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The radii of atomic nuclei are of the order of 5.0 × 1015 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 × 1015 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: ΔxΔp2where Δx is the uncertainty in position, approximately the radius of the nucleus, 5.0×1015 m, and 1.05×1034 J·s.Thus, the minimum uncertainty in momentum (Δp) is: Δp2Δx=1.05×1034Js2×5.0×1015m1.05×1020kgm/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.,p1.05×1020kgm/s
03

Calculate Kinetic Energy Using Relativistic Relationship

Use the relativistic energy-momentum relation:E2=(pc)2+(m0c2)2where E is the total energy, p is momentum, and m0 for electron is 9.11×1031 kg. Assuming m0c2 is negligible, kinetic energy KE can be estimated as:Kpc=1.05×1020kgm/s×3×108m/s3.15×1012J
04

Calculate Magnitude of Coulomb Potential Energy

The Coulomb potential energy between a proton and an electron separated by distance (r=5.0×1015 m) is given by:U=ke2rwhere: - k=8.99×109Nm2/C2 - e=1.60×1019CHence,U=8.99×109Nm2/C2×(1.60×1019C)25.0×1015m4.6×1014J
05

Compare Energies and Conclude

Comparing the kinetic energy (3.15×1012J) to the Coulomb potential energy (4.6×1014J): 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 × 1015 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 ΔxΔp2, where Δx is the uncertainty in position, and Δp 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 m0 of a particle with the speed of light c: E2=(pc)2+(m0c2)2This 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 (m0c2)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=ke2rwhere:
  • k is Coulomb's constant, valued at 8.99 × 109 Nm2/C2
  • e stands for the elementary charge, approximately 1.60 × 1019 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) What is the smallest amount of energy in electron volts that must be given to a hydrogen atom initially in its ground level so that it can emit the Hα line in the Balmer series? (b) How many different possibilities of spectral-line emissions are there for this atom when the electron starts in the n = 3 level and eventually ends up in the ground level? Calculate the wavelength of the emitted photon in each case.

In the second type of helium-ion microscope, a 1.2-MeV ion passing through a cell loses 0.2 MeV per μm of cell thickness. If the energy of the ion can be measured to 6 keV, what is the smallest difference in thickness that can be discerned? (a) 0.03 μm; (b) 0.06 μm; (c) 3 μm; (d) 6 μm.

An electron is moving with a speed of 8.00 × 106 m/s. What is the speed of a proton that has the same de Broglie wavelength as this electron?

A beam of alpha particles is incident on a target of lead. A particular alpha particle comes in "head-on" to a particular lead nucleus and stops 6.50 × 1014 m away from the center of the nucleus. (This point is well outside the nucleus.) Assume that the lead nucleus, which has 82 protons, remains at rest. The mass of the alpha particle is 6.64 × 1027 kg. (a) Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in joules and in MeV. (b) What initial kinetic energy (in joules and in MeV) did the alpha particle have? (c) What was the initial speed of the alpha particle?

(a) Using the Bohr model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels. (c) The average lifetime of the first excited level of a hydrogen atom is 1.0 × 108 s. In the Bohr model, how many orbits does an electron in the n = 2 level complete before returning to the ground level?

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