Intro Physics Ii - Studypalooza Solutions

Studypalooza Solutions Thermodynamics and statistics (1)Thermodynamics Critical temperature is given by 0 = Δ𝐺 = Δ𝐻 − 𝑇Δ𝑆 → 𝑇 = $% $& = (23"##*,/567) ("#*,/567// = 310)𝐾 Below critical temperature the low-enthalpy state dominates (A); above critical temperature the high-entropy state dominates (B) (2)Free energy A. Scaling up from the standard molar volume gives Δ𝑉 = (−22.4)𝐿) ∗ (298)𝐾 ⁄273)𝐾) = − 24.5)𝐿 = −0.0245)𝑚! . There are lots of other ways to get this. B. 𝑝)Δ𝑉 = (1.01 × 10" )𝑃𝑎)( −0.0245)𝑚! ) = −2.47)𝑘𝐽. You can also get this by scaling up from the standard molar energy 𝑅𝑇#. Δ𝐻 = Δ𝑈 + 𝑝Δ𝑉 → Δ𝑈 = Δ𝐻 − 𝑝Δ𝑉 = (−170)𝑘𝐽) + (2.47)𝑘𝐽 ) = −168)𝑘𝐽 C. Δ𝐺 = Δ𝐻 − 𝑇Δ𝑆 = (−170)𝑘𝐽) − (298)𝐾)( −0.152 𝑘𝐽 ⁄𝐾 ) = ) −125)𝑘𝐽. dG<0, so YES. $% '()#*+, D. Setting Δ𝐺 = Δ𝐻 − 𝑇Δ𝑆 = 0 and solving for T gives 𝑇 = = '#.(".*+, ⁄/ = 1100)𝐾 $& (3)Boltzmann Factor A. p0 = e^0.05/0.025 / (e^0.05/0.025 + e^0.025/0.025) = e^2/(e^2 + e^1) = 0.731 -> 731 molecules in the ground state; 269 in the excited state B. p0/p1 = 0.9/.1 = 9 = e^-(G0-G1)/kT = e^0.025/kT -> kT = 0.0114 eV -> T = 132 K C. p0/p1 = .1/.9 = 0.11 = e^-(G0-G1)/kT = e^0.025/kT -> kT = -0.0114 eV -> T = -132 K -> impossible! D. 1000; 0 E. 500; 500 Electrostatics and Circuits (1) E, F, U, and V A. B. C. D. E. (2) Potential/energy A1. Use conservation of energy: 2 F. 𝑚𝑣 . H = 𝑘 ( 9! : , so 𝑣 = I5: . +9 ! A2. mass 1 charge 1 mass 2 charge 2 distance 𝑣" is smaller 𝑚 −𝑒 𝑚 −𝑒 𝑑 ◯ 𝑚 # 𝑒 $ 𝑑 ◯ X 𝑚 # 𝑒 $ 𝑚 2𝑒 𝑑 ◯ 2𝑚 𝑒 2𝑚 2𝑒 𝑑 ◯ 𝑚 𝑒 𝑚 𝑒 2𝑑 ◯ X 𝑚 𝑒 𝑚 2𝑒 2𝑑 ◯ 𝑚 𝑒 𝑣" is unchanged 𝑣" is larger X ◯ ◯ X ◯ X ◯ ◯ ◯ X ◯ ◯ ◯ ◯ ◯ ◯ . Use the formula from the previous part to figure out the scaling. In particular, increasing mass decreases final speed. Only the product 𝑞( 𝑞. matters (not individual charges). Increasing separation reduces final speed. B1. Conservation of momentum: 𝑚)𝑣( = (2𝑚)𝑣. so 𝑣( = 2𝑣. . Conservation of energy: F . 𝑚𝑣(. H + ) F . (2𝑚)𝑣. . H = 𝑘 ( ( expression for 𝑣( from above to give: F. 𝑚(2𝑣. H + ) F ( ( ). 9! 9! 𝑘 :, so +9 ! B +9 ! 2+9 ! 