a)-8,78.10^11 B)8.78.10^11 NS)7.89.10^21 D)8.10^-19 TO)-7,89.10^21 According to the figure, what is the acceleration of the electron in m/s^2 (me=9.111.10^-31 kg)?2)9e= -16-10one -

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Answered: According to the figure, what is the…

An Introduction to Physical Science

14th Edition

ISBN:9781305079137

Author:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres

Publisher:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres

Chapter9: Atomic Physics

Section: Chapter Questions

Problem 12FIB

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Which of the following are not allowed? State which rule is violated for any that are not allowed. (a) 1p3(b) 2p8(c) 3g11(d) 4f2
Cosmic rays are high-energy charged particles produced by astronomical objects. Many of the cosmic rays that make their way to the Earth are trapped by the Earths magnetic field and never reach the surface. These trapped cosmic rays are found in the Van Allen beltsdonut-shaped zones over the Earths equator (Fig. 30.34). These cosmic rays are mostly protons with energies of about 30 MeV. The inset in the figure shows a cosmic ray proton as it is about to enter the Earths magnetic field. The cosmic rays velocity is initially perpendicular to the field. Three students discuss what happens to the incoming cosmic ray. Decide which student or students are correct. Figure 30.34 The Van Allen belts are donut-shaped zones of trapped cosmic rays above the Earths surface. Inset: What happens to this cosmic ray as it enters the Earths magnetic field? Shannon: The velocity is perpendicular to the magnetic field, so the cosmic ray just passes through the field and hits the Earths atmosphere. Avi: What you are saying is that the magnetic field exerts no force on the cosmic ray. Actually, it exerts a huge force because the velocity is perpendicular to the magnetic field. The force will be into the page. Cameron: Avi is right. The cosmic ray proton is going to feel a huge magnetic force. Because it is positively charged, it will be pushed upward along the magnetic field lines. Shannon: I never said the force was zero. There is a force, but the force is perpendicular to the magnetic field lines. In this case, thats to the lefttoward the Earth. Avi: The force is perpendicular to the magnetic field, but it also has to be perpendicular to the velocity. Because B and v are both in the plane of the page, the force must be perpendicular to the page.
Integrated Concepts In a Millikan oil-drop experiment using a setup like that in Figure 30.9, a 500-V potential difference is applied to plates separated by 2.50 cm. (a) What is the mass of an oil drop having two extra electrons that is suspended motionless by the field between the plates? (b) What is the diameter of the drop, assuming it is a sphere with the density of olive oil?
In 1911, Ernest Rutherford and his assistants Geiger and Marsden conducted an experiment in which they scattered alpha particles (nuclei of helium atoms) from thin sheets of gold. An alpha particle, having charge +2e and mass 6.64 10-27 kg, is a product of certain radioactive decay's. The results of the experiment led Rutherford to the idea that most of an atoms mass is in a very small nucleus, with electrons in orbit around it. (This is the planetary model of the atom, which well study in Chapter 42.) Assume an alpha particle, initially very far from a stationary gold nucleus, is fired with a velocity of 2.00 107 m/s directly toward the nucleus (charge +79e). What is the smallest distance between the alpha particle and the nucleus before the alpha particle reverses direction? Assume the gold nucleus remains stationary.
In the classical model of a hydrogen atom, an electron orbits a proton with a kinetic energy of +13.6 eV and an electric potential energy of 27.2 eV. (a) Use the kinetic energy to calculate the classical orbital speed. (b) Use the electric potential energy to calculate the classical orbital radius.
Where in an atom would you expect to find electrons? Protons? Neutrons?
Calculate the speed of (a) an electron and (b) a proton with a kinetic energy of 1.00 electron volt (eV). (c) Calculate the average translational kinetic energy in cV of a 3.00 102 K ideal gas particle. (Recall from Topic 10 that 12mv2 = 32kBT.)
The force on an electron is “negative the gradient of the potential energy function.” Use this knowledge and Equation 8.1 to show that the force on the electron in a hydrogen atom is given by Coulomb’s force law. Ur=ke2r(8.1)
In the Millikan oil-drop experiment illustrated in Figure 15.21, an atomizer (a sprayer with a fine nozzle) is used to introduce many tiny droplets of oil between two oppositely charged parallel metal plates. Some of the droplets pick up one or more excess electrons. The charge on the plates is adjusted so that the electric force on the excess electrons exactly balances the weight of the droplet. The idea is to look for a droplet dial has the smallest electric force and assume it has only one excess electron. This strategy lets the observer measure the charge on the electron. Suppose we are using an electric field of 3 104 N/C. The charge on one electron is about 1.6 1019 C. Estimate the radius of an oil drop of density 858 kg/m5 for which its weight could be balanced by the electric force of this field on one electron. (Problem 42 is courtesy of E.F. Redish. For more problems of this type, visit www.physics.umd.cdu/pcrg/.)
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For an electron in a three-dimensional metal, show that the average energy is given by E=1N0EFEg(E)dE=35EF , Where N is the total number electrons in the metal.

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According to the figure, what is the acceleration of the electron in m/s^2 (me=9.111.10^-31 kg)? 2) 9e: -16-10