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Use the Heisenberg uncertainty principle to make a reasonable argument against the hypothesis that free electrons can be present in a nucleus. Use relativistic expressions for the momentum and energy, and include appropriate assumptions and approximations.

Short Answer

Expert verified
The extreme level of energy that a free electron must have to reside within the nucleus, as dictated by the Heisenberg uncertainty principle, is considerably more than the electron's rest energy. Thus, it's extremely improbable that free electrons exist in a nucleus.

Step by step solution

01

Write Down the Heisenberg Uncertainty Principle

The Heisenberg uncertainty principle is given by the expression \( \Delta x \Delta p \geq \hbar/2 \), where \( \Delta x \) is the uncertainty in position, \( \Delta p \) is the uncertainty in momentum, and \( \hbar \) is the reduced Planck’s constant.
02

Relativistic Momentum Expression

The relativistic momentum expression is given by the equation \( p = \gamma mv \), where \( m \) is the mass of the electron, \( v \) is the velocity of the electron, and \( \gamma = 1/ \sqrt{1 - v^2/c^2} \) is the Lorentz factor. Here, \( c \) is the speed of light.
03

Approximation of Position Uncertainty

The position uncertainty of an electron in the nucleus can be approximated as the size of the nucleus, which is about \( 10^{-15} \) meters.
04

Computing Uncertainty in Momentum

Substitute the position uncertainty into the Heisenberg uncertainty principle to compute the minimum uncertainty in the momentum. \( \Delta p \geq \hbar / (2 \Delta x) \).
05

Calculating Energy

Obtain energy by substituting this minimum momentum uncertainty in the relativistic energy equation, \( E = \sqrt{(pc)^2 + (mc^2)^2} \), where \( E \) is the energy.
06

Comparison with Electron's Rest Energy

Compare the energy obtained in Step 5 with the rest energy of an electron, \( mc^2 \). This comparison allows us to analyze the feasibility of a free electron being in a nucleus.

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Key Concepts

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

Relativistic Momentum
Relativistic momentum is an essential part of understanding how particles like electrons behave at speeds close to the speed of light. Traditional momentum is calculated by multiplying mass and velocity. However, when particles move at very high speeds, nearly approaching the speed of light, we use the concept of relativistic momentum to account for these speeds. The formula for relativistic momentum is given by \( p = \gamma mv \), where \( m \) is the particle's mass, \( v \) is its velocity, and \( \gamma \) is the Lorentz factor. The Lorentz factor is defined as \( \gamma = 1/ \sqrt{1 - v^2/c^2} \), where \( c \) is the speed of light.
  • This factor accounts for the way time and space are perceived differently at speeds close to the speed of light.
  • As velocity \( v \) approaches \( c \), \( \gamma \) increases substantially, making the relativistic momentum significantly larger than classical momentum.
This principle is key for studying high-energy particles in both particle physics and nuclear physics.
Nuclear Physics
Nuclear physics delves into the core components and interactions inside an atomic nucleus. It is crucial for understanding the stability and types of available particles within a nucleus, such as protons and neutrons. Electrons, being free particles, generally orbit the nucleus rather than residing within it.A key idea in nuclear physics is the equivalent diameter of the nucleus. This size is approximately \( 10^{-15} \) meters. Keeping this small scale in mind helps to comprehend how tightly bound nuclear particles are and why electrons might not fit comfortably within this space.
  • Nuclear physics allows us to explore forces, like the strong nuclear force, that hold the nucleus together.
  • It informs whether certain hypotheses, like electrons being inside the nucleus, make practical sense given size and energy constraints.
The dimension and energy scale considerations within nuclear physics make it improbable for free electrons to exist in such tiny spaces.
Electron in Nucleus
The idea of electrons existing within the nucleus requires examining the scale and size of the nucleus. The Heisenberg Uncertainty Principle helps explain why free electrons are unlikely to inhabit nuclei.The principle asserts that there is a fundamental limit to the precision with which certain pairs of physical properties, like position and momentum, can be known. The formula is \( \Delta x \Delta p \geq \hbar/2 \). When considering a nucleus-sized space, \( \Delta x \approx 10^{-15} \) meters, we find a significant uncertainty for momentum.An electron within such a small region would necessitate a very high momentum, translating into vast energies when considered relativistically. Thus, based on energy constraints:
  • The relativistic energy \( E = \sqrt{(pc)^2 + (mc^2)^2} \) suggests unreasonable energy levels when an electron is presumed to exist in a nucleus at zero velocity.
  • This exceeds the electron’s rest energy \( mc^2 \), reinforcing the unlikelihood of free electrons being nuclear inhabitants.
Quantum Mechanics
Quantum Mechanics is the branch of physics that describes the strange and often non-intuitive behavior of particles at microscopic scales, where classical physics fails. The Heisenberg Uncertainty Principle is a fundamental part of quantum mechanics. It implies that precise knowledge of a particle's position and momentum simultaneously is impossible. For an electron supposedly housed within a nucleus, applying the uncertainty principle would mean:
  • The electron’s position is highly constrained, leading to very high momentum uncertainty.
  • This results in exorbitant energy requirements when evaluatively using relativistic momentum.
Quantum mechanics emphasizes that classical physics alone cannot describe micro-scale phenomena adequately. Uncertainty and wave-particle dual characteristics of electrons are better understood within this framework. Thus, these tendencies suggest free electrons cannot simply reside in a nucleus due to quantum mechanical limits.

