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Chapter 31: Problem 32

In 9.0 days the number of radioactive nuclei decreases to one-eighth the number present initially. What is the half-life (in days) of the material?

Short Answer

Expert verified
The half-life of the material is 3 days.

Step by step solution

01

Understanding the Problem

We are given that in 9.0 days, the number of radioactive nuclei decreases to one-eighth of its initial amount. We need to find the half-life of this material. The half-life is the time it takes for half of the radioactive nuclei to decay.
02

Exponential Decay Formula

Radioactive decay can be expressed through the formula: \[ N(t) = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}} \]where \(N(t)\) is the number of nuclei at time \(t\), \(N_0\) is the initial number of nuclei, and \(T_{1/2}\) is the half-life. In this case, \(N(t) = \frac{1}{8}N_0\) at \(t = 9\) days.
03

Plug In the Values

Substitute the known values into the formula:\[ \frac{1}{8}N_0 = N_0 \left( \frac{1}{2} \right)^{9/T_{1/2}} \]Divide both sides by \(N_0\) to simplify:\[ \frac{1}{8} = \left( \frac{1}{2} \right)^{9/T_{1/2}} \]
04

Solve for Half-Life

Take the logarithm of both sides to solve for \(T_{1/2}\):\[ \log\left( \frac{1}{8} \right) = \frac{9}{T_{1/2}} \log\left( \frac{1}{2} \right) \]Since \(\log\left( \frac{1}{8} \right) = -3\log(2)\) and \(\log\left( \frac{1}{2} \right) = -\log(2)\), we have:\[ -3\log(2) = \frac{9}{T_{1/2}} (-\log(2)) \]Cancel out the \(-\log(2)\) on both sides:\[ 3 = \frac{9}{T_{1/2}} \]Solve for \(T_{1/2}\):\[ T_{1/2} = \frac{9}{3} = 3 \text{ days} \]
05

Verification

Verify the calculation by checking if after 9 days the material should indeed reach \(\frac{1}{8}\) of the initial amount using the calculated half-life of 3 days. Using 3-day cycles, the material will halve to \(0.5\) by 3 days, \(0.25\) by 6 days, and \(0.125\) by 9 days, confirming the solution.

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Key Concepts

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

Radioactive Decay
Radioactive decay is a natural process where unstable atomic nuclei lose energy by emitting radiation. This process leads to the transformation of a radioactive element into a more stable one, often changing the atomic structure. Imagine it like a slow, constant release of energy from radioactive materials:
  • Not all nuclei decay at the same speed; different materials have different rates.
  • The rate of decay is random for each individual atom, but predictable for a large collection.
  • With decay, the substance changes into another element or isotope over time.
Understanding this process is vital in fields like archaeology for carbon dating and in medicine for cancer treatments. Core to this is knowing how quickly a given radioactive material reduces to half its original amount, known as the half-life.
Exponential Decay Formula
The exponential decay formula helps us model the way quantities decrease over time, particularly in processes like radioactive decay. The formula is often given as:\[ N(t) = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}} \]Here:
  • \(N(t)\) represents the quantity remaining after time \(t\).
  • \(N_0\) is the initial quantity of the substance.
  • \(T_{1/2}\) symbolizes the half-life of the substance.
In scenarios involving radioactive decay, this formula captures how a substance gradually decreases in quantity over evenly spaced intervals (half-lives). This formula shows that each half-life marks a reduction to half of the previous amount. This predictable pattern makes it easy to model the future quantity of radioactive materials, essential in planning medical doses and dating ancient objects.
Radioactive Nuclei
Radioactive nuclei are the core of atoms that are unstable and undergo decay. Their instability lies in the combination of protons and neutrons, which create an imbalance of forces:
  • Over time, these nuclei emit particles to reach a more stable state.
  • This emission is what we identify as radioactive decay.
  • During decay, the nucleus can convert into a completely different element.
For example, uranium-238 decays into thorium while releasing alpha particles. Understanding radioactive nuclei helps scientists predict how long a material will remain active or dangerous. This knowledge is not only crucial for scientific research but also for practical applications, such as nuclear energy, where controlling the stability of materials is key.
Logarithms in Physics
Logarithms are a powerful mathematical tool used to simplify calculations involving exponential growth and decay, like those in radioactive processes. When dealing with exponential equations, such as the decay formula, logarithms become invaluable:
  • They transform multiplicative processes into additive ones, simplifying calculations.
  • In the context of decay, logarithms help solve for unknowns, such as half-life or time.
  • Using logarithms, we can derive values from the decay formula by taking the log of both sides.
In physics, and particularly in nuclear science, logarithms facilitate the analysis of growth and decay patterns by reducing complex multiplicative relationships into straightforward calculations. This simplification is crucial when working with exponential decay scenarios, making it possible to quickly and accurately solve for unknown variables, as seen in the calculation of half-lives.

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Most popular questions from this chapter

The largest stable nucleus has a nucleon number of \(209,\) and the smallest has a nucleon number of 1. If each nucleus is assumed to be a sphere, what is the ratio (largest/smallest) of the surface areas of these spheres?

A copper penny has a mass of \(3.0 \mathrm{g}\). Determine the energy \((\mathrm{in}\) MeV) that would be required to break all the copper nuclei into their constituent protons and neutrons. Ignore the energy that binds the electrons to the nucleus and the energy that binds one atom to another in the structure of the metal. For simplicity, assume that all the copper nuclei are \({ }_{29}^{63} \mathrm{Cu}\) (atomic mass \(=62.939598 \mathrm{u})\).

When a sample from a meteorite is analyzed, it is determined that \(93.8 \%\) of the original mass of a certain radioactive isotope is still present. Based on this finding, the age of the meteorite is calculated to be \(4.51 \times 10^{9} \mathrm{yr}\). What is the half-life (in yr) of the isotope used to date the meteorite?

An unknown nucleus contains 70 neutrons and has twice the volume of the nickel \({ }_{28}^{60} \mathrm{Ni}\) nucleus. Identify the unknown nucleus in the $$ \text { form } \frac{A}{Z} X $$. Use the periodic table on the inside of the back cover as needed.

(a) Energy is required to separate a nucleus into its constituent nucleons, as Interactive Figure 31.3 indicates; this energy is the total binding energy of the nucleus. In a similar way one can speak of the energy that binds a single nucleon to the remainder of the nucleus. For example, separating nitrogen \({ }^{14}{ }_{7} \mathrm{N}\) into nitrogen \({ }^{13}{ }_{7} \mathrm{N}\) and a neutron takes energy equal to the binding energy of the neutron, as shown below: Find the energy (in MeV) that binds the neutron to the \({ }^{14}, \mathrm{N}\) nucleus by considering the mass of \({ }_{7}^{13} \mathrm{N}\) (atomic mass \(=13.005738 \mathrm{u}\) ) and the mass of \({ }_{0}^{1}\) n (atomic mass \(=1.008665 \mathrm{u}\) ), as compared to the mass of \({ }_{7}^{14} \mathrm{N}\) (atomic mass \(=\) \(14.003074 \mathrm{u}) .\) (b) Similarly, one can speak of the energy that binds a single proton to the \({ }^{14}{ }_{7} \mathrm{N}\) nucleus: Following the procedure outlined in part (a), determine the energy (in MeV) that binds the proton (atomic mass \(=1.007825 \mathrm{u}\) ) to the \({ }_{7}^{14} \mathrm{N}\) nucleus. The atomic mass of carbon \({ }_{6}^{13} \mathrm{C}\) is \(13.003355 \mathrm{u}\). (c) Which nucleon is more tightly bound, the neutron or the proton?
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