Connect with LearnSmart for Krauskopf: The Physical Universe, 16e
16th Edition
ISBN: 9781259663895
Author: KRAUSKOPF, Konrad B.
Publisher: Mcgraw-hill Higher Education (us)
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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
Connect with LearnSmart for Krauskopf: The Physical Universe, 16e
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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 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_forward
- A) 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_forwardThe diameter of an atomic nucleus is about 10 fm (1 fm = 10-15 m). What is the kinetic energy, in MeV, of a proton with a de Broglie wavelength of 10 fm?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_forward
- The 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_forwardIn 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_forward
- A 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_forwardThe electron beam in a scanning electron microscope (SEM) can have an energy ranging from 0.2 keV to 40 keV. (Note: an eV is an electronvolt, the measure of an amount of kinetic energy for a single electron. 1 eV = 1.60218×10−19 J and 1 J = 1 kg m2 s−2). What is the de Broglie wavelength of the electron beam in the SEM when the electrons are accelerated to kinetic energy of 6.6 keV?arrow_forwardAn electron has a kinetic energy of 4.1 × 10–¹³ J. 46. What is the electron's speed? (a) 0.960c (b) 0.968c (c) 0.972c (d) 0.980c (e) 0.986c 47. What is the de Broglie wavelength of the electron described? (a) 4.09 × 10-13 m (b) 4.94 × 10-13m (c) 5.42 × 10-13 m (d) 6.25 × 10-13 m (e) 7.81 x 10-13marrow_forward
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