Physics 43 HW 7 T: E: 7, 15, 17 P: 27, 30, 36, 40

38.7. Model: The electron volt is a unit of energy and is defined as the kinetic energy gained by an electron or proton if it accelerates through a potential difference of 1 volt. Solve: (a) Converting electron volts to joules, 1.60× 10−19 J 100 eV =× 100 eV =×1.60 10−17 J 1 eV

1 2 Using the definition of kinetic energy K = 2 mv ,

−17 2K 21.6010()× J v == =×5.9 106 m/s m 9.11× 10−31 kg (b) Likewise, the speed of the neutron is ()(2 5.0 MeV ) 2() 8.0× 10−13 J v == =3.1× 107 m/s 1.67×× 10−−27 kg 1.67 1027 kg (c) The mass of the particle is 23.3410× 13 J 2K 22.09() MeV ( ) −27 m ==2 22 = =6.69 × 10 kg v ()1.0×× 1077 m/s() 1.0 10 m/s The mass of the particle is the same as the mass of four protons (or two protons and two neutrons). It is an alpha particle.

38.15. Model: For a neutral atom, the number of electrons is the same as the number of protons, which is the atomic number Z. An atom’s mass number A is A = Z + N, where N is the number of neutrons. Solve: (a) This is a neutral atom with the symbol 10Be. (b) This is a doubly positively charged ion with A = 6 + 5 = 11. We have 6 protons, but there are only 4 electrons. The symbol is 11C++.

38.17. Model: For a neutral atom, the number of electrons is the same as the number of protons, which is the atomic number Z. An atom’s mass number is A = Z + N, where N is the number of neutrons. Solve: (a) For a 197Au atom, Z = 79. So, N = 197 – 79 = 118. A neutral 197Au atom contains 79 electrons, 79 protons, and 118 neutrons. (b) Assuming that the neutron rest mass is the same as the proton rest mass, the density of the gold nucleus is 197 1.67× 10−27 kg 197mproton ( ) 17 3 ρ nucleus ==33 =2.29× 10 kg/m 44ππ−−15 15 33()7.0×× 10 m() 7.0 10 m (c) The nuclear density in part (b) is 2.0 × 1013 times the density of lead. Assess: The mass of the matter is primarily in the nuclei and the volume of the matter is essentially due to the electrons around the nuclei.

38.27. Model: Use the relativistic expression for the total energy. Solve: (a) The energy of the proton is

2 1.67×× 10−27 kg 3.0 108 m/s −19 2 ()( ) 9 1.60× 10 J Emc==γ p 500 GeV = =××500 10 eV 1− vc22 1 eV ⇒−1vc22 = 1.879 × 10−3 ⇒=v 0.999998 c (b) Likewise for the electron, 2 (9.11×× 10−31 kg)( 3.0 108 m/s) E ==2.0 GeV 1− vc22

⇒−1vc22 = 2.562 × 10−4 ⇒=v 0.99999997 c

38.30. Model: Mass and energy are equivalent. Solve: The energy released as kinetic energy is

22−−7812 11 eV KE=Δ =() Δ mc = (0.185 )() 1.66× 10 kg() 3.0× 10 m/s = 2.76 × 10 J ×−19 = 173 MeV 1.6× 10 J

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

−14 3 −10 3 4π ⎛⎞ 1.0× 10 m −42 3 4π ⎛⎞ 1.2× 10 m −31 3 Vnucleus ==⎜⎟0.524× 10 m Vatom ==⎜⎟9.05× 10 m 32⎝⎠ 32⎝⎠

−42 3 Vnucleus 0.524× 10 m −13 ⇒=−31 3 ≈×5.8 10 Vatom 9.05× 10 m 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%.

38.40. Model: The nucleus of an atom is very small and it contains protons and neutrons. Solve: (a) The repulsive electric force between two protons in the nucleus is

922− 192 1 e2 (8.99×× 10 N m /C)( 1.60 10 C) FE == =58 N 4πε 22−15 0 ()2.0 fm ()2.0× 10 m (b) The attractive gravitational force between two protons in the nucleus is 2 2 6.67×× 10−−11 N m 2 /kg 2 1.67 1027 kg Gm ()( ) −35 FG ==22=×4.7 10 N ()2.0 fm ()2.0× 10−15 m

Because FG << FE, gravitational force could not be the force to hold two protons together. (c) The nuclear force must be very strong to overcome FE and it must be independent of charge because both protons and neutrons are held in the nucleus very tightly. Furthermore, nuclear force is a very short range force since it is not felt outside the nucleus.