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Estimate the energy width (energy uncertainty) of the \(\psi\) if its mean lifetime is \(7.6 \times 10^{-21}\) s. What fraction is this of its rest energy?

Short Answer

Expert verified
The energy uncertainty, or energy width, of the particle is at least 43.44 eV. However, due to the missing information, it's impossible to calculate the fraction of this energy uncertainty over the rest energy of the particle.

Step by step solution

01

Apply Heisenberg's uncertainty principle

According to Heisenberg's uncertainty principle, the product of the uncertainties in energy and time is at least as large as a well-known physical constant. Mathematically, this expressed as: \(\Delta E \Delta t \geq \frac{1}{2} \hbar\), where \(\Delta E\) is the uncertainty in energy, \(\Delta t\) is the uncertainty in time, and \(\hbar\) is the reduced Planck constant. Given that \(\Delta t = 7.6 \times 10^{-21}\) s (the mean lifetime of the particle), we can rearrange the inequality to give us the minimum possible \(\Delta E\). \(\Delta E \geq \frac{1}{2} \frac{\hbar}{\Delta t}\)
02

Find the minimum energy uncertainty

Plugging in the given lifetime and the value of the reduced Planck constant (\(\hbar = 1.0545718 \times 10^{-34}\) m² kg / s), we find an estimate for the energy uncertainty: \(\Delta E \geq \frac{1}{2} \frac{1.0545718 \times 10^{-34}\) m² kg / s}{7.6 \times 10^{-21}\) s = 6.96 \times 10^{-15}\) kg m² / s². Since 1 kg m² / s² = 1 joule (J), we find \(\Delta E \geq 6.96 \times 10^{-15}\) J
03

Convert the minimum energy uncertainty to eV

For consistency with typical atomic physics measurements, we could convert the energy uncertainty to electronvolts (eV). 1 eV = 1.602176634 × 10^-19 J. Therefore, \[\Delta E \geq \frac{{6.96 \times 10^{-15}\)J}}{{1.602176634 × 10^-19 J/eV}} = 43.44\ eV.\]
04

Calculate the rest energy

Rest energy, \(E_0\), of a particle can be obtained from its mass by Einstein’s equation \(E_0 = mc^2\), where, \(m\) is the mass of the particle and \(c\) is the speed of light. The question does not provide the mass of the particle. Thus, it's impossible to calculate the rest energy and find the fraction of the energy uncertainty over the rest energy.

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Key Concepts

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

Energy Uncertainty
When talking about energy uncertainty in the context of particles, we often refer to Heisenberg's Uncertainty Principle. This principle states that there is a fundamental limit to how precisely we can know certain pairs of physical properties of a particle, such as its position and momentum, or energy and time. For energy and time, this is expressed with the inequality: \[\Delta E \Delta t \geq \frac{1}{2} \hbar\]Here, \(\Delta E\) represents the energy uncertainty, and \(\Delta t\) represents the time uncertainty, which in this case is the particle's mean lifetime. Using this principle, we can calculate the smallest possible energy uncertainty for a particle using its mean lifetime. - Substitute the given mean lifetime into the inequality - Solve for \(\Delta E\), using the reduced Planck constant. By doing this, we can get an idea of how spread out the energy levels of the particle could be, which is critical in advanced fields like quantum mechanics, where such small variations have significant implications.
Mean Lifetime
The mean lifetime of a particle, designated by \(\Delta t\), is fundamentally important in determining how long a particle exists before decaying. In the context of our problem, this mean lifetime is \(7.6 \times 10^{-21}\) s. Understanding this concept is crucial, not just for calculating energy uncertainty, but also for studying how unstable particles behave over time.- A shorter lifetime generally means a larger energy uncertainty.- A longer lifetime suggests a more stable particle with a smaller energy uncertainty.The mean lifetime essentially describes the "average" time duration a particle can exist in a particular state before transitioning into something else. This is intrinsically linked to the energy uncertainty since, according to Heisenberg's principle, limited information about a particle's lifetime translates to more significant uncertainty in its energy levels.
Reduced Planck Constant
The reduced Planck constant, often denoted as \(\hbar\), is a fundamental constant in quantum mechanics. It is crucial in the equations that describe quantum-level systems, and in this exercise, it provides the scale for the uncertainty limit in energy-time measurements.The value of \(\hbar\) is \(1.0545718 \times 10^{-34}\) m² kg / s. This constant is derived from the Planck constant \(h\), divided by \(2\pi\).Its role is central to:
  • Setting the minimum bound for the uncertainty in measurements.
  • Facilitating calculations in spherical waveforms or systems.
Having \(\hbar\) dictates that there's a trade-off between the precision of time and energy measurements: when one becomes more precise, the other becomes less so. By applying it in the uncertainty principle, it helps us understand and predict phenomena within the quantum world, leading to discoveries such as energy levels and spectral lines in atoms.

