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What is the likely mass number of a spherical nucleus with a radius of \(3.6 \mathrm{fm}\) as measured by electron-scattering methods?

Short Answer

Expert verified
The likely mass number is 27.

Step by step solution

01

Understanding the Formula for Nuclear Radius

The radius of a nucleus is given by the formula \( R = R_0 A^{1/3} \), where \( R \) is the radius, \( R_0 \) is a constant approximately equal to 1.2 fm, and \( A \) is the mass number we are looking for.
02

Solving for the Mass Number

We need to solve the equation \( R = R_0 A^{1/3} \) for \( A \). Substitute \( R = 3.6 \mathrm{fm} \) and \( R_0 = 1.2 \mathrm{fm} \) into the equation, giving us \( 3.6 = 1.2 A^{1/3} \).
03

Isolating \( A^{1/3} \)

Divide both sides of the equation \( 3.6 = 1.2 A^{1/3} \) by 1.2 to isolate \( A^{1/3} \), resulting in \( A^{1/3} = 3 \).
04

Solving for \( A \)

Cube both sides to solve for \( A \): \( A = 3^3 \). Therefore, \( A = 27 \).

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Key Concepts

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

Nuclear Radius
The concept of nuclear radius is fundamental in nuclear physics. It is related to how large the nucleus of an atom is. The nucleus is incredibly small compared to the whole atom, but it's significant because it contains nearly all the atom's mass. The commonly used formula to calculate the nuclear radius is given by \( R = R_0 A^{1/3} \). Here, \( R \) represents the nuclear radius, \( R_0 \) is a constant, often taken to be around 1.2 femtometers (fm), and \( A \) is the mass number of the atom, which is the number of protons and neutrons in the nucleus.
This formula shows that the nuclear radius increases with the cube root of the mass number. Some key aspects of nuclear radius are:
  • It helps us understand the size of the nucleus and, by extension, the structure of the atom.
  • It varies relatively slowly with changes in the mass number due to the \( A^{1/3} \) relationship.
  • This concept is also useful when studying nuclear stability and reactions.
Understanding the nuclear radius is crucial for grasping many concepts in nuclear physics.
Mass Number Calculation
Calculating the mass number (\( A \)) is important when analyzing nuclear properties. The mass number represents the total count of protons and neutrons in the nucleus. In our exercise, we used the nuclear radius formula to find the mass number. When a nucleus has been given a specific radius, one can use the formula \( R = R_0 A^{1/3} \) and rearrange it to solve for \( A \).
To do this, the equation needs to be reorganized to isolate \( A \):
  • Substitute the known values, such as \( R_0 = 1.2 \mathrm{fm} \) and the given radius \( R = 3.6 \mathrm{fm} \).
  • Rearranging the equation, you divide both sides by \( R_0 \) and solve for \( A^{1/3} \), which gives you \( A^{1/3} = 3 \).
  • By cubing both sides, you find \( A = 27 \).
This calculated mass number gives an idea of the nucleus's composition, helping to determine isotopic stability and nuclear reactions.
Electron Scattering Method
The electron scattering method is a powerful technique used to investigate the size and structure of atomic nuclei. This method involves firing electrons at high velocities toward the atomic nuclei and observing how these electrons scatter. Due to their small mass and charge, electrons are ideal probes for this purpose. They interact weakly with the nucleus, ensuring they can penetrate without disturbing the nucleus significantly.
By analyzing how electrons scatter, scientists gain insights into the nuclear radius and structure. The technique relies on the interaction between the incoming electrons and the electric field surrounding the nucleus, enabling measurements without direct physical disturbance. Important points about this method include:
  • It provides precise measurements of nuclear size.
  • It helps in mapping out the charge distribution within the nucleus.
  • This method is non-invasive, minimizing changes to the nucleus during measurement.
The electron scattering method is vital in nuclear physics, aiding in the understanding of both fundamental atomic structures and interactions.

