Physical Universe
16th Edition
ISBN: 9780077862619
Author: KRAUSKOPF, Konrad B. (konrad Bates), Beiser, Arthur
Publisher: Mcgraw-hill Education,
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Chapter 9, Problem 24E
To determine
The experiment used to differentiate an electron and gamma ray of same wavelength.
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Chapter 9 Solutions
Physical Universe
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- What is the de Brogue wavelength of a proton whose kinetic energy is 2.0 MeV? 10.0 MeV?arrow_forwardThe 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.)arrow_forwardWhat is the de Broglie wavelength of a proton whose kinetic energy is 2.0 MeV?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_forwardWhat is the de Broglie wavelength of a proton whose kinetic energy is 2.0 MeV? 10.0 MeV?arrow_forwardA) Calculate the de Broglie wavelength of a neutron (mn = 1.67493×10-27 kg) moving at one six hundredth of the speed of light (c/600). (Enter at least 4 significant figures.) B) Calculate the velocity of an electron (me = 9.10939×10-31 kg) having a de Broglie wavelength of 230.1 pm.arrow_forward
- The diameter of the nucleus is about 12.1fm. What is the kinetic energy of a proton with a de Broglie wavelength of 12.1fm?arrow_forwardThe 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 energy conservation principle that applies to the photoelectric experiment is Ephoton = (KE)electron + W, where W is the “work function” for the metal. (The work function is the minimum energy required to eject an electron from the metal surface.) The work function for calcium metal is 4.60 10−19 J. If calcium is irradiated with 400-nm photons, what is the de Broglie wavelength of the resulting photoelectron beam?arrow_forward
- 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.arrow_forwardDetermine the de Broglie wavelength of a neutron (m = 1.67 × 10^–27 kg) that has a speed of 5.0 m/s.arrow_forwardA certain metal has a work function of 233.0 kJ per mole of e_. What wavelength of electromagnetic radiation (in nm) must be directed at the surface in order for electrons to be ejected with a de Broglie wavelength of 14.18 angstroms (Å) 1 Å = 1×10-10 metersarrow_forward
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