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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

What is the ground-state electron configuration of nickel (Ni, atomic number 28 )?
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.
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