SampleEx2S - Chem 6 sample Exam 2 solutions 1(a(1/2)mv2 = h...

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Chem 6 sample Exam 2 solutions 1 . (a) (1/2)mv 2 = h ν φ . Since 274 nm is the maximum wavelength that will eject electrons, E = hc/ λ = φ = (6.626x10 –34 Js) (3x10 8 ms –1 )/(274 nm) (1m/10 9 nm) = 7.25x10 –19 J Convert to kJ/mol (7.25x10 –19 J) (1kJ/1000J) (6.02x10 23 /mol) = 437 kJ/mol (b) (1/2)mv 2 = h ν φ = hc/ λ φ = (6.626x10 –34 Js) (3x10 8 ms –1 )/(100 nm) (1m/10 9 nm) = 1.99x10 –18 J – 7.25x10 –19 J = 1.26x10 –18 J (c) KE = 1/2mv 2 = 1.26x10 –18 J v = [2 (1.26x10 –18 J)/m e ] 1/2 where m e = electron mass = 9.11x10 –31 kg so v = 1.66x10 6 ms –1 This is less than the speed of light, as it should be, which provides a useful check on the magnitude of the answer. 2 . The Heisenberg Uncertainty Principle states that x p h/4 π In this problem, the uncertainty in position, x, is 2 times the radius of the nucleus, 2x10 –15 m. Then p (h/4 π )(1/2x10 –15 m) = 2.64x10 –20 kgms –1 Find the uncertainty in the kinetic energy of the electron E E ( p) 2 /2m Plug in the electron mass m e = 9.11x10 –31 kg and the value for p to get
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E 3.82x10 –10 J This uncertainty in the energy is greater than the potential energy barrier supposedly holding the electron inside the nucleus! Therefore, the electron cannot really be held inside the nucleus. [Note: another way of looking at this problem is to find the uncertainty in the electron's velocity, v. Since p=mv, then p = m v or v = ( p)/m = (2.64x10 –20 kgms –1 )/9.11x10 –31 kg = 2.9x10 10 ms –1 . The uncertainty in the velocity is thus larger than the speed of light, which is again physically unreasonable. Thus, same conclusion as above -- the electron can't be stuck inside the nucleus.] 3 . This is the photoelectric effect applied to an atom instead of a metal surface.
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This note was uploaded on 05/01/2008 for the course CHEM 5/6 taught by Professor Gribble during the Spring '08 term at Dartmouth.

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SampleEx2S - Chem 6 sample Exam 2 solutions 1(a(1/2)mv2 = h...

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