A bone sample containing strontium-90 \(\left(t_{1 / 2}=29 \mathrm{yr}\right)\) emits \(7.0 \times 10^{4} \beta^{-}\) particles per month. How long will it take for the emission to decrease to \(1.0 \times 10^{4}\) particles per month?

Understanding the concept of half-life is crucial in solving any radioactive decay problem. A half-life \(\t_{1/2}\) is the time required for half of the radioactive substance to decay. For instance, if you start with 100 particles, after one half-life, only 50 of those particles will remain. The half-life of strontium-90, as given in this problem, is 29 years. Knowing the half-life helps you determine how the substance decreases over time. This calculation informs us how long it would take for the initial amount to reduce by half. In this exercise, you will note that the problem implicitly requires us to keep dividing the sample's emissions by two until we reach the final value.

The radioactive decay formula is an essential mathematical tool for such problems. The decay formula is: \[ A = A_0 \times \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]. Here, \(A\) is the final activity, \(A_0\) is the initial activity, \(t\) is the time, and \(t_{1/2}\) is the half-life. The formula shows how the initial activity decreases over time based on the half-life. To apply this formula, you input the initial activity and the desired final activity, then solve for the time \(t\), which tells you how long it will take to decay from the initial to the final activity level. Remember to keep units consistent for meaningful results.

Logarithmic functions are vital when solving for time \(t\) in decay problems. After using the decay formula, you'll often end up with an expression where the unknown variable is in an exponent, such as: \[ (\frac{1}{2})^{\frac{t}{29}} = \frac{1}{7} \]. To solve for \(t\), you can take the natural logarithm (ln) of both sides. This process simplifies the equation, allowing you to isolate the variable. The logarithmic technique transforms exponential decay problems into linear ones. The process is as follows: \[ \text{ln}((\frac{1}{2})^{\frac{t}{29}}) = \text{ln}(\frac{1}{7}) \], which simplifies to \[ \frac{t}{29} \times \text{ln}(\frac{1}{2}) = \text{ln}(\frac{1}{7}) \]. This approach helps solve for \(t\) efficiently.

Initial and final activities are key to understanding decay problems. Initial activity \(A_0\) represents the rate of decay at the start. In our problem, it’s given as \(7.0 \times 10^4\) particles/month. This initial activity helps set up your decay formula. Final activity \(A\) is the decay rate you want to find out when it reaches a certain value, here \(1.0 \times 10^4\) particles/month. Knowing these values lets you directly apply them to the decay formula: \[ 1.0 \times 10^4 = 7.0 \times 10^4 \times \left(\frac{1}{2}\right)^{\frac{t}{29}} \]. By solving this equation, you can determine the time frame for the activity to decrease to the desired level. Initial and final activities provide the essential boundary conditions for your calculations.