# Model assume that the electrons transfer all their

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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

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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 = = ×
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%.

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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
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