FUNDAMENTALS OF PHYSICS (LLF)+WILEYPLUS
11th Edition
ISBN: 9781119459132
Author: Halliday
Publisher: WILEY
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Chapter 38, Problem 56P
To determine
To find:
a) The de Broglie wavelength
b) Whether the wave nature of incident alpha particles should have been taken into account in interpreting given experiments
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The existence of the atomic nucleus was discovered in 1911 by Ernest Rutherford, who properly interpreted some experiments in which a beam of alpha particles was scattered from a metal foil of atoms such as gold. (a) If the alpha particles had a kinetic energy of 7.5 MeV, what was their de Broglie wavelength? (b) Explain whether the wave nature of the incident alpha particles should have been taken into account in interpreting these experiments. The mass of an alpha particle is 4.00 u (atomic mass units), and its distance of closest approach to the nuclear center in these experiments was about 30 fm. (The wave nature of matter was not postulated until more than a decade after these crucial experiments were first performed.)
Suppose that the uncertainty in position of an electron is equal to the radius of the n=1n=1 Bohr orbit, about 0.529×10−10m0.529×10−10m.
A) Calculate the minimum uncertainty in the corresponding momentum component.
Express your answer in kilogram meters per second.
B) Compare this with the magnitude of the momentum of the electron in the n=1n=1 Bohr orbit.
Compare this with the magnitude of the momentum of the electron in the Bohr orbit.
a) This is greater than the magnitude of the momentum of the electron in the n=1n=1 Bohr orbit.
b) This is the same as the magnitude of the momentum of the electron in the n=1n=1 Bohr orbit.
c) This is less than the magnitude of the momentum of the electron in the n=1n=1 Bohr orbit.
In a Rutherford scattering experiment a target nucleus has a diameter of 1.4×10-14 m. The incoming has a mass of 6.64×10-27 kg. What is the kinetic energy of an particle that has a de Broglie wavelength equal to the diameter of the target nucleus? Ignore relativistic effects.
Chapter 38 Solutions
FUNDAMENTALS OF PHYSICS (LLF)+WILEYPLUS
Ch. 38 - Prob. 1QCh. 38 - Prob. 2QCh. 38 - Prob. 3QCh. 38 - Prob. 4QCh. 38 - Prob. 5QCh. 38 - Prob. 6QCh. 38 - Prob. 7QCh. 38 - Prob. 8QCh. 38 - Prob. 9QCh. 38 - Prob. 10Q
Ch. 38 - Prob. 11QCh. 38 - Prob. 12QCh. 38 - Prob. 13QCh. 38 - Prob. 14QCh. 38 - Prob. 15QCh. 38 - Prob. 16QCh. 38 - Prob. 1PCh. 38 - Prob. 2PCh. 38 - Prob. 3PCh. 38 - Prob. 4PCh. 38 - Prob. 5PCh. 38 - Prob. 6PCh. 38 - Prob. 7PCh. 38 - Prob. 8PCh. 38 - Prob. 9PCh. 38 - Prob. 10PCh. 38 - Prob. 11PCh. 38 - Prob. 12PCh. 38 - Prob. 13PCh. 38 - Prob. 14PCh. 38 - Prob. 15PCh. 38 - Prob. 16PCh. 38 - Prob. 17PCh. 38 - Prob. 18PCh. 38 - Prob. 19PCh. 38 - Prob. 20PCh. 38 - Prob. 21PCh. 38 - Prob. 22PCh. 38 - Prob. 23PCh. 38 - Prob. 24PCh. 38 - Prob. 25PCh. 38 - Prob. 26PCh. 38 - Prob. 27PCh. 38 - Prob. 28PCh. 38 - Prob. 29PCh. 38 - Prob. 30PCh. 38 - Prob. 31PCh. 38 - Prob. 32PCh. 38 - Prob. 33PCh. 38 - Prob. 34PCh. 38 - Prob. 35PCh. 38 - Prob. 36PCh. 38 - Prob. 37PCh. 38 - Prob. 38PCh. 38 - Prob. 39PCh. 38 - Prob. 40PCh. 38 - Prob. 41PCh. 38 - Prob. 42PCh. 38 - Prob. 43PCh. 38 - Prob. 44PCh. 38 - Prob. 45PCh. 38 - Prob. 46PCh. 38 - Prob. 47PCh. 38 - Prob. 48PCh. 38 - Prob. 49PCh. 38 - Prob. 50PCh. 38 - Prob. 51PCh. 38 - Prob. 52PCh. 38 - Prob. 53PCh. 38 - Prob. 54PCh. 38 - Prob. 55PCh. 38 - Prob. 56PCh. 38 - Prob. 57PCh. 38 - Prob. 58PCh. 38 - Prob. 59PCh. 38 - Prob. 60PCh. 38 - Prob. 61PCh. 38 - Prob. 62PCh. 38 - Prob. 63PCh. 38 - Prob. 64PCh. 38 - Prob. 65PCh. 38 - Prob. 66PCh. 38 - Prob. 67PCh. 38 - Prob. 68PCh. 38 - Prob. 69PCh. 38 - Prob. 70PCh. 38 - Prob. 71PCh. 38 - Prob. 72PCh. 38 - Prob. 73PCh. 38 - Prob. 74PCh. 38 - Prob. 75PCh. 38 - Prob. 76PCh. 38 - Prob. 77PCh. 38 - Prob. 78PCh. 38 - Prob. 79PCh. 38 - Prob. 80PCh. 38 - Prob. 81PCh. 38 - Prob. 82PCh. 38 - Prob. 83PCh. 38 - Prob. 84PCh. 38 - Prob. 85PCh. 38 - Prob. 86PCh. 38 - Prob. 87PCh. 38 - Prob. 88PCh. 38 - Prob. 89PCh. 38 - Prob. 90P
