Problem 80

Question

The half-life period of radium is 1580 years. It remains \(1 / 16\) after how many years? (a) 1580 years (b) 3160 years (c) 4740 years (d) 6320 years

Step-by-Step Solution

Verified
Answer
6320 years (option d).
1Step 1: Understand the Concept
The problem is about the half-life of a radioactive substance, radium. Half-life is the time required for a quantity to reduce to half its initial amount. We are given the half-life of radium and asked to find out how long it will take for radium to reduce to \(\frac{1}{16}\) of its original amount.
2Step 2: Determine Exponent Required for Reduction
Since the substance reduces to half in each half-life period, we need to determine how many half-lives \( n \) are required to reach \(\frac{1}{16}\). Each half-life divides the substance amount by 2, so:\[\left( \frac{1}{2} \right)^n = \frac{1}{16}.\]
3Step 3: Solve for the Exponent
Recognize that \(\frac{1}{16} = \frac{1}{2^4}\), which means:\[\left( \frac{1}{2} \right)^n = \left( \frac{1}{2} \right)^4.\]Thus, \( n = 4 \). This tells us that radium needs to go through 4 half-life periods to become \(\frac{1}{16}\) of the initial amount.
4Step 4: Calculate Total Time
Multiply the number of half-life periods by the duration of one half-life period. The half-life of radium is 1580 years:\[4 \times 1580 = 6320 \text{ years}.\]
5Step 5: Select the Correct Answer
Compare the calculated time with the options provided: (a) 1580 years (b) 3160 years (c) 4740 years (d) 6320 years The correct choice is (d) 6320 years.

Key Concepts

Understanding Half-LifeRadium Decay in ContextExponential Decay Explained
Understanding Half-Life
In the realm of radioactive decay, half-life is a central concept. Half-life refers to the time taken for a substance to reduce to half of its initial amount. When we talk about a radioactive element, it doesn't decay completely in one go. Instead, it gradually loses its effectiveness by half every half-life period.
This means after one half-life,
  • the quantity becomes half of what it was initially.
  • After two half-lives, it is one-quarter of the starting amount.
This cycle continues. For example, if radium has a half-life of 1580 years, then in another 1580 years, only half will remain. Understanding this pattern is crucial to solving related problems.
To calculate how much would be left after several half-lives, we use the formula:\[ \left( \frac{1}{2} \right)^n \]where \( n \) is the number of half-lives.
Radium Decay in Context
When we discuss radium decay, we're delving into how this specific radioactive element decreases over time. Radium, known for its historical uses in luminous paints but now understood for its dangers, provides a textbook example of half-life concepts. It boasts a notably long half-life of 1580 years.
This characteristic decay pattern gives insights into the timescale of radioactive decay processes.
For instance:
  • After 1580 years, 50% of radium remains.
  • After another 1580 years (total 3160 years), only 25% (\(\frac{1}{4}\)) remains.
  • This continues until, after 6320 years, just \(\frac{1}{16}\) of the original sample is still there.
These calculations help scientists and researchers estimate how long radium, and similarly radioactive substances, will continue to emit radiation. Understanding this helps in fields like archaeology, environmental science, and nuclear medicine.
Exponential Decay Explained
Radioactive decay follows an exponential decay model, a faster reduction in the beginning that slows down over time. The mathematical basis is the same for various processes that diminish rapidly at first and continue reducing at a progressively slower rate.
This concept applies beyond radioactivity, too.
The core equation for exponential decay is:\[ y(t) = y_0 \left( \frac{1}{2} \right)^{t/T} \]where:
  • \( y(t) \) is the remaining quantity at time \( t \).
  • \( y_0 \) is the initial amount.
  • \( T \) represents the half-life.
This formula shows how quickly the amount of substance decreases over time. Using it, you can see why radium reduces to just \(\frac{1}{16}\) of its original amount over four half-life periods.