Model assume that the electrons transfer all their

Info icon This preview shows pages 34–38. Sign up to view the full content.

View Full Document Right Arrow Icon
Model: Assume that the electrons transfer all their energy to the foil. Solve: (a) From Chapter 17, the rise in thermal energy of the foil is ( ) ( )( ) 6 th 10 10 kg 385 J/kg K 6.0 C 0.0231 J E mc T Δ = Δ = × ° = This energy is provided by the electron impacts. Each electron that strikes the foil has been accelerated through a potential difference of 2000 V. Its kinetic energy when it strikes the foil is K = e Δ V = 2000 eV = 3.2 × 10 –16 J. The number of electrons that strike the foil in 10 seconds is 13 th 16 7.2 10 3.2 10 J E Δ = × × (b) The current of the electron beam is ( )( ) 13 19 7.2 10 1.6 10 C 1.2 A 10 s q t μ × × Δ = = Δ
Image of page 34

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
38.35. Model: Model the Li atom as a single valence electron orbiting a sphere with net charge q = + 1 e due to the 3 protons and 2 inner electrons. A sphere of charge acts like a point charge with the total charge concentrated at the center of the sphere. Visualize: Solve: The electron’s energy is both kinetic and potential: E = 1 2 mv 2 + kq (– e )/ r . To say that the energy needed to ionize the atom is 5.14 eV means that you would need to increase the electron’s energy by 5.14 eV to remove it from the atom, taking it to r . Since E = 0 for charged particles that are infinitely separated, the energy of the atom must be E = –5.14 eV = –8.224 × 10 –19 J. Negative energy indicates that the system is bound, and the absolute value is the binding energy . Thus from energy considerations we learn that 2 2 19 1 2 8.224 10 J Ke E mv r = = − × The Coulomb force on the electron provides the centripetal acceleration of circular motion. The force equation is 2 2 2 2 1 2 2 2 r r Ke mv Ke F ma mv r r r = = = = Substitute this expression for the kinetic energy into the energy equation: 2 2 2 2 2 19 1 2 9 2 2 19 2 10 19 8.224 10 J 2 2 (8.99 10 N m /C )(1.60 10 C) 1.40 10 m = 0.140 nm 2(8.224 10 J) Ke Ke Ke Ke mv r r r r r = = − = − × × × = = × × With the radius now known, we can use the result of the force equation to find that 2 6 1.34 10 m/s Ke v mr = = ×
Image of page 35
38.36. Model: The mass of an atom is concentrated almost entirely in the nucleus. Solve: The volumes of the nucleus and the atom are 3 14 42 3 nucleus 4 1.0 10 m 0.524 10 m 3 2 V π × = = × 3 10 31 3 atom 4 1.2 10 m 9.05 10 m 3 2 V π × = = × 42 3 13 nucleus 31 3 atom 0.524 10 m 5.8 10 9.05 10 m V V × = × × Thus 5.8 × 10 –11 % = 0.000000000058% of the atom’s volume contains all the mass. The percent of empty space in an atom is 99.999999999942%.
Image of page 36

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
38.37. Solve: (a) Try relating the wavelengths to the number 125.00. We get 2 2 2 2 125.00 125.00 125.00 125.00 125.00 125.00 31.25 13.89 7.81 5.00 1 2 3 4 5 = = = = = Thus, the series can be represented by the formula ( ) 2 125.00 nm n λ = , n = 1, 2, 3, . . . (b) Notice that the common denominator of each of the wavelengths is 75. The six wavelengths are 5 75, 12 75, 21 75, 32 75, 45 75, and 60 75 nm. The series 5, 12, 21, 32, 45, 60 is increasing faster than linear, but these aren’t squares of integers. However, a little thought finds that each of these is 4 less than the square of an integer. That is, 5 = 3 2 4, 12 = 4 2 – 4, 21 = 5 2 – 4, and so on. That is, the series of multipliers is ( n 2 – 4) with n = 3, 4, 5, . . . Thus the
Image of page 37
Image of page 38
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern