CHEM-6 Midterm: Chem 6 Dartmouth Spring Exam 2S
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Chem 6 sample exam 2 solutions: (a) (1/2)mv2 = h . Since 274 nm is the maximum wavelength that will eject electrons, e = hc/ = . = (6. 626x10 34js) (3x108ms 1)/(274 nm) (1m/109 nm) = 7. 25x10 19j. Convert to kj/mol (7. 25x10 19j) (1kj/1000j) (6. 02x1023/mol) = 437 kj/mol (b) (1/2)mv2 = h . = hc/ = (6. 626x10 34js) (3x108ms 1)/(100 nm) (1m/109 nm) = 1. 99x10 18j 7. 25x10 19j = 1. 26x10 18j (c) ke = 1/2mv2 = 1. 26x10 18j v = [2 (1. 26x10 18j)/me]1/2 where me = electron mass = 9. 11x10 31kg so v = 1. 66x106ms 1. This is less than the speed of light, as it should be, which provides a useful check on the magnitude of the answer: the heisenberg uncertainty principle states that. In this problem, the uncertainty in position, x, is 2 times the radius of the nucleus, 2x10 15m. Find the uncertainty in the kinetic energy of the electron e.
