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Chapter 28: Problem 26

Nuclei have energy levels just as atoms do. An excited nucleus can make a transition to a lower energy level by emitting a gamma-ray photon. The lifetime of a typical nuclear excited state is about 1 ps. What is the uncertainty in the energy of the gamma-rays emitted by a typical nuclear excited state? [Hint: Use the energy-time uncertainty principle, Eq. \((28-3) .]\)

Short Answer

Expert verified
The uncertainty in the energy is approximately \(5.275 \times 10^{-23} \text{ J}\).

Step by step solution

01

Understanding the Energy-Time Uncertainty Principle

The energy-time uncertainty principle is given by the equation \(\Delta E \Delta t \geq \frac{\hbar}{2}\), where \(\Delta E\) is the uncertainty in energy, \(\Delta t\) is the uncertainty in time (lifetime of the state), and \(\hbar\) is the reduced Planck's constant (\(\hbar = \frac{h}{2\pi}\), with \(h = 6.626 \times 10^{-34} \text{ J s}\)).
02

Substituting the Given Values

We know \(\Delta t = 1 \text{ ps} = 1 \times 10^{-12} \text{ s}\). Substituting \(\Delta t\) and \(\hbar = 1.055 \times 10^{-34} \text{ J s}\) into the inequality \(\Delta E \Delta t \geq \frac{\hbar}{2}\), we have: \(\Delta E \times 1 \times 10^{-12} \geq \frac{1.055 \times 10^{-34}}{2}\).
03

Solving for the Energy Uncertainty

Rearrange the inequality to solve for \(\Delta E\): \(\Delta E \geq \frac{1.055 \times 10^{-34}}{2 \times 10^{-12}}\). Calculate \(\Delta E\): \(\Delta E \geq 5.275 \times 10^{-23} \text{ J}\).
04

Concluding the Calculation

The minimum uncertainty in the energy of the gamma-ray emitted by this nuclear transition is \(5.275 \times 10^{-23} \text{ J}\). This result reflects the inherent uncertainty due to the short lifetime of the excited state.

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Key Concepts

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

Nuclear Energy Levels
Nuclear energy levels are conceptually similar to electron energy levels in atoms. Just like electrons, nuclei also exist in discrete energy levels. These levels represent the various energetic states a nucleus can occupy, and transitions between these levels involve absorption or emission of energy in a very specific form.
Atomic nuclei are composed of protons and neutrons, bounded by the nuclear force. When a nucleus absorbs energy, it gets excited to a higher energy level. Conversely, if it releases energy, it typically does so by emitting radiation, often as gamma rays.
  • Energy absorption and emission happen in quantized steps, each representing a specific level change.
  • A nucleus in an excited state has more energy than in its ground (or most stable) state.
  • When nuclei transition to lower energy levels, they emit energy in the form of photons, often gamma rays.
Gamma-Ray Emission
Gamma-ray emission is a process where an excited nucleus returns to a lower energy state, releasing the excess energy in the form of gamma photons. These photons are highly energetic and belong to the electromagnetic spectrum.
Gamma rays have very short wavelengths and high frequencies, making them the most energetic form of light. This energy is what allows gamma rays to penetrate materials that block other forms of electromagnetic radiation.
  • Gamma-ray emission is a common observation in nuclear transitions.
  • Because of their high energy, gamma rays can cause ionization, making them quite penetrative and useful in medical applications like cancer treatment.
  • They can also be observed in astronomical phenomena, giving insights into nuclear processes happening in the universe.
Planck's Constant
Planck's constant is a fundamental physical constant that is crucial in the field of quantum mechanics. It relates the energy of a photon to the frequency of its electromagnetic wave.
This constant serves as the cornerstone of quantum theory, illustrating the discrete nature of energy in the micro-world.
  • Planck's constant (h) has a value of approximately \(6.626 \times 10^{-34} \text{ J s}\).
  • The reduced Planck's constant (\hbar\u007f) is often used in quantum mechanics and is defined as \(\frac{h}{2\pi}\).
  • It plays a crucial role in the energy-time uncertainty principle, helping determine the uncertainty in energy over time.
  • This principle reveals the inherent uncertainties and probabilistic nature of quantum states.
Energy Uncertainty Calculation
The calculation of energy uncertainty is based on the energy-time uncertainty principle. According to this principle, there is a fundamental limit to how precisely the energy and the lifetime (\Delta t\u007f) of a quantum state can be known simultaneously.
In practice, this is represented by the inequality:
  • \(\Delta E \cdot \Delta t \geq \frac{\hbar}{2}\)
  • Here, \(\Delta E\) represents the uncertainty in energy and \(\Delta t\) the uncertainty (or typical time) for the system to remain in the excited state.
  • Given \(\Delta t\) and \(\hbar\), the energy uncertainty \(\Delta E\) can be calculated.
  • For a nuclear transition with a lifetime of 1 picosecond (\(1 \times 10^{-12} \text{ s}\)), inserting \(\hbar = 1.055 \times 10^{-34}\) Js into the inequality gives us an energy uncertainty of \(\Delta E \geq 5.275 \times 10^{-23} \text{ J}\).
The result emphasizes the quantum mechanical nature of nuclei, and how short lifetimes translate to greater uncertainty in energy measurements.