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Most popular questions from this chapter

A proton and an antiproton annihilate, producing two photons. Find the energy, frequency, and wavelength of each photon (a) if the \(\mathrm{p}\) and \(\overline{\mathrm{p}}\) are initially at rest and \((\mathrm{b})\) if the \(\mathrm{p}\) and \(\overline{\mathrm{p}}\) collide head-on, each with an initial kinetic energy of \(620 \mathrm{MeV}\).

In which of the following reactions or decays is strangeness conserved? In each case, explain your reasoning. (a) \(\mathrm{K}^{+} \rightarrow \mu^{+}+\nu_{\mu}\) (b) \(n+K^{+} \rightarrow p+\pi^{0}\) (c) \(\mathrm{K}^{+}+\mathrm{K} \rightarrow \pi^{0}+\pi^{0}\) (d) \(p+K^{-} \rightarrow \Lambda^{0}+\pi^{0}\)

One proposed proton decay is \(\mathrm{p}^{+} \rightarrow \mathrm{e}^{+}+\pi^{0}\), which violates both baryon and lepton number conservation, so the proton lifetime is expected to be very long. Suppose the proton half-life were \(1.0 \times 10^{18} \mathrm{y} .\) (a) Calculate the energy deposited per kilogram of body tissue (in rad) due to the decay of the protons in your body in one year. Model your body as consisting entirely of water. Only the two protons in the hydrogen atoms in each \(\mathrm{H}_{2} \mathrm{O}\) molecule would decay in the manner shown; do you see why? Assume that the \(\pi^{0}\) decays to two \(\gamma\) rays, that the positron annihilates with an electron, and that all the encrgy produced in the primary decay and these secondary decays remains in your body. (b) Calculate the equivalent dose (in rem) assuming an RBE of 1.0 for all the radiation products, and compare with the 0.1 rem due to the natural background and the 5.0 rem guideline for industrial workers. Based on your calculation, can the proton lifetime be as short as \(1.0 \times 10^{18} \mathrm{y} ?\)

The strong nuclear force can be crudely modeled as a Hooke's-law spring force, increasing linearly with the quark scparation distance. The energy stored in this "spring" corresponds to the energy content of the gluon field. In this picture, as quarks are further separated, increasing energy is stored between them. At a critical separation distance, the energy converts to matter, and a new quark-antiquark pair is generated, guaranteeing that there can never be a free quark. (a) A proton has a diameter of about \(1.5 \mathrm{fm}\). Fstimate the repulsive Coulomb force between two up quarks separated by \(0.5 \mathrm{fm}\). (b) Model the strong force as \(F_{\mathrm{s}}=k s,\) where \(s\) is the distance between two quarks. If this force balances the electrostatic repulsion between two up quarks when \(s=0.5 \mathrm{fm},\) what is the effective spring constant \(k,\) in \(\mathrm{SI}\) units? (c) Convert \(k\) into units of \(\mathrm{MeV} / \mathrm{fm}^{2}\). (d) How much energy is stored in the gluon field when \(s=0.5 \mathrm{fm} ?\) The mass of an up quark is thought to be about \(2.3 \mathrm{MeV} / \mathrm{c}^{2}\). (e) How much energy is needed to produce an up quark and an antiup quark? (f) How far would two up quarks need to be separated so that the gluon energy \(\frac{1}{2} k s^{2}\) matches the rest energy of an up-antiup quark pair?

Bubble-chamber photographs taken with a \(0.40 \mathrm{~T}\) magnetic field show at one point a positron moving perpendicular to the field in a \(15 \mathrm{~cm}\) circle. What is the magnitude of the positron's momentum at that point?

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