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jill rodriguez

jill r.

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The Experimental Condensed Matter Research Group in the School of Physics is doing experiments on paramagnetic phosphorus atoms implanted into an ultra-pure, isotopically enriched silicon-28 crystal. The silicon-38 isotope of silicon has zero nuclear spin and all electrons are in pairs of opposite spin orientation so there are no background magnetic moments to perturb the paramagnetic phosphorus atoms, apart from a small magnetic field from the phosphorus nucleus that can be neglected. At low temperature, the phosphorus atoms each have an unpaired electron of spin s = 1/2 and are distinguishable from their location in the crystal. The unpaired electrons do not interact with each other and are therefore an ensemble of paramagnets. A large background magnetic field, B, is imposed by an external electromagnet controlled by the experimenters. Relative to the background magnetic field, each electron can exist in two states: "up" or "down" and can therefore occupy two energy levels because of the interaction of the electron magnetic moment with the background magnetic field. The energy gap between these levels is ɛ. The system is in equilibrium with a large thermal heat bath provided by a cryogenic refrigerator at temperature T. In terms of the number of phosphorus atoms in the crystals, N, and the internal energy, U, due to the interaction with the external magnetic field, we define U/N to be the average energy per atom in the limit of large N (i.e. N → ∞). (a) What is the maximum possible value of U/N for the system? (b) If the crystal is placed in equilibrium with a heat bath, what is U/N for the most probable configuration of the system? Be sure to justify your answer. (c) Engineered paramagnetic phosphorus atoms in an isotopically pure silicon crystal have potential applications in quantum computer technology. To initialise the state of the system, the phosphorus electron spins must be in the low energy state. When a single phosphorus atom in the crystal changes state by flipping from the high energy state to the low energy state and emits a photon of energy & into the heat bath, the population changes from n₁ → n₁ + 1 and n₂ → n2 1 where n₁ and n₂ are the populations in the low and high energy states respectively. In terms of n2/n₁, determine the entropy change of the crystal for the single photon emission process. Explain the significance of the sign of your result. Be sure to state any approximations you make. (d) Determine the entropy change of the heat bath for the absorption of the photon. (e) Assume the photon emission process is reversible and that a photon emitted from the bath can return the system to the original configuration. Hence determine the population ratio of n2/11 as a function of the bath temperature T.

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The name of the alkane with seven carbons connected with single bonds in a chain is hexane.

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Select the single best answer. The compound CH3COO− is a Bronsted acid. a Bronsted base. both a Bronsted acid and base.

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Part II: Short Answer: 1. Complete the reactions below by DRAWING AND NAMING all products AND reactants. [14 marks/A 2 marks each] a) ethylbutanoate + water ? b) ? 2-pentanone c) butanoic acid + [R] ? d) ? propoxypropane e) 2-chloropropane + ammonia ?

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During which phase of the cell cycle does the cell accumulate the raw materials (building blocks) to create chromosomal DNA? G1 G2 Prophase Cytokinesis

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Find the integral.\\ $\int \sin^3 x \cos^4 x \, dx$

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Determine the amplitude, period, any vertical translation, and any phase shift of the given graph y = -4 cos \left( x + \frac{\pi}{12} \right)

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fluid ounces 14. Be Precise A 100-watt solar power panel usually costs $146. They are on sale for 0.75 of the regular price. Multiply 146 by 0.75 to find the sale price.

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I want an equivalent circuit diagram and phasor diagram for this...ASAP. Reactive power, Q = IX Active power, Hence, R = 100.03. We know that the secondary current, I = Es/Ztoa power input to the transformer. Problem 2-9/8: P = IR X = 103.58. We know that Ztoad = 14446. I need S (Lowfide- b). Given, Load impedance, Zioad = 14446. We know that the turn ratio, a = VLs/VHs = (45.83/103.58) = (45.83/100.03) = 100.03 + 103.58j. Given primary voltage, Ep = 220 V and find primary voltage, Es. (d) Input impedance, Zin = a^2 Ztoad = (0.033333)^2(14446) = 0.16046 impedance at primary terminals of transformer. (e) Active, reactive, and apparent = 217558.28 var = 210101.9W. He Trats is = 45.83 - 46A = 240V/7200V = 0.033333. = 217.6kvar = 210.1kW phatr.

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What volume of mercury (Specific Gravity = 13.54) in liters would weigh the same as $0.020m^3$ of fuel oil, if the weight density of fuel oil is $9.42kN/m^3$? *\ a. 20\ b. 2.656\ c. 8.03\ d. 1.42

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