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Chapter 13: Problem 11

Obtain approximately the ratio of the nuclear radii of the gold isotope \({ }_{79}^{197}\) Au and the silver isotope \({ }_{47}^{107} \mathrm{Ag}\).

Short Answer

Expert verified
The approximate ratio of the nuclear radii of gold to silver is 1.23.

Step by step solution

01

Understanding the Formula

The nuclear radius of an atom can be approximated using the formula \( R = R_0 A^{1/3} \), where \( R_0 \) is a constant approximately equal to 1.2 fm, and \( A \) is the mass number of the isotope.
02

Calculate the Radius of Gold

To find the radius of the gold isotope \({}_{79}^{197}\mathrm{Au}\), use the formula \( R_{Au} = R_0 imes 197^{1/3} \). This simplifies to \( R_{Au} = 1.2 imes 197^{1/3} \approx 1.2 imes 5.82 \approx 6.98 \) fm.
03

Calculate the Radius of Silver

Now, apply the radius formula to the silver isotope \({ }_{47}^{107} \mathrm{Ag}\) as \( R_{Ag} = R_0 imes 107^{1/3} \). This becomes \( R_{Ag} = 1.2 imes 107^{1/3} \approx 1.2 imes 4.74 \approx 5.69 \) fm.
04

Compute the Radius Ratio

To find the ratio of the nuclear radii, divide the radius of gold by the radius of silver: \( \frac{R_{Au}}{R_{Ag}} = \frac{6.98}{5.69} \approx 1.23 \).

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Key Concepts

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

Understanding Isotopes
Isotopes are atoms of the same element that have different numbers of neutrons. They share the same number of protons but vary in atomic mass. For example, in the exercise, we examined isotopes of gold and silver. Gold typically has an isotope with a mass number of 197, while silver has one with a mass number of 107. This difference in mass numbers is due to the differing numbers of neutrons.

Isotopes have significant applications and implications in various fields:
  • Medical Field: Isotopes are used in diagnostic imaging and radiation therapy for cancer treatment.
  • Carbon Dating: Radioactive isotopes like carbon-14 are used in determining the age of archaeological findings.
  • Nuclear Energy: Specific isotopes like Uranium-235 are used as fuel in nuclear reactors due to their ability to sustain nuclear fission.
Understanding isotopes helps in grasping the concept of atomic nuclei's differing properties, including their nuclear radii.
Exploring Mass Number
Mass number is a crucial concept in nuclear physics. It is defined as the sum of the number of protons and neutrons within an atom's nucleus. For isotopes, the mass number helps differentiate between their forms. For instance, the gold isotope mentioned in the problem set has a mass number of 197, consisting of 79 protons and 118 neutrons.

The mass number is essential for several reasons:
  • Nuclear Reactions: Knowing the mass number helps in predicting the outcomes of nuclear reactions.
  • Isotope Identification: Since isotopes are distinguished based on their mass numbers, identifying isotopes correctly is key in scientific research and applications.
  • Stability: The ratio of protons to neutrons, which relates to the mass number, influences the stability of the nucleus.
By understanding mass numbers, one can gain deeper insights into nuclear structure and the properties of various elements and isotopes.
Nuclear Physics Fundamentals
Nuclear physics is the branch of physics that studies the constituents and interactions of atomic nuclei. It involves understanding fundamental concepts such as nuclear forces, the behavior of neutrons and protons, and nuclear reactions.

In the context of the exercise, nuclear physics provides the basis for calculating nuclear radii:
  • Nuclear Radius Formula: The formula \( R = R_0 A^{1/3} \) allows scientists to estimate the size of a nucleus based on its mass number.
  • Binding Energy: This concept describes the energy required to separate a nucleus into its constituent protons and neutrons, enlightening how tightly bound a nucleus is.
  • Radioactivity: As a result of nuclear decay, elements like isotopes may transform into different elements, releasing energy in the process.
Nuclear physics not only helps understand the atomic structures but also enables practical applications in energy production, medical technology, and understanding cosmic phenomena.

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Most popular questions from this chapter

A \(1000 \mathrm{MW}\) fission reactor consumes half of its fuel in \(5.00 \mathrm{y}\). How much \({ }_{9}^{235} \mathrm{U}\) did it contain initially? Assume that the reactor operates \(80 \%\) of the time, that all the energy generated arises from the fission of \({ }_{92}^{25} \mathrm{U}\) and that this nuclide is consumed only by the fission process.

The radionuclide \({ }^{11} \mathrm{C}\) decays according to \({ }_{6}^{11} \mathrm{C} \rightarrow{ }_{5}^{11} \mathrm{~B}+e^{+}+v: \quad T_{1 / 2}=20.3 \mathrm{~min}\) The maximum energy of the emitted positron is \(0.960 \mathrm{MeV}\). Given the mass values: \(m\left({ }_{6}^{11} \mathrm{C}\right)=11.011434 \mathrm{u}\) and \(m\left({ }_{6}^{11} \mathrm{~B}\right)=11.009305 \mathrm{u}\) calculate \(Q\) and compare it with the maximum energy of the positron emitted.

In a periodic table the average atomic mass of magnesium is given as \(24.312 \mathrm{u} .\) The average value is based on their relative natural abundance on earth. The three isotopes and their masses are \({ }_{12}^{24} \mathrm{Mg}\) \((23.98504 \mathrm{u}),{ }_{12}^{25} \mathrm{Mg}(24.98584 \mathrm{u})\) and \({ }_{12}^{26} \mathrm{Mg}(25.98259 \mathrm{u})\). The naturalabundance of \({ }_{12}^{24} \mathrm{Mg}\) is \(78.99 \%\) by mass. Calculate the abundances of other two isotopes.

Find the \(\mathrm{g}\) -value and the kinetic energy of the emitted \(\alpha\) -particle in the \(\alpha\) -decay of (a) \(28 \mathrm{Ra}\) and (b) \({ }_{86}^{220} \mathrm{Rn}\). Given \(m\left({ }_{88}^{226} \mathrm{Ra}\right)=226.02540 \mathrm{u}\) \(m\left(\frac{22}{86} \mathrm{Rn}\right)=222.01750 \mathrm{u}\) \(m\left(_{86}^{22} \mathrm{Rn}\right)=220.01137 \mathrm{u}\) \(m\left({ }_{84}^{216} \mathrm{Po}\right)=216.00189 \mathrm{u}\)

Suppose, we think of fission of a \({ }_{26}^{56} \mathrm{Fe}\) nucleus into two equal fragments, \({ }_{13}^{28} \mathrm{Al} .\) Is the fission energetically possible? Argue by working out \(g\) of the process. Given \(m\left({ }_{26}^{56} \mathrm{Fe}\right)=55.93494 \mathrm{u}\) and \(m\left(\begin{array}{c}28 \\\ 13\end{array} \mathrm{Al}\right)=27.98191 \mathrm{u}\)
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