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Most popular questions from this chapter

A sample of radioactive material is said to be carrier-free when no stable isotopes of the radioactive element are present. Calculate the mass of strontium in a carrierfree \(5-\mathrm{mCi}\) sample of \({ }^{90} \mathrm{Sr}\) whose half-life is \(28.8 \mathrm{yr}\).

A laboratory stock solution is prepared with an initial activity due to \({ }^{24} \mathrm{Na}\) of \(2.5 \mathrm{mCi} / \mathrm{mL}\), and \(10 \mathrm{~mL}\) of the stock solution is diluted (at \(t_{0}=0\) ) to a working solution with a total volume of \(250 \mathrm{~mL}\). After \(48 \mathrm{~h}\), a \(5-\mathrm{mL}\) sample of the working solution is monitored with a counter. What is the measured activity? (Note: \(1 \mathrm{~mL}=\) 1 milliliter, and the half-life of \({ }^{24} \mathrm{Na}\) is \(15.0 \mathrm{~h}\).)

A radioactive nucleus with decay constant \(\lambda\) decays to a stable daughter nucleus. (a) Show that the number of daughter nuclei, \(N_{2}\), increases with time according to the expression $$ N_{2}=N_{01}\left(1-e^{-\lambda t}\right) $$ where \(N_{01}\) is the initial number of parent nuclei. (b) Starting with \(10^{6}\) parent nuclei at \(t=0\), with a halflife of \(10 \mathrm{~h}\), plot the number of parent nuclei and the number of daughter nuclei as functions of time over the interval 0 to \(30 \mathrm{~h}\).

(a) Find the radius of the 12 C nucleus. (b) Find the force of repulsion between a proton at the surface of a \({ }^{12}{6}\) C nucleus and the remaining five protons. (c) How much work (in MeV) must be done to overcome this electrostatic repulsion and put the last proton into the nucleus? \((\mathrm{d})\) Repeat \((\mathrm{a}),(\mathrm{b})\), and (c) for \({ }_{92}^{238} \mathrm{U}\).

What specific activity (see Problem 36 ), in disintegrations \(/ \mathrm{min} \cdot \mathrm{g}\), would be expected for carbon samples from 2000 -year-old bones? (Note that the ratio of \({ }^{14} \mathrm{C}\) to \({ }^{12} \mathrm{C}\) in living organisms is \(1.3 \times 10^{-12}\) and the halflife of \({ }^{14} \mathrm{C}\) is \(5730 \mathrm{yr}\).)

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