Solutions to General Chemistry (Linus Pauling)/The Electron the Nuclei of Atoms and the Photon

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  1. An ordinary electric light bulb is operated under conditions such that one ampere of current (1 C s-1) is passing through the filament. How many electrons pass through the filament each second? (Remember that the charge of the electron is -0.160 × 10-18 C.)
    6.25 × 1018 electrons are passing through the filament each second. The calculation should be fairly self-evident.
  2. According to the law of gravitation, the gravitational force of attraction between two particles between two particles with masses m1 and m2 a distance r apart is Gm1m2 / r2, where G, the constant of gravitation has been found by experiment to have the value 0.6673 × 10-10 N m2 kg -2; that is, the force of attraction between two particles each with mass 1 kg and the distance 1 m apart is 0.6673 × 10-10 N
    1. Calculate the force of electrostatic attraction between an electron and a proton 10 Å apart.
      The appropriate rule to apply here is (the scalar form of) Coulomb's Law. To the reader's ire, Coulomb's law is given using Stoney units, while the charges of electrons and protons are typically given in coulombs. The constant required to calculate the electrostatic force is given later on in the chapter and is 8.9876 × 109. That being said, the electrostatic force between the two particles is 8.9876 × 109 × -1.60217646 × 10-19 C × 1.60217646 × 10-19 C / (10-9)2, or about -2.30708943 × 10-10 N. Note the sign; it's important. A negative sign implies an attractive force, while a positive sign implies a repulsive force. Because the electron and proton are of opposite polarity, they attract.
    2. Calculate the force of gravitational attraction between an electron and a proton 10 Å apart. What is the relationship of the electrostatic attraction to the gravitational attraction at this distance?
      Here the formula is given directly and is straightforward once the masses are known. The mass of an electron is 9.10938188 × 10-31 kg; the mass of a proton, 1.67262158 × 10-27 kg. The gravitational attraction then is 1.0167349 × 10-49 N, according to the law of gravitation.[1]
    3. What is the dependence on distance of the ratio of electrostatic attraction and gravitational attraction of an electron and proton?
      This question seems vague, but the author appears to be referring to only the ratio of electrostatic and gravitational force given the same 10 Å distance. It is 2.269116 × 1039.
  3. Calculate the velocity with which the electrons would move in the apparatus used by J.J. Thomson, operated at an accelerating voltage of 6000 V. Assume that each electron has kinetic energy equal to eV, where e is the charge of the electron in coulombs and V is the accelerating potential in volts.
    The kinetic energy of the electron is given by:
    eV = \frac{1}{2}mv^2
    Inserting known values:
    1.60217646 \cdot 10^{-19} \cdot 6000 = \frac{1}{2} \cdot 9.10938188 \cdot 10^{-31} \cdot v^2
    Accordingly, the velocity is about 45,941,095 meters per second. Is it necessary to use relativity here?

Notes[edit]

  1. This example presents a good opportunity to demonstrate the abilities of Google's scientific calculator