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Most popular questions from this chapter

Before the discovery of the neutron, one theory of the nucleus proposed that the nucleus contains protons and electrons. For example, the helium-4 nucleus would contain 4 protons and 2 electrons instead of - as we now know to be true- 2 protons and 2 neutrons. (a) Assuming that the electron moves at nonrelativistic speeds, find the ground-state energy in mega-electron- volts of an electron confined to a one-dimensional box of length $5.0 \mathrm{fm}\( (the approximate diameter of the \)^{4} \mathrm{He}$ nucleus). (The electron actually does move at relativistic speeds. See Problem \(80 .)\) (b) What can you conclude about the electron-proton model of the nucleus? The binding energy of the \(^{4} \mathrm{He}\) nucleus - the energy that would have to be supplied to break the nucleus into its constituent particles-is about \(28 \mathrm{MeV} .\) (c) Repeat (a) for a neutron confined to the nucleus (instead of an electron). Compare your result with (a) and comment on the viability of the proton-neutron theory relative to the electron-proton theory.
(a) What are the electron configurations of the ground states of fluorine \((Z=9)\) and chlorine \((Z=17) ?\) (b) Why are these elements placed in the same column of the periodic table?
What is the de Broglie wavelength of a basketball of mass \(0.50 \mathrm{kg}\) when it is moving at \(10 \mathrm{m} / \mathrm{s} ?\) Why don't we see diffraction effects when a basketball passes through the circular aperture of the hoop?

Suppose the electron in a hydrogen atom is modeled as an electron in a one- dimensional box of length equal to the Bohr diameter, \(2 a_{0} .\) What would be the ground-state energy of this "atom"? How does this compare with the actual ground-state energy?
In the Davisson-Germer experiment (Section \(28.2),\) the electrons were accelerated through a \(54.0-\mathrm{V}\) potential difference before striking the target. (a) Find the de Broglie wavelength of the electrons. (b) Bragg plane spacings for nickel were known at the time; they had been determined through x-ray diffraction studies. The largest plane spacing (which gives the largest intensity diffraction maxima) in nickel is \(0.091 \mathrm{nm} .\) Using Bragg's law [Eq. ( \(25-15\) )], find the Bragg angle for the first-order maximum using the de Broglie wavelength of the electrons. (c) Does this agree with the observed maximum at a scattering angle of \(130^{\circ} ?\) [Hint: The scattering angle and the Bragg angle are not the same. Make a sketch to show the relationship between the two angles.]
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