Question

Graph A sample of polonium-214 originally has a mass of 1.0 g. Express the mass remaining as a percent of the original sample after a period of one, two, and three half-lives. Graph the percent remaining versus the number of half- lives. Approximately how much time has elapsed when 20% of the original sample remain?

Short answer

The mass of polonium-214 remaining after one, two, and three half-lives is 0.5 g (50%), 0.25 g (25%), and 0.125 g (12.5%), respectively. Graphing the percentage remaining against the number of half-lives yields a decreasing exponential function. Approximately 2.32 half-lives have elapsed when 20% of the original sample remains. If the half-life of polonium-214 is 163 microseconds, then approximately 378 microseconds have elapsed when 20% of the original sample remains.

Step-by-step solution

Calculate the mass remaining after each half-life

To calculate the mass remaining after each half-life, we can use the formula: \(Mass \, remaining = Initial \, mass \, \times \, (\frac{1}{2})^n\) where n is the number of half-lives. Since the initial mass is 1.0 g, we can plug in the values to find the mass remaining after each half-life: - After one half-life, n = 1: \(Mass \, remaining = 1.0 \, \times \, (\frac{1}{2})^1 = 1.0 \, \times \, \frac{1}{2} = 0.5 \, g\) - After two half-lives, n = 2: \(Mass \, remaining = 1.0 \, \times \, (\frac{1}{2})^2 = 1.0 \, \times \, \frac{1}{4} = 0.25 \, g\) - After three half-lives, n = 3: \(Mass \, remaining = 1.0 \, \times \, (\frac{1}{2})^3 = 1.0 \, \times \, \frac{1}{8} = 0.125 \, g\)

Calculate the percentage remaining after each half-life

Now, we can calculate the percentage of the initial mass remaining after each half-life using the formula: \(Percentage \, remaining = \frac{Mass \, remaining}{Initial \, mass} \times 100\%\) - After one half-life: \(Percentage \, remaining = \frac{0.5 \, g}{1.0 \, g} \times 100\% = 50\%\) - After two half-lives: \(Percentage \, remaining = \frac{0.25 \, g}{1.0 \, g} \times 100\% = 25\%\) - After three half-lives: \(Percentage \, remaining = \frac{0.125 \, g}{1.0 \, g} \times 100\% = 12.5\%\)

Graph the percentage remaining versus the number of half-lives

Now, we can graph the percentage remaining against the number of half-lives using the following data points: - (0, 100%): At the beginning (0 half-lives), the entire sample (100%) is present - (1, 50%): After one half-life, 50% of the sample remains - (2, 25%): After two half-lives, 25% of the sample remains - (3, 12.5%): After three half-lives, 12.5% of the sample remains The graph is a decreasing exponential function.

Find the approximate time when 20% of the original sample remains

To find the approximate time when 20% of the original sample remains, we can use the half-life formula and solve for n: \(Percentage \, remaining = \frac{Mass \, remaining}{Initial \, mass} \times 100\%\) \(20\% = \frac{Mass \, remaining}{1.0 \, g} \times 100\% = (\frac{1}{2})^n\) Divide by 100% to find \(0.2 = (\frac{1}{2})^n\). Now, take the logarithm base 0.5 of both sides to solve for n: \(n = log_{0.5}(0.2)\) \(n \approx 2.32\) This means that approximately 2.32 half-lives have elapsed when 20% of the original sample remains. To find the elapsed time, we need to know the half-life of polonium-214, which is 163 microseconds. Therefore, Elapsed time ≈ 2.32 half-lives × 163 microseconds/half-life ≈ 378 microseconds. So, approximately 378 microseconds have elapsed when 20% of the original sample of polonium-214 remains.

Key concepts

Understanding Radioactive Decay

Radioactive decay is a fascinating natural process observed in unstable atomic nuclei. This process involves the spontaneous transformation of an unstable atomic nucleus into a lighter nucleus. What's enthralling about this process is that it emits radiation in the form of particles or electromagnetic waves.

Think of it like a ticking time bomb on a molecular level—except instead of an explosion, the atom releases energy to become more stable. This stability is achieved through ALPHA PARTICLES, BETA PARTICLES, or GAMMA RAYS' emission depending on the decay process.

Each radioactive element has a characteristic 'half-life', a term that brings us to the crux of the process. The half-life is the time required for half of the radioactive substance to undergo decay. It's like having 100 balls, and every minute half of them turn blue—the half-life is how long you wait for 50 of them to change color. This does not mean all elements take the same amount of time; each has its unique half-life that can range from microseconds to billions of years.

When solving problems involving radioactive decay, it's crucial to grasp two things: the concept of half-life itself and the mathematical relationship that represents the decay—this is where exponential functions come into play.

Exponential Functions and Decay

Trying to describe radioactive decay without exponential functions would be like trying to describe a spiral staircase without steps—they're the mathematical building blocks that make it all make sense. In the simplest terms, an exponential function is a mathematical expression where the variable in question is the exponent.

If we're dealing with a half-life problem, the function has the form y = a(1/2)^x where 'a' is the initial amount, '1/2' is the decay factor (since half-life is involved), and 'x' is the number of half-lives that have passed. Every time you increment 'x', the amount 'y' is cut in half.

Let's say you started with 1.0 g of a radioactive isotope. After one half-life, you'd be left with 0.5 g. After two half-lives, it'd be 0.25 g, and so on. The fascinating aspect here is that regardless of how much you start with, the process of cutting it in half behaves the same—a consistent, predictable decay over time. Understanding this function is pivotal when trying to predict the remaining mass of a substance after a certain duration or number of half-lives.

Applying Logarithms to Radioactive Decay

When you want to find unknowns in the realm of exponential decay, logarithms come to your rescue like a mathematical superhero. They are essentially the inverses of exponential functions and help us unravel the exponent when the base and result are known.

Considering our previous radiation problem, we used a logarithm to find out how many half-lives had passed to leave us with a certain percentage of the material. It is like working backward: given the percentage standing in for 'y', and knowing the initial amount and base of the decay rate, we utilize logs to solve for 'x'.

As per our calculation steps, you can see how this plays out: with 20% of the substance remaining, we did some algebraic manipulation to isolate \( (1/2)^n \) and then applied the logarithm to extricate 'n', the variable which represented the number of half-lives that had lapsed. Logarithms might seem daunting at first, but with practice, they become an essential and handy tool in your math arsenal, especially when dealing with problems in radioactive decay and other practices involving exponential growth or decay.