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6 Electronic Structure of Atoms Solutions to Exercises The ground state energy for hydrogen-like particles is: E = RHZ². (By definition, n = 1 for the ground state of a one-electron particle.) (c) Z = 6 for C⁵⁺. E = -2.18 J = -7.85 6.88 Plan. Change keV to J/electron. Calculate V from kinetic energy. = h/mv. Solve. 18.6 keV 1000 keV 96.485 mol 1000 6.022 1 mol electrons = 2.980 10⁻¹⁵ = 2.98 10⁻¹⁵ J/electron V 8.089 10⁷ = 8.09 10⁷ m/s = h/mv = 9.1094 10⁻³¹ 6.626 kg 10⁻³⁴ 8.089 10⁷ m/s = 8.99 10⁻¹² m = 8.99 pm 6.89 Heisenberg postulated that the dual nature of matter places a limitation on how precisely we can know both the position and momentum of an object. This limitation is significant at the subatomic particle level. The Star Trek transporter (presumably) disassembles humans into their protons, neutrons and electrons, moves the particles at high speed (possibly the speed of light) to a new location, and reassembles the particles into the human. Heisenberg's uncertainty principle indicates that if we know the momentum (mv) of the moving particles, we can't precisely know their position (x). If a few of the subatomic particles don't arrive in exactly the correct location, the human would not be reassembled in their original form. So, the "Heisenberg compensator" is necessary to make sure that the transported human arrives at the new location intact. 6.90 (a) (b) n and (c) (d) 6.91 (a) Probability density, is the probability of finding an electron at a single point at distance r from the nucleus. The radial probability function, is the probability of finding an electron at any point on the sphere defined by radius r. (b) The term explains the differences in plots of the two functions. Plots of the probability density, for S orbitals shown in Figure 6.21 each have their maximum value at r = 0, with smaller maxima at greater values of r. The plots of radial probability, P(r), for the same S orbitals shown in Figure 6.18 have values of zero at r = 0 and the size of the maxima increases. P(r) is the product of and At r = 0, the value of is finite and large, but the value of 4πr2 is zero, so the value of P(r) is zero. As r increases, the values of vary as shown in Figure 6.21, but the values of 4πr2 increase continuously, leading to the increasing size of P(r) maxima as r increases. 160