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Most popular questions from this chapter

A periodic table might list the average atomic mass of magnesium as being \(24.312 \mathrm{u}\), which is the result of weighting the atomic masses of the magnesium isotopes according to their natural abundances on Earth. The three isotopes and their masses are \({ }^{24} \mathrm{Mg}(23.98504 \mathrm{u}),{ }^{25} \mathrm{Mg}(24.98584 \mathrm{u})\), and \({ }^{26} \mathrm{Mg}(25.98259 \mathrm{u})\) The natural abundance of \({ }^{24} \mathrm{Mg}\) is \(78.99 \%\) by mass (that is, \(78.99 \%\) of the mass of a naturally occurring sample of magnesium is due to the presence of \({ }^{24} \mathrm{Mg}\) ). What is the abundance of (a) \({ }^{25} \mathrm{Mg}\) and (b) \({ }^{26} \mathrm{Mg}\) ?

An organic sample of mass \(4.00 \mathrm{~kg}\) absorbs \(2.00 \mathrm{~mJ}\) via slow neutron radiation \((\mathrm{RBE}=5)\). What is the dose equivalent (mSv)?

co The isotope \({ }^{40} \mathrm{~K}\) can decay to either \({ }^{40} \mathrm{Ca}\) or \({ }^{40} \mathrm{Ar}\); assume both decays have a half-life of \(1.26 \times 10^{9} \mathrm{y}\). The ratio of the Ca produced to the Ar produced is \(8.54 / 1=8.54\). A sample originally had only \({ }^{40} \mathrm{~K}\). It now has equal amounts of \({ }^{40} \mathrm{~K}\) and \({ }^{40} \mathrm{Ar} ;\) that is, the ratio of \(\mathrm{K}\) to Ar is \(1 / 1=1\). How old is the sample? (Hint: Work this like other radioactive-dating problems, except that this decay has two products.)

The radioactive nuclide \({ }^{99} \mathrm{Tc}\) can be injected into a patient's bloodstream in order to monitor the blood flow, measure the blood volume, or find a tumor, among other goals. The nuclide is produced in a hospital by a "cow" containing \({ }^{99}\) Mo, a radioactive nuclide that decays to \({ }^{99} \mathrm{Tc}\) with a half-life of \(67 \mathrm{~h} .\) Once a day, the cow is "milked" for its \({ }^{99} \mathrm{Tc}\), which is produced in an excited state by the \({ }^{99} \mathrm{Mo} ;\) the \({ }^{99} \mathrm{Tc}\) de-excites to its lowest energy state by emitting a gamma-ray photon, which is recorded by detectors placed around the patient. The de-excitation has a half- life of \(6.0 \mathrm{~h}\). (a) By what process does \({ }^{99}\) Mo decay to \({ }^{99} \mathrm{Tc}\) ? (b) If a patient is injected with an \(8.2 \times 10^{7}\) Bq sample of \({ }^{99} \mathrm{Tc}\), how many gamma-ray photons are initially produced within the patient each second? (c) If the emission rate of gamma-ray photons from a small tumor that has collected \({ }^{99} \mathrm{Tc}\) is 38 per second at a certain time, how many excitedstate \({ }^{99} \mathrm{Tc}\) are located in the tumor at that time?

High-mass radionuclides, which may be either alpha or beta emitters, belong to one of four decay chains, depending on whether their mass numher \(A\) is of the form \(4 n, 4 n+1,4 n+2\), or \(4 n+3\) where \(n\) is a positive integer. (a) Justify this statement and show that if a nuclide belongs to one of these families, all its decay products belong to the same family. Classify the following nuclides as to family: (b) \({ }^{235} \mathrm{U},(\mathrm{c})^{236} \mathrm{U}\) (d) \(^{238} \mathrm{U},(\mathrm{e})^{239} \mathrm{Pu},(\mathrm{f})^{240} \mathrm{Pu},(\mathrm{g})^{245} \mathrm{Cm}\), (h) \(^{246} \mathrm{Cm}\), (i) \({ }^{249} \mathrm{Cf}\), and \((\mathrm{j})^{253} \mathrm{Fm}\).

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