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# Hwksolns11_Publishers - 38.5(a The atomic number of an...

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38.5. (a) The atomic number of an element, represented by Z, is the number of electrons in a neutral atom and the number of protons in the nucleus. The atomic number of hydrogen is Z = 1, helium is Z = 2, and lithium is Z = 3. However, the masses of these three elements do not progress in this linear manner. The mass of helium is four times the mass of hydrogen and the mass of lithium is seven times the mass of hydrogen. If a nucleus contains only Z protons, then helium should be just twice as massive as hydrogen and lithium three times as massive as hydrogen. So, there is something else in the nucleus. (b) Thomson and his student Aston found that many elements consist of atoms of differing mass. For example, neon (Z = 10) atoms are found to have integral masses 20 u, 21 u, and 22 u. Potassium (Z = 19) atoms are found to have integral masses 39 u, 40 u, and 41 u. Because the masses of the different isotopes differ by 1 u, the mass of neutron must be 1 u, the same as the mass of a proton.

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38.6. Larger than because the force exerted by a magnetic field on the charged particle to deflect it is proportional to the amount of charge on the particle. An alpha particle has more charge than a cathode-ray particle.
38.7. Scientists at the time could not imagine the extremely high density of the tiny nucleus. They also had no idea what would hold protons together in a nucleus when there was a known repulsive force between protons.

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38.8. Alpha particles scattered through large angles because they had near collisions with very massive and highly charged particles. Only small deflections were expected for an alpha particle passing through a Thomson atom.
38.9. (a) 6 Li + (b) 13 C

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38.2. Model: Assume the fields between the electrodes are uniform and that they are zero outside the electrodes. Visualize: Please refer to Figures 38.7 and 38.8. Solve: (a) The speed with which a particle can pass without deflection is 3 7 3 / 600 V/5.0 10 m 6.0 10 m/s 2.0 10 T EV d v BB Δ× == = = × × (b) The radius of cyclotron motion in a magnetic field is 31 7 19 3 9.11 10 kg 6.0 10 m 0.17 m 17 cm 1.60 10 C mv r eB −− ⎛⎞ ×× ⎜⎟ ⎝⎠
38.3. Model: Assume the fields between the electrodes are uniform and that they are zero outside the electrodes. Visualize: Without the external magnetic field B , the electrons will be deflected up toward the positive electrode. The magnetic field must therefore be directed out of the page to exert a balancing downward force on the negative electron. Solve: In a crossed-field experiment, the magnitudes of the electric and magnetic forces on the electron are given by Equation 38.4. The magnitude of the magnetic field is 3 3 6 / 200 V/8.0 10 m 5.0 10 T 5.0 10 m EV d B vv Δ× == = = × × Thus G B = (5.0 × 10 –3 T, out of page).

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38.4. Model: Assume the electric field ( E = Δ V / d ) between the plates is uniform.
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Hwksolns11_Publishers - 38.5(a The atomic number of an...

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