/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 The isotope \({ }_{88}^{224}\) R... [FREE SOLUTION] | 91Ó°ÊÓ

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The isotope \({ }_{88}^{224}\) Ra of radium has a decay constant of \(2.19 \times 10^{-6} \mathrm{~s}^{-1}\). What is the halflife (in days) of this isotope?

Short Answer

Expert verified
The half-life of the isotope is approximately 3.66 days.

Step by step solution

01

Understanding decay constant and half-life relationship

The decay constant \( \lambda \) is related to the half-life \( t_{1/2} \) of a radioactive substance by the formula \( t_{1/2} = \frac{\ln(2)}{\lambda} \). This equation derives from the exponential decay law and the definition of half-life, which is the time taken for the substance to reduce to half its initial amount.
02

Calculate the half-life in seconds

Substitute the given decay constant \( \lambda = 2.19 \times 10^{-6} \mathrm{~s}^{-1} \) into the formula. \[t_{1/2} = \frac{\ln(2)}{2.19 \times 10^{-6}}\]Calculate \( \ln(2) \approx 0.693 \) and solve:\[t_{1/2} = \frac{0.693}{2.19 \times 10^{-6}} \approx 3.163 \times 10^5 \text{ seconds}\]
03

Convert seconds to days

Now convert the half-life from seconds to days. There are 86400 seconds in one day (60 seconds per minute, 60 minutes per hour, 24 hours per day).\[t_{1/2} = \frac{3.163 \times 10^5}{86400} \approx 3.66 \text{ days}\]
04

Final Calculation and Solution

The half-life of the isotope \({ }_{88}^{224}\) Ra is calculated to be approximately 3.66 days.

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Key Concepts

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

Half-life Calculation
A core concept in radioactive decay is calculating the half-life of an isotope, which is crucial for understanding the time it takes for half of a radioactive substance to transform. The half-life is symbolized as \( t_{1/2} \) and can be determined using the formula:
  • \( t_{1/2} = \frac{\ln(2)}{\lambda} \)
This formula links the half-life to the decay constant, \( \lambda \). \( \ln(2) \) is the natural logarithm of 2, approximately 0.693, which reflects the property of the exponential function describing the decay.
To calculate the half-life in seconds, you plug the decay constant given into the formula. For example, with a decay constant of \( 2.19 \times 10^{-6} \) s\(^{-1}\), inputting this value provides the half-life in seconds.
Finally, converting seconds to more manageable units like days helps illustrate how quickly or slowly a substance decays, particularly if it involves human timescales.
Decay Constant
The decay constant, represented as \( \lambda \), is a vital parameter in the mathematics of radioactive decay. It reveals the probability per unit time that a nucleus will decay. A higher decay constant means the isotope decays more quickly. The decay constant is measured in units of inverse time, such as s\(^{-1}\).
Here's how it's used:
  • Linked to half-life by the formula \( t_{1/2} = \frac{\ln(2)}{\lambda} \)
  • Provides insight into the stability of the isotope, with larger constants indicating rapid decay.
In practical applications, it helps predict how long it takes for a certain amount of radioactive material to lose its radioactivity. Understanding \( \lambda \) is essential, especially when dealing with nuclear processes or any field where radioactive materials are utilized.
Exponential Decay
Exponential decay is a mathematical concept used to describe processes where quantities decrease at a rate proportional to their current value. In the context of radioactivity, it is used to illustrate how radioactive substances reduce over time. This process can be expressed with the equation:
  • \( N(t) = N_0 e^{-\lambda t} \)
Here:
  • \( N(t) \) is the quantity of substance that still remains after time \( t \)
  • \( N_0 \) is the initial quantity
  • \( e \) is the base of the natural logarithm
  • \( \lambda \) is the decay constant
This equation shows how the number of undecayed nuclei decreases over time. The concept of half-life fits into this model as it represents the time needed for \( N(t) \) to reduce to \( N_0/2 \). Understanding exponential decay is crucial in fields such as nuclear physics and environmental science, where it predicts the behavior of unstable atoms and informs measures for safety and usage.

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Most popular questions from this chapter

A sample of ore containing radioactive strontium \({ }_{38}^{90} \mathrm{Sr}\) has an activity of \(6.0 \times 10^{5} \mathrm{~Bq}\) The atomic mass of strontium is \(89.908 \mathrm{u},\) and its half-life is 29.1 yr. How many grams of strontium are in the sample?

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?

Both gold \(\frac{198}{79} \mathrm{Au}\left(T_{1 / 2}=2.69\right.\) days \()\) and iodine \(\frac{131}{53} \mathrm{I}\left(T_{1 / 2}=8.04\right.\) days \()\) are used in diagnostic medicine related to the liver. At the time laboratory supplies are monitored, the activity of the gold is observed to be five times greater than the activity of the iodine. How many days later will the two activities be equal?

Find the energy released when lead \(\underset{82}^{211} \mathrm{~Pb}\) (atomic mass \(=210.988735 \mathrm{u}\) ) undergoes \(\beta^{-}\) decay to become bismuth \(\underset{83}{211} \mathrm{Bi}\) (atomic mass \(\left.=210.987255 \mathrm{u}\right)\).

Concept Questions (a) Physically, what does the binding energy of a nucleus represent? (b) What is the relationship between the binding energy of a nucleus and its mass defect? (c) How is the mass defect of a nucleus related to the mass of the intact nucleus and the masses of the individual nucleons? (d) A piece of metal, such as a coin, contains a certain number of atoms, and, hence, nuclei. How much energy would be required to break all the nuclei into their constituent protons and neutrons? Give your answer in terms of the number of atoms in the coin and the binding energy of each nucleus. (e) How is the number of atoms in a piece of metal related to its mass? Express your answer in terms of the atomic mass of the atoms and Avogadro's number. Problem A copper penny has a mass of \(3.0 \mathrm{~g}\). Determine the energy (in \(\mathrm{MeV}\) ) 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 \(\frac{63}{29} \mathrm{Cu}\) (atomic mass \(=62.939598 \mathrm{u}\) ).

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