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The radii of atomic nuclei are of the order of 5.0 \(\times\) 10\(^{-15}\) m. (a) Estimate the minimum uncertainty in the momentum of an electron if it is confined within a nucleus. (b) Take this uncertainty in momentum to be an estimate of the magnitude of the momentum. Use the relativistic relationship between energy and momentum, Eq. (37.39), to obtain an estimate of the kinetic energy of an electron confined within a nucleus. (c) Compare the energy calculated in part (b) to the magnitude of the Coulomb potential energy of a proton and an electron separated by 5.0 \(\times\) 10\(^{-15}\) m. On the basis of your result, could there be electrons within the nucleus? (\(Note\): It is interesting to compare this result to that of Problem 39.72.)

Short Answer

Expert verified
No, electrons cannot exist in the nucleus due to high kinetic energy exceeding potential energy.

Step by step solution

01

Estimate Minimum Uncertainty in Momentum

Use the Heisenberg Uncertainty Principle: \[ \Delta x \Delta p \geq \frac{\hbar}{2} \]where \( \Delta x \) is the uncertainty in position, approximately the radius of the nucleus, \(5.0 \times 10^{-15}\) m, and \( \hbar \approx 1.05 \times 10^{-34}\) J·s.Thus, the minimum uncertainty in momentum (\( \Delta p \)) is: \[ \Delta p \geq \frac{\hbar}{2 \Delta x} = \frac{1.05 \times 10^{-34} \, \text{J}\cdot\text{s}}{2 \times 5.0 \times 10^{-15} \, \text{m}} \approx 1.05 \times 10^{-20} \, \text{kg}\cdot\text{m/s} \]
02

Estimate the Magnitude of the Momentum

According to part (b) of the problem, take the uncertainty in momentum calculated as the approximate magnitude of the momentum, i.e.,\[ p \approx 1.05 \times 10^{-20} \, \text{kg}\cdot\text{m/s} \]
03

Calculate Kinetic Energy Using Relativistic Relationship

Use the relativistic energy-momentum relation:\[ E^2 = (pc)^2 + (m_0 c^2)^2 \]where \(E\) is the total energy, \(p\) is momentum, and \(m_0\) for electron is \(9.11 \times 10^{-31}\) kg. Assuming \(m_0c^2\) is negligible, kinetic energy \(K\approx E\) can be estimated as:\[ K \approx pc = 1.05 \times 10^{-20} \, \text{kg}\cdot\text{m/s} \times 3 \times 10^8 \, \text{m/s} \approx 3.15 \times 10^{-12} \, \text{J} \]
04

Calculate Magnitude of Coulomb Potential Energy

The Coulomb potential energy between a proton and an electron separated by distance (\( r = 5.0 \times 10^{-15} \) m) is given by:\[ U = \frac{k e^2}{r} \]where: - \( k = 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \) - \( e = 1.60 \times 10^{-19} \, \text{C} \)Hence,\[ U = \frac{8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \times (1.60 \times 10^{-19} \, \text{C})^2}{5.0 \times 10^{-15} \, \text{m}} \approx 4.6 \times 10^{-14} \, \text{J} \]
05

Compare Energies and Conclude

Comparing the kinetic energy (\(3.15 \times 10^{-12} \, \text{J}\)) to the Coulomb potential energy (\(4.6 \times 10^{-14} \, \text{J}\)): the kinetic energy is significantly higher.Since this high kinetic energy far exceeds the potential energy binding an electron to a proton, electrons within a nucleus cannot exist under these circumstances.

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Key Concepts

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

Atomic Nucleus
Atomic nuclei are incredibly tiny structures at the heart of every atom. Their size is often around 5.0 \(\times\) 10\(^{-15}\) meters, which is approximately a million times smaller than the atom itself. Inside this compact space reside protons and neutrons, collectively known as nucleons. The smallest radius of nuclei makes them densely packed with mass and energy.

The Heisenberg Uncertainty Principle plays a key role when estimating the behavior of particles within the nucleus. According to this principle, it is impossible to know precisely both the position and momentum of a particle. The principle is commonly expressed by the relation \(\Delta x \Delta p \geq \frac{\hbar}{2}\), where \(\Delta x\) is the uncertainty in position, and \(\Delta p\) is the uncertainty in momentum. For electrons confined within a nucleus, this principle helps calculate a minimum uncertainty in momentum due to limited space.