𝑣 = I !5: and 𝑣( = 2𝑣. = 2I = I !5: !5: C . (4) Circuits A. : . Substitute in the ( . (2𝑚)𝑣. . H = 3𝑚𝑣. . = B. 1) I1 = I2 + I3 2) 12V -2ΩI1 – 1ΩI2 - 1ΩI2 - 4ΩΙ1 = 0 3) 6Ω – 1ΩΙ2 - 1ΩΙ2 + 3ΩΙ3 = 0 C. Ι1 = 1.33 Α, Ι2 = 2Α, Ι3 = -0.67Α Magnetism (1) Lorentz Force (1) 𝒗L×𝑩L points up; 𝒒𝒗L×𝑩L points down. (2) 𝑰L ×𝑩L points into the page. (3) 𝒗L×𝑩L points right; 𝒒𝒗L× 𝑩L points right. (4) 𝒗L×𝑩L points out of the page; 𝒒𝒗L×𝑩L points out of the page. (5). No force. (2) Current of Magnets Field due to horizontal wire point out of the page at p, so the current in the vertical wire must go upward to make its field point into the page. To match magnitudes: From , since point p is half the distance to the vertical wire as to the horizontal wire, we only need half as much current (5A). SHM and Waves (1) Simple Harmonic Oscillator A. B. C. D. E. (2) Standing waves A. 2/5*3m = 1.2 m B. 𝑣 = 𝜆𝑓 = (1.2𝑚)( 200)𝐻𝑧) = 240)𝑚/𝑠 C. From 𝑣 = U𝑇/𝜇, 𝑇 = 𝜇𝑣 . = ; < D. 𝜆( = 2𝐿 = 6)𝑚) and 𝑓( = @ = ? E. 𝜆( = 𝐿 = 3)𝑚) and 𝑓( = ' ? @' = 𝑣. = "×(#%& *+> !*5 .2#*5/A 3*5 .2#*5/A !*5 (240)𝑚/𝑠). = 96)𝑁 = 40)𝐻𝑧 = 80)𝐻𝑧 Ray Optics (1) Ray Reflection Everything is symmetric on the way in and on the way out, so 𝜃6 = 𝜃B = 40°. You can calculate all the angles if you want, but you’ll always end up with this result (no matter the value of 𝜃B ) (2) Ray Tracing I think it’s easier to solve this using ray tracing, but if you really want to use the thin lens equation that works too: From A + ( ( A( = , solve for 𝑠′: 𝑠 D = . Then calculate the magnification 𝑚 = − A = − A'C . Plug A'C C A( AC ( in a particular multiple of 𝑓 that fits each of the ranges: C 𝑠 = 0.5𝑓 → 𝑚 = − = +2 (upright, larger) '#."C C 𝑠 = 1.5𝑓 → 𝑚 = − #."C = −2 (inverted, larger) 𝑠 = 2.5𝑓 → 𝑚 = − (."C = − (inverted,smaller) C . ! (3) Compound lenses A. Spring 2020 Final B. position: 948 cm, magnification: -0.42 Wave Optics and Spectroscopy (1) Double slits (2) Particle in a box ) 𝐸(. = 3𝐸( ; )𝐸(! = 8𝐸( so 𝐸(! = ! 𝐸(. → 𝜆(! = E (3) Spectroscopy + Boltzmann Factor 𝑇=0 𝑇=∞ 𝑇 = 290,K N1 1000 333 689 N2 0 333 309 𝜆 E (. ! = ! E (720)nm) = 270)nm C N3 0 333 2 total 1000 1000 1000 At T=0 all are in the lowest state. At T=inf all are equally distributed. The Boltzmann factors for T=290K are: exp(−𝐸B /(0.025) ) = 2981, 1339,)and)7.4. The fraction in each state are then .FE( (!!F ).2 𝑝( = (FE(G(!!FG) = 0.689, 𝑝. = (FE(G(!!FG) = 0.309, and 𝑝! = (FE(G(!!FG) = 0.002. (4) Spectroscopy Α. Level Energy (eV) 0 0.00 1 1.00 2 2.00 3 2.50 4 - 5 - Wavelength (nm) 496 620 827 1240 2480 Β. 827 nm; Now we can also absorb from the n=1 state. This give another 1 ev = 1240 n, (which is redundant) plus a new 1.5 eV = 827 nm. C.