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Chapter 30: Problem 8

What is the de Broglie wavelength of a proton with kinetic energy $1.0 \mathrm{TeV} ?$

Short Answer

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Question: Find the de Broglie wavelength of a proton with a kinetic energy of 1.0 TeV. Answer: The de Broglie wavelength of a proton with a kinetic energy of 1.0 TeV is approximately \(2.054 \times 10^{-16}\) meters.

Step by step solution

01

Convert the kinetic energy to Joules

Given the kinetic energy of the proton as 1.0 TeV, we need to convert it to Joules. First, we convert TeV to eV and then to Joules using the conversion factors: 1 TeV = \(10^{12}\) eV, and 1 eV = \(1.6 \times 10^{-19}\) J So, \(1.0 \text{ TeV} = 1.0\cdot10^{12} \text{ eV} = 1.0\cdot10^{12} \cdot 1.6\cdot10^{-19} \text{ J} = 1.6\cdot10^{-7} \text{ J} \)
02

Calculate the momentum of the proton

Now we need to find the momentum of the proton using the kinetic energy formula: \(E_K = \frac{p^2}{2m}\), where \(E_K\) is the kinetic energy, \(p\) is the momentum, and \(m\) is the mass of the proton. The mass of a proton is roughly \(m = 1.67 \times 10^{-27} \text{ kg}\). Rearrange the formula to solve for the momentum \(p\): \(p = \sqrt{2mE_K} = \sqrt{2 \cdot 1.67 \times 10^{-27} \text{ kg} \cdot 1.6 \times 10^{-7} \text{ J}} \approx 3.226 \times 10^{-19} \text{ kg m/s}\)
03

Calculate the de Broglie wavelength

Now we can find the de Broglie wavelength using de Broglie's equation: \(\lambda = \frac{h}{p}\), where \(\lambda\) is the wavelength, \(h\) is the Planck's constant, and \(p\) is the momentum. Plugging in the values for Planck's constant (\(h \approx 6.626 \times 10^{-34} \text{ Js}\)) and the momentum calculated in step 2, we have: \(\lambda = \frac{6.626 \times 10^{-34} \text{ Js}}{3.226 \times 10^{-19} \text{ kg m/s}} \approx 2.054 \times 10^{-16} \text{ m}\) So, the de Broglie wavelength of a proton with 1.0 TeV kinetic energy is approximately \(2.054 \times 10^{-16}\) meters.

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Most popular questions from this chapter

At the Stanford Linear Accelerator, electrons and positrons collide together at very high energies to create other elementary particles. Suppose an electron and a positron, each with rest energies of \(0.511 \mathrm{MeV},\) collide to create a proton (rest energy \(938 \mathrm{MeV}\) ), an electrically neutral kaon \((498 \mathrm{MeV}),\) and a negatively charged sigma baryon \((1197 \mathrm{MeV}) .\) The reaction can be written as: $$\mathrm{e}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{p}^{+}+\mathrm{K}^{0}+\Sigma^{-}$$ (a) What is the minimum kinetic energy the electron and positron must have to make this reaction go? Assume they each have the same energy. (b) The sigma can decay in the reaction \(\Sigma^{-} \rightarrow \mathrm{n}+\pi^{-}\) with rest energies of \(940 \mathrm{MeV}\) (neutron) and \(140 \mathrm{MeV}\) (pion). What is the kinetic energy of each decay particle if the sigma decays at rest?
A proton in Fermilab's Tevatron is accelerated through a potential difference of 2.5 MV during each revolution around the ring of radius \(1.0 \mathrm{km} .\) In order to reach an energy of 1 TeV, how many revolutions must the proton make? How far has it traveled?
When a proton and an antiproton annihilate, the annihilation products are usually pions. (a) Suppose three pions are produced. What combination(s) of \(\pi^{+}, \pi^{-},\) and \(\pi^{0}\) are possible? (b) Suppose five pions are produced. What combination(s) of \(\pi^{+}, \pi^{-},\) and \(\pi^{0}\) are possible? (c) What is the maximum number of pions that could be produced if the kinetic energies of the proton and antiproton are negligibly small? The mass of a charged pion is \(0.140 \mathrm{GeV} / c^{2}\) and the mass of a neutral pion is \(0.135 \mathrm{GeV} / c^{2}.\)
The \(K^{0}\) meson can decay to two pions: \(K^{0} \rightarrow \pi^{+}+\pi^{-}\) The rest energies of the particles are: \(K^{0}=497.7 \mathrm{MeV}\) \(\pi^{+}=\pi^{-}=139.6 \mathrm{MeV} .\) If the \(K^{0}\) is at rest before it decays, what are the kinetic energies of the \(\pi^{+}\) and the \(\pi^{-}\) after the decay?
Estimate the magnetic field strength required at the LHC to make 7.0-TeV protons travel in a circle of circumference \(27 \mathrm{km}\). Start by deriving an expression, using Newton's second law, for the field strength \(B\) in terms of the particle's momentum \(p,\) its charge \(q,\) and the radius \(r\). Even though derived using classical physics, the expression is relativistically correct. (The estimate will come out much lower than the actual value of 8.33 T. In the LHC, the protons do not travel in a constant magnetic field; they move in straight-line segments between magnets.)
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