For example, using the nucleus's radius as the uncertainty in position, physicists can estimate the momentum's uncertainty. It's fascinating to consider how such quantum principles influence our understanding of these invisible yet fundamental components of matter.
Relativistic Energy-Momentum Relation
The relativistic energy-momentum relation is essential for understanding the energy of particles moving at high speeds. This equation links the total energy \(E\), momentum \(p\), and rest mass \(m_0\) of a particle with the speed of light \(c\): \[ E^2 = (pc)^2 + (m_0 c^2)^2 \]This formula illuminates the fact that as particles approach the speed of light, their kinetic energy increases significantly, potentially surpassing their rest mass energy.

When studying an electron within an atomic nucleus, where the estimates for momentum are exceedingly high, using this relation provides insight into its kinetic energy. In many cases, the rest mass energy term \((m_0c^2)^2\) becomes negligible compared to \((pc)^2\). Thus, the kinetic energy \(K\) can be approximated by \(pc\), yielding a straightforward calculation for electrons confined in tiny spaces.

This high kinetic energy of electrons calculated through the relativistic energy-momentum equation often exceeds the binding energies found in nuclear environments, indicating that such electrons would not be stably confined within the nucleus. Understanding this concept bridges the gaps between special relativity and quantum mechanics.
Coulomb Potential Energy
Coulomb potential energy describes the energy of interaction between two charged particles. It is given by the formula:\[ U = \frac{k e^2}{r} \]where:
  • \(k\) is Coulomb's constant, valued at 8.99 \(\times\) 10\(^9\) N\(\cdot\)m\(^2\)/C\(^2\)
  • \(e\) stands for the elementary charge, approximately 1.60 \(\times\) 10\(^{-19}\) C
  • \(r\) represents the separation distance between the charges
When considering a proton and an electron separated by a nuclear-sized distance, this equation helps estimate the potential energy binding these particles together.

In many cases, this potential energy is much smaller than the kinetic energies calculated for confined particles, such as electrons within an atomic nucleus. This disparity means that the binding forces due to electrostatic attraction are insufficient to keep particles like electrons stably bound at such close interactions in the nucleus.

By comparing calculated kinetic and potential energies, we gain insights into why electrons do not exist within the nucleus. The conditions required for such existence would demand unrealistically low kinetic energies, making stable electron presence impossible within such a dense space.

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Most popular questions from this chapter

(a) A nonrelativistic free particle with mass \(m\) has kinetic energy \(K\). Derive an expression for the de Broglie wavelength of the particle in terms of \(m\) and \(K\). (b) What is the de Broglie wavelength of an 800-eV electron?

A pesky 1.5-mg mosquito is annoying you as you attempt to study physics in your room, which is 5.0 m wide and 2.5 m high. You decide to swat the bothersome insect as it flies toward you, but you need to estimate its speed to make a successful hit. (a) What is the maximum uncertainty in the horizontal position of the mosquito? (b) What limit does the Heisenberg uncertainty principle place on your ability to know the horizontal velocity of this mosquito? Is this limitation a serious impediment to your attempt to swat it?

A CD-ROM is used instead of a crystal in an electrondiffraction experiment. The surface of the CD-ROM has tracks of tiny pits with a uniform spacing of 1.60 \(\mu{m}\). (a) If the speed of the electrons is 1.26 \(\times\) 10\(^4\) m/s, at which values of \(\theta\) will the \(m\) = 1 and \(m\) = 2 intensity maxima appear? (b) The scattered electrons in these maxima strike at normal incidence a piece of photographic film that is 50.0 cm from the CD-ROM. What is the spacing on the film between these maxima?

The negative muon has a charge equal to that of an electron but a mass that is 207 times as great. Consider a hydrogenlike atom consisting of a proton and a muon. (a) What is the reduced mass of the atom? (b) What is the ground-level energy (in electron volts)? (c) What is the wavelength of the radiation emitted in the transition from the \(n\) = 2 level to the \(n\) = 1 level?

An alpha particle (\(m = 6.64 \times 10^{-27} kg\)) emitted in the radioactive decay of uranium-238 has an energy of 4.20 MeV. What is its de Broglie wavelength?

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