PS1-08-Solution - Chem 442 Problem Set #1 - Solutions DUE:...

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Chem 442 Problem Set #1 - Solutions DUE: Wednesday, Nov. 5, 2008 1) (McQ&S 1-8) 3 max 3 10 max 7 10 max Using the Wien displacement law (eqn 1.4) or Planck's law (1.5): 2.90 10 2.90 10 2.90 10 10 Considering sig figs, we have 3 10 0.3 This is in the x-ray region. Tm K mK m K mn m λ ×⋅ == × = 2a) (McQ&S 1-19) 7 9 8 15 7 Using equation 1.6, we have where is the work function. We must convert wavelength into frequency using . 1 Example calc: 100 1.00 10 10 2.998 10 / 31 0 1.00 10 It will KE hv cv m nm m nm ms vH z m φ = = ×= × × × × 18 18 also be convenient to convert the energy into J. 1J Example calc: 10.1eV 1.62 10 6.241506 10 Now plot KE vs . The plot is shown on the next page. The linear regression line is (to one Sig Fig) J eV v × × 34 19 : 71 0 41 0 where KE has units of J and has units of Hz. KE v v −− −× -34 34 -19 Slope = h = 7 10 / 7 10 Intercept=Work function = 4 10 JH z J S J × ×
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Problem 2a - KE vs. Freq KE = 7E-34v - 4E-19 0 2E-19 4E-19 6E-19 8E-19 1E-18 1.2E-18 1.4E-18 1.6E-18 1.8E-18 0 5E+14 1E+15 1.5E+15 2E+15 2.5E+15 3E+15 3.5E+15 Freq (Hz) KE (J) 2b) 18 -19 -19 6.241506 10 Work function = 4 10 2.49 2 (to one sig fig) or if you carry more sig figs into the calculation: Work function = 3.59 10 2.24 2 (to one sig fig - same as above) This is eV Je V e V J V e V × ×× = = ×= = not the same as the ionization potential for Na, nor should it be. The work function represents the minimum energy of radiation to remove an electron from the surface of a solid block of Na. The ionization potential represents the energy required to remove an electron from a gaseous Na atom.
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This note was uploaded on 02/12/2011 for the course CHEMISTRY 442 taught by Professor Lizzy during the Spring '10 term at Hanoi University of Technology.

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PS1-08-Solution - Chem 442 Problem Set #1 - Solutions DUE:...

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