Midterm_Fa 11_solutions

Midterm_Fa 11_solutions - MatE 153 Fall 2011 Midterm...

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Unformatted text preview: MatE 153 Fall 2011 Midterm Solutions ____ / 80 Total Points = _____% 1.) A tungsten wire (W) heats up to 80ºC during use. a. What is the resistivity of the wire at 80ºC? (10 points) ρ = ρ o [1 + α o (T − T0 )] T = 80 + 273 = ⎡ ⎣ ρ = 50nΩ ⋅ m ⎢1 + 1 (353 − 273)⎤ = 69.8nΩ ⋅ m ⎥ 202 ⎦ b. What is the electron mobility at 80ºC? (10 points) n= # valenceelectron / atom ⋅ density ⋅ N A = MA g atoms 6.02 × 10 23 3 mol = 2.53 × 10 23 electrons = 2.53 × 10 29 electr cm = g cm 3 m3 183.84 mol 2 σ 1 1 −4 m μ= = = = 3.53 × 10 electr V A⋅ s en enρ V ⋅s ⋅ 69.8 × 10 −9 Ω ⋅ m 1.6 × 10 −19 C ⋅ 2.53 × 10 29 3 A⋅Ω C m 4electron / atom ⋅ 19.3 2.) In a photoelectric effect experiment on silver (Ag) target, the measure stopping voltage was 7 eV. The intensity of light was 30 mW/m2. The work function of Ag is 4.26 eV. a. What was the wavelength of light used in the experiment? (10 points) KE = Elight − Φ Elight = 7eV + 4.26eV = 11.26eV 1.6 × 10 −19 J hc = 1.80 × 10 −18 J = eV λ m 6.626 × 10 −34 Js ⋅ 3 × 10 8 hc s = 1.1 × 10 −7 m = 110nm λ= = Elight 1.80 × 10 −18 J Elight = 11.26eV b. What was the momentum of the light? (5 points) kg ⋅ m 2 −34 6.626 × 10 J ⋅ s h J ⋅ s 2 = 6 × 10 − 27 kg ⋅ m p= = −7 λ s 1.1 × 10 m c. What was the photon flux {Γph in photons/(m2 s)} (5 points) 1 W N ph Intensity Intensity m 2 J = 1.67 × 1016 photons Γ photon = = = = t⋅A hυ light Elight m2s 1.80 × 10 −18 J Ws d. Why does current saturate in the photoelectric effect? (5 points) The measured current in the photoelectric effect comes from electrons that are freed from the target. The electrons are freed because a photon collides with the atom and exchanges enough energy with the electron to break the electron free. This is a one to one relationship- one photon collides with one electron. So, when the photons are exhausted (per unit time), no more electrons can be free and the measured current levels off (saturates). 30 × 10 −3 3.) Calculate the size of a cube (box with all sides equal) in which blue light (475 nm) excites an electron from the ground state to the next highest energy level. (10 points) In a cube, the next highest energy level is degenerate (Ψ112, Ψ211, and Ψ121 all have the same energy). m 6.626 × 10 −34 J ⋅ s ⋅ 2.998 × 10 8 hc s = 4.18 × 10 −19 J = ΔE Elight = = −9 λ 475 × 10 m 2 2 2 2 2 h ⎛ n1 n2 n3 ⎞ h2 ⎜ 2 + 2 + 2⎟= (12 + 12 + 12 ) = 3h 2 (12 + 12 + 12 ) E1sr = E111 = 8m e ⎜ a a a ⎟ 8m e a 2 8me a ⎝ ⎠ E 2 nd = E 211 = E121 = E112 ΔE = E 2 nd − E1st = (6.626 × 10 h2 = 8m e (6 − 3)h 2 2 2 ⎛ n12 n2 n3 ⎞ h2 6h 2 2 2 2 ⎜ 2 + 2 + 2⎟= 1 + 2 +1 = ⎜a a a ⎟ 8m e a 2 8m e a 2 ⎝ ⎠ ( ) 8m e a 2 ) J ⋅ s (6 − 3) kg ⋅ m 2 a= = 6.57 × 10 −10 m = 6.57 Ansgtroms −31 −19 2 8 ⋅ 9.1 × 10 kg ⋅ 4.18 × 10 J J ⋅ s Note: those of you who did 1-D coincidentally got the right answer. Check the solutions over! −34 2 4.) Very briefly describe how the potential (V) is different for an electron in a box versus a H atom. (5 points) For an electron in a box (or a well), the outside potential on the electron is ∞ (V= ∞). Note, a lot of you said V=0. This is not the potential acting on the electron (the V used in Schrodinger’s equation. This V is only 0 for the free electron solution.) For an electron in a H atom, the potential acting on the electron is the Coulombic potential from the nucleus. 5.) What does degeneracy mean? (5 points) Degeneracy refers to states (electrons) that have the same energy. 2 6.) For Se (Z=34) a. Write the electron configuration. (5 points) 1s22s22p63s23p64s23d104p4 b. For the outermost sub-shell, show how the electrons would fill the sub-shell (using arrows as the electrons). (5 points) This is an application of Hund’s rule: first all three p orbitals fill with an up arrow, the fourth electron goes in the first sub-shell with an arrow down. _↑↓_ __↑___ __↑___ c. Write the quantum numbers for the very last electron in your picture in part b (only for one electron). (5 points) I had meant the last to fill which if the down arrow one: n=4, l=1, ml= 1 ms = -1/2 However, if you did the far right in the picture: that was fine n=4, l=1, ml= -1 ms = +1/2 Note: that any of the ml of -1, 0, 1 would be accepted as they are just different perspectives on the x, y, and z lobes. 3 ...
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This note was uploaded on 03/13/2012 for the course MATE 153 taught by Professor Staff during the Fall '08 term at San Jose State University .

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