Polonium-210 is an alpha emitter and has a half-life of 138 days. (a) Write the equation for the radioactive decay of polonium-210. (b) How long will it take before only \(5.00 \%\) of the original amount of \({ }^{210}\) Po in a sample remains?

The radioactive decay of polonium-210 (symbol: Po, atomic number: 84) can be represented as: \[ {}^{210}\text{Po} \rightarrow {}^{206}\text{Pb} + {}^{4}\text{He} + \gamma \] Here, polonium-210 (Po) decays to lead-206 (Pb), releasing an alpha particle (He) and gamma radiation (γ).

To determine how long it will take for 5% of the original polonium-210 to remain, first, we need to calculate the decay constant (λ). The half-life (T) of polonium-210 is given as 138 days. The decay constant can be determined using the equation: \[ T = \frac{\ln{2}}{\lambda} \] Rearrange the equation to solve for λ: \[ \lambda = \frac{\ln{2}}{T} \] Now, we can plug in the half-life of polonium-210 (T = 138 days) and calculate λ: \[ \lambda = \frac{\ln{2}}{138} \approx 0.00502 \: \text{days}^{-1} \]

To find how long it takes for only 5% of the original amount of polonium-210 to remain, we can use the exponential decay model: \[ N(t) = N_{0}e^{-\lambda t} \] Where: \(N(t)\) is the amount of polonium-210 remaining after time t, \(N_{0}\) is the initial amount of polonium-210, and \(e\) is the base of the natural logarithm (approximately equal to 2.718). First, we will express the percentage of remaining polonium-210 in terms of the initial amount: \[ N(t) = 0.05N_{0} \] Next, we will plug this expression into our decay equation, isolating the variable t: \[ 0.05N_{0} = N_{0}e^{-\lambda t} \] Now, we can divide both sides by \(N_{0}\): \[ 0.05 = e^{-\lambda t} \] Take the natural logarithm of both sides: \[ \ln{0.05} = -\lambda t \] Finally, isolate t and plug in the value of λ we calculated earlier: \[ t = \frac{\ln{0.05}}{-0.00502 \: \text{days}^{-1}} \approx 598.5 \: \text{days} \] So, it takes approximately 598.5 days for only 5% of the original amount of polonium-210 in a sample to remain.