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- The work function for potassium is 2.26 eV. What is the cutoff frequency when this metal is used as photoelectrode? What is the stopping potential when for the emitted electrons when this photo electrode is exposed to radiation of frequency 1200 THz?arrow_forwardWhat is the de Brogue wavelength of a proton whose kinetic energy is 2.0 MeV? 10.0 MeV?arrow_forwardAt what velocity will an electron have a wavelength of 1.00 m?arrow_forward
- What is the de Brogue wavelength of an electron that is accelerated from rest through a potential difference of 20 keV?arrow_forwardThe de Brogue wavelength of a neutron is 0.01 nm. What is the speed and energy of this neutron?arrow_forwardThe neutron has a mass of 1.67 × 10-27 kg. Neutrons emitted in nuclear reactions can be slowed down by collisions with matter. They are referred to as thermal neutrons after they come into thermal equilibrium with the environment. The average kinetic energy (3/2 kBT) of a thermal neutron isapproximately 0.04 eV. (a) Calculate the de Broglie wavelength of a neutron with a kinetic energy of 0.040 0 eV. (b) How does your answer compare with the characteristic atomic spacing in a crystal? (c) Explain whether you expect thermal neutrons to exhibit diffraction effects when scattered by a crystal.arrow_forward
- The neutron has a mass of 1.67 ✕ 10-27 kg. Neutrons emitted in nuclear reactions can be slowed down via collisions with matter. They are referred to as thermal neutrons once they come into thermal equilibrium with their surroundings. The average kinetic energy (3kB T/2) of a thermal neutron is approximately 0.04 eV.Calculate the de Broglie wavelength of a neutron with a kinetic energy of 0.0980 eV.arrow_forwardCompare the de Broglie wavelength of a bullet moving at 700 miles per hour (313 m/s) to that of an alpha particle moving at 3.40×107 miles per hour (1.52×107 m/s) and a baseball with a speed of 90.0 miles per hour (40.2 m/s) . Particle Mass (kg) Velocity (m/s) Region Wavelength bullet 0.00190 313 alpha particle 6.64*10^-27 1.52*10^7 baseball 0.140 40.2arrow_forwardOn average, people in Adelaide have a mass of m would have a de Broglie wavelength of A N 3.66 x 10 36 85.5 kg and walk at a speed of v = 2.12 ms, If we were to consider such a person to be a particle with mass m and speed v, then they m when walking. Explain why the average Adelaidean is not observed to diffract when walking through a doorway.arrow_forward
- The de Broglie wavelength of an electron has to do with spatial resolution of an electron microscope, which is often expressed in the unit of length Å (Angstrom). The 1 V potential difference causes an electron to gain kinetic energy EK of 1 electron Volt (eV). In the SI units, kinetic energy in eV must be converted to Joules. The conversion factor is 1 eV = 1.6 x 10-19 Joule. And, the formula for the wavelength is: λ = h / p = h / √(2 m EK) where m is electron mass. Calculate the de Broglie wavelength of an electron (in Å) when the electron is accelerated from rest through a potential difference of: a) 1 kV = 1,000 V (a low resolution setting of microscope), b) 10 kV = 10,000 V (intermediate resolution). c) 100 kV = 100,000 V (high resolution),arrow_forwardThe wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which(a) an electron, and(b) a neutron, would have the same de Broglie wavelengtharrow_forwardAn alpha particle (m = 6.64 × 10−27 kg) emitted in the radioactive decay of Uranium-238 has an energy of 4.20 MeV. What is its de Broglie wavelength?arrow_forward
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