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The half life of radioactive element is 160 days. The time for \(15 / 16\) disintegration is a. 280 days b. 640 days c. 560 days d. 2240 days

Short Answer

Expert verified
640 days.

Step by step solution

01

Understanding the Problem

We are tasked with finding the time it takes for a radioactive element to undergo \(\frac{15}{16}\) disintegration. This means we need to determine how long it takes for only \(\frac{1}{16}\) of the original sample to remain.
02

Applying the Half-Life Concept

The half-life of a radioactive element is the time required for half of the radioactive nuclei in a sample to decay. For this problem, the half-life is given as 160 days.
03

Determine the Ratio of Remaining Substance

Since we need \(\frac{15}{16}\) of the original amount to decay, this implies that \(\frac{1}{16}\) of the original amount is remaining. This helps us find how many half-lives are needed for this to occur.
04

Calculate the Number of Half-Lives Needed

The formula to determine the remaining amount is \((\frac{1}{2})^n = \frac{1}{16}\), where \(n\) is the number of half-lives. Solving this equation, we get \(n = \log_{\frac{1}{2}}{\frac{1}{16}}\).
05

Solving for \(n\)

Rewriting the equation, we have \( (\frac{1}{2})^n = (\frac{1}{2})^4 \). Equating exponents gives \(n = 4\). Thus, 4 half-lives are required for \(\frac{1}{16}\) of the radioactive element to remain.
06

Calculate the Total Time

Since each half-life is 160 days, and 4 half-lives are needed, the total time is \(4 \times 160 = 640\) days.
07

Conclusion

The time required for \(\frac{15}{16}\) of the radioactive element to disintegrate is **640 days**.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-Life Calculation
Radioactive decay can often be a complex concept. But with a fundamental understanding of half-life, we can simplify the problem. Half-life is the term used to describe the time it takes for half of the radioactive atoms in a sample to decay. In other words, it's the time it takes for a substance to reduce to 50% of its initial radioactivity.
For instance, if we have a radioactive element with a half-life of 160 days, this means that in 160 days, half of the original amount will have decayed. If we want to find out how much of the substance has decayed after multiple half-lives, we can use this pattern of reduction.
  • Every half-life reduces the substance to half of its previous amount.
  • For our example, after one half-life (160 days), only 50% remains.
  • After another half-life (320 days total), only 25% remains.
This process continues until we've reached the needed fraction. It's a powerful tool to calculate how long it takes for a certain fraction of the substance to decay.
Exponential Decay
Radioactive decay is a process that can be characterized as exponential decay. In this type of decay, quantities diminish at a rate proportional to their current value.
Let's take a deeper dive into how this applies to radioactive substances.
  • We can visualize exponential decay as a rapid decrease initially, which slows over time, never fully reaching zero.
  • The mathematical foundation for this relies on determining how many times the substance halves within a given period known as 'half-life.'
  • This results in the equation: \(A = A_0 \, (\frac{1}{2})^n\) where:
    • \(A\) is the remaining amount of substance,
    • \(A_0\) is the initial amount,
    • \(n\) is the number of half-lives elapsed.
Through exponential decay, we see how the process reflects a consistent, predictable pattern over time, allowing us to accurately predict remaining quantities and decay periods.
Logarithmic Equations
When working with exponential decay, especially involving half-lives, logarithmic equations become incredibly useful.
They enable us to solve for unknown quantities such as time or the number of half-lives.
  • The core idea is to express an equation in a form that allows us to utilize logarithms to resolve it.
  • For example, if we have the equation \( (\frac{1}{2})^n = \frac{1}{16} \), we can express this in logarithmic form:
  • \( n = \log_{(\frac{1}{2})}(\frac{1}{16}) \) reads as finding how many times we need to apply the base \(\frac{1}{2}\) for it to become \(\frac{1}{16}\).
  • This approach uncovers the exact number of half-lives, yielding \( n = 4 \) in our specific case.
Mastering logarithmic equations provides an efficient method for not only understanding decay patterns but also predicting future behaviors.

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Most popular questions from this chapter

Which of the following reaction represents \({ }_{22} \mathrm{Ti}^{44}\) decay by electron capture? a. \({ }_{22} \mathrm{~T}_{1}^{-44}+{ }_{-1} \mathrm{e}^{0} \rightarrow{ }_{21} \mathrm{Sc}^{44}\) b. \(\quad, T_{1}^{i 4}+, \mathrm{e}^{0} \rightarrow\) c. \(_{22} \mathrm{Ti}^{44}+{ }_{0} \mathrm{e}^{1} \rightarrow{ }_{22} \mathrm{Ti}^{45}\) d. \({ }_{22} \mathrm{Ti}^{44}+{ }_{0} \mathrm{e}^{-1} \rightarrow{ }_{22} \mathrm{Ti}^{43}\)

The radioisotope, tritium \(\left({ }_{3}^{1} \mathrm{H}\right)\) has a half life of \(12.3\) years. If the initial amount of tritium is \(32 \mathrm{mg}\), how many milligrams of it would remain after \(49.2\) years? a. \(4 \mathrm{mg}\) b. \(8 \mathrm{mg}\) c. \(1 \mathrm{mg}\) d. \(2 \mathrm{mg}\)

A positron is emitted from \(_{11} \mathrm{Na}^{23} .\) The ratio of the atomic mass and atomic number of the resulting nuclide is a. \(22 / 10\) b. \(22 / 11\) c. \(23 / 10\) d. \(23 / 12\)

Which of the following statement is/are incorrect? a. the end nuclide formed in thorium (4n) series is \({ }_{83} \mathrm{Bi}^{200}\) b. \(\mathrm{N}^{15}\) and \({ }_{8} \mathrm{O}^{16}\) are isobars. c. \({ }_{20} \mathrm{Ca}^{40}\) has magic number of protons and magic number of neutrons d. The radius \((\mathrm{R})\) of a nuclide of mass number \(\mathrm{A}\) is given by the equation \(\quad \mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 2}\) \(\left(\mathrm{R}_{0}=\right.\) constant \()\)

If uranium (mass number 238 and atomic number 92 ) emits an \(\alpha\)-particle, the product has mass number and atomic number a. 234 and 90 b. 236 and 92 c. 238 and 90 d. 236 and 90

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