm{y}}\) after manufacture?

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

Given a radioactive sample's beginning activity,\({{\rm{R}}_{\rm{0}}}\), at an initial time\({t_0} = 0\), the activity\({\rm{R}}\)is given by,

\(R = {R_0}{e^{ - \lambda t}}\) …. (1)

where\({\rm{t}}\)represents time,\(\lambda = \frac{{0.693}}{{{t_{1/2}}}}\)represents the decay constant, and\({{\rm{t}}_{{\rm{1/2}}}}\)represents the half-life. The activity for a sample of N nuclei is,

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

We solve for the number of nuclei N as a function of activity using this equation.

\(N = \frac{{{t_{1/2}}R}}{{0.693}}\) … (3)

\({t_{1/2}} = 12.33\)years are \( = 388.84 \times {10^6}\,{\rm{s}}\)for tritium. In the equation (3), substituting

\(\begin{array}{}N &= \frac{{\left( {388.84 \times {{10}^6}\,\;{\rm{s}}} \right)\left( {(15.0)3.70 \times {{10}^{10}}\,{{\rm{s}}^{{\rm{ - 1}}}}} \right)}}{{0.693}}\\ & = 311.41 \times {10^{18}}\end{array}\) ….. (4)

\(1\,{\rm{Ci}} = 3.70 \times {10^{10}}\)decays per second. The number of nuclei in a given mass is equal to.

\(N = \frac{{{N_A}\left( {\;{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}} \right)}}{{A(\;{\rm{g/mol}})}}m(\;{\rm{g}})\) …..(5)

\(\begin{array}{c}m &= \frac{{A(\;{\rm{g/mol}})}}{{{N_A}\left( {\;{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}} \right)}}N\\ &= \left( {\frac{{3\;\,{\rm{g/mol}}}}{{6.02 \times {{10}^{23}}\,{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}}}} \right)311.41 \times {10^{18}}\\ &= 1.55 \times {10^{ - 3}}\,{\rm{g}}.{\rm{ }}\\m &= 1.55 \times {10^{ - 3}}\,{\rm{g}}\end{array}\) …(6)

Therefore, the mass of the tritium is\(m = 1.55 \times {10^{ - 3}}\,{\rm{g}}\)

b) Using the equation (1),

\(\begin{array}{c}R &= 15.0\,{\rm{Ci}} \times {e^{ - \frac{{0.693}}{{12.93\,{\rm{y}}}}5.00\,{\rm{y}}}}\\ &= 11.32\,{\rm{Ci }}\\R &= 11.32\,{\rm{Ci}}\end{array}\) …. (7)

Therefore, after 5 years, the activity is \(R = 11.32\,{\rm{Ci}}\).