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Chapter 28: Problem 25

The omega particle \((\Omega)\) decays on average about \(0.1 \mathrm{ns}\) after it is created. Its rest energy is 1672 MeV. Estimate the fractional uncertainty in the \(\Omega\) 's rest energy \(\left(\Delta E_{0} / E_{0}\right)\) [Hint: Use the energy-time uncertainty principle, Eq. \((28-3) .]\)

Short Answer

Expert verified
Answer: The fractional uncertainty in the Omega particle's rest energy is approximately \(1.97 \times 10^{-16}\).

Step by step solution

01

Determine relevant quantities and constants

We are given the average decay time, which we will consider as the time uncertainty in this calculation: \(\Delta t = 0.1\, \mathrm{ns}\). Additionally, we have the rest energy: \(E_0 = 1672\, \mathrm{MeV}\). We also need the reduced Planck constant: \(\hbar = 6.582 \times 10^{-16}\, \mathrm{eV}\,\mathrm{s}\).
02

Solve for energy uncertainty using the energy-time uncertainty principle

Using the energy-time uncertainty principle formula (\(\Delta E \cdot \Delta t \gtrsim \hbar / 2\)), we can solve for the energy uncertainty: \(\Delta E = \frac{\hbar}{2\Delta t}\) Note that we will consider the equality for simplicity. Now, we can plug in the known values: \(\Delta E = \frac{6.582 \times 10^{-16}\, \mathrm{eV}\,\mathrm{s}}{2(0.1 \times 10^{-9}\, \mathrm{s})}\)
03

Calculate the energy uncertainty

Now we can calculate the energy uncertainty: \(\Delta E = \frac{6.582 \times 10^{-16}\, \mathrm{eV}\,\mathrm{s}}{2\times0.1 \times 10^{-9}\, \mathrm{s}} = 3.291\times 10^{-7}\, \mathrm{eV}\) Note that to compare this with the rest energy in MeV, we need to convert it to MeV: \(\Delta E = 3.291 \times 10^{-7}\, \mathrm{eV} \times \frac{1\, \mathrm{MeV}}{10^6\, \mathrm{eV}} = 3.291 \times 10^{-13}\, \mathrm{MeV}\)
04

Determine the fractional uncertainty

Finally, we can determine the fractional uncertainty using the energy uncertainty and the rest energy: \(\frac{\Delta E_{0}}{E_{0}} = \frac{3.291 \times 10^{-13}\, \mathrm{MeV}}{1672\, \mathrm{MeV}}\)
05

Calculate the fractional uncertainty

Now we can calculate the fractional uncertainty: \(\frac{\Delta E_{0}}{E_{0}} = \frac{3.291 \times 10^{-13}\,\mathrm{MeV}}{1672\, \mathrm{MeV}} \approx 1.97 \times 10^{-16}\) The fractional uncertainty in the Omega particle's rest energy is approximately \(1.97 \times 10^{-16}\).

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Most popular questions from this chapter

At a baseball game, a radar gun measures the speed of a 144-g baseball to be \(137.32 \pm 0.10 \mathrm{km} / \mathrm{h} .\) (a) What is the minimum uncertainty of the position of the baseball? (b) If the speed of a proton is measured to the same precision, what is the minimum uncertainty in its position?
An electron is confined to a one-dimensional box. When the electron makes a transition from its first excited state to the ground state, it emits a photon of energy 1.2 eV. (a) What is the ground-state energy (in electronvolts) of the electron? (b) List all energies (in electronvolts) of photons that could be emitted when the electron starts in its second excited state and makes transitions downward to the ground state either directly or through intervening states. Show all these transitions on an energy level diagram. (c) What is the length of the box (in nanometers)?
What is the maximum possible value of the angular momentum for an outer electron in the ground state of a bromine atom?
The particle in a box model is often used to make rough estimates of energy level spacings. Suppose that you have a proton confined to a one-dimensional box of length equal to a nuclear diameter (about \(10^{-14} \mathrm{m}\) ). (a) What is the energy difference between the first excited state and the ground state of this proton in the box? (b) If this energy is emitted as a photon as the excited proton falls back to the ground state, what is the wavelength and frequency of the electromagnetic wave emitted? In what part of the spectrum does it lie? (c) Sketch the wave function \(\psi\) as a function of position for the proton in this box for the ground state and each of the first three excited states.
(a) What are the electron configurations of the ground states of lithium \((Z=3),\) sodium \((Z=11),\) and potassium \((Z=19) ?\) (b) Why are these elements placed in the same column of the periodic table?
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