This preview shows page 1. Sign up to view the full content.
Unformatted text preview: S117 Homework 2 Out of 40 Points Chapter 12 (not to turn in) 37. a. False; it takes less energy to ionize the electron from an excited state. b. True c. False; the energy difference is greater between and than between and , so the wavelength of light emitted by an to transition is longer than that emitted by an to transition. d. True e. False; the first excited state corresponds to . 44. The units of are . The units of are . Thus, the units are the same in either expression of the Heisenberg uncertainty principle. 45. The Heisenberg uncertainty principle is the mass is known, , a. b. c. According to Figure 12.38, the atomic radius of hydrogen is , so the diameter is . Clearly, the uncertainty is position of the electron in part a. is much greater than the size of a hydrogen atom. d. A baseball is at least several centimeters across. Its wavelength under these conditions is extraordinarily minute. In fact, even compared to the hydrogen atom, it is incredibly small. 55. a. This is not allowed because l can never be as great as n. b. This is allowed. c. This is allowed. d. This is not allowed because ms must be . e. This is not allowed because l cannot be negative. f. This is not allowed because ml cannot be greater than l. 56. Three orbitals can have the designation 5p. Only one orbital can have the designation . Five orbitals can have the designation 4d. orbitals can have the designation . These correspond to one s orbital, three p orbitals, five d orbitals, seven f orbitals, and nine g orbitals. orbitals can have the designation . ; thus, . In the case that 57. No electrons can have the designation 1p. Two electrons can have the designation . Fourteen electrons can have the designation 4f. Two electrons can have the designation 7py. Two electrons can have the designation 2s. electrons can have the designation . Chapter 12 (to turn in) 46. (10 points) An electron in a one-dimensional box has energy energy levels is then requires . This corresponds to light of wavelength . 50. (5 points) The wave function for a particle in a box at finding the particle at some x is is ; thus, the probability of . The change in energy between to in a box of length . To transition from . The plot of this is given below, where the yaxis is 2 and the x-axis is in units of L. Thus, the plot shows the probability of finding the particle at every point in the box. Between and , one half of one "hill" exists. Each "hill" consists of one third of the total probability, so the probability of finding the particle between and is one sixth. , but there are many Alternatively, integration can solve this problem. steps involved in that integration. 62. (5 points) The only part of this function which can ever go to 0 is the 3s orbital of the hydrogen atom are given by the solutions to and , and . Now , so ; therefore, the nodes in . . 121. (5 points) Removing the third electron requires much more energy than removing the first two electrons, so the third electron is apparently a core electron. This means that the element is likely in group 2A, the alkaline earth metals. 129. (10 points) a. Since the lines increase in wavelength going from left to right on the spectrum, they corresponding to decreasing energy. (Another way of knowing this is that, for one-electron species, the energy level spacings decrease as n increases. Since the lines on the left of the spectrum are closer together, they correspond to higher values of n.) Since all of these lines correspond to transitions to the state, the right-most line must correspond to a transition from to . Thus, line B corresponds to the to transition, and line A corresponds to the to transition. b. For a one-electron atom or ion, the following relationship is true: Z is the atomic number, RH is the Rydberg constant, and n1 and n2 are principal quantum numbers. This equation can be rearranged to . The wavelength of line B is , where 142.5nm; one of the values of n is 5, and one of the values of n is 3. Substituting values into the previous equation, . The wavelength of A can be solved by using the formula in another rearranged form, . Here, Z has been calculated to be 3; one of the values of n is 6, and one of the values of n is 3. Substituting in, . 133. (5 points) a. The three combinations of quantum numbers which minimize the energy in the equation are and , and , and and , corresponding to the ground state, the first excited state, and the second excited state, respectively. b. The change in energy when the particle is promoted from the first to the second excited state is given by , corresponding to a wavelength of . ...
View Full Document
- Fall '11