ps02soln

# ps02soln - 1 Physics 316 Solution for homework 2 Spring...

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Unformatted text preview: 1 Physics 316 Solution for homework 2 Spring 2005 Cornell University Steve Drasco I. EXERCISE 1 A. Given the knowledge that atomic diameters are of the order of 1 or 2 ˚ A , estimate the order of magnitude of the number of atoms in a human being and in the whole earth (use 40000km as circumference) We will take the atomic radius to be r A ∼ 10- 10 m, where the symbol ∼ means that we will keep things only to one digit. The atomic volume is V A = (4 / 3) πr 3 A ∼ 4 × 10- 30 m 3 . If we take the mass of a human to be 55 kg, and the density to be that of water, then the volume of a human is approximately V H = (55 kg) µ L kg ¶µ 10- 3 m 3 L ¶ ∼ 6 × 10- 2 m 3 . (1.1) There must then be about V H /V A ∼ 2 × 10 28 atoms in a human. Using the given circumference C ⊕ of the earth, its volume is approximately V ⊕ = 4 3 π µ C ⊕ 2 π ¶ 3 ∼ 10 21 m 3 . (1.2) There must then be about V ⊕ /V A ∼ 3 × 10 50 atoms in the earth. B. (Bonus) Assume that all molecules of Plato’s last breath are still in the atmosphere, how many of them are on average contained in every breath you take? For a rough approximation, assume that the atmosphere is 5km thick, that the mol volume is 22.4L/mol in all of the atmosphere, and that a typical breath contains about 1L. If the earth’s atmosphere is approximated to extend Δ r = 5 × 10 3 m beyond its surface (there is of course no strict boundary for the atmosphere, but this approximation will include about half of its mass) then its volume is V atm = 4 3 π £ ( r ⊕ + Δ r ) 3- r 3 ⊕ / ∼ 4 πr 2 ⊕ Δ r ∼ 3 × 10 18 m ∼ 3 × 10 21 L . (1.3) The number of molecules in a single breath, of volume V breath ∼ 1 L, is given by N breath ∼ (1 L) µ mol 22 . 4 L ¶ ∼ 3 × 10 22 . (1.4) If we assume that the molecules contained in Plato’s last breath are evenly distributed in the atmosphere today, then the number N Plato of those molecules in any single breath today is given by N Plato = V breath V atm N breath ∼ 10 . (1.5) II. EXERCISE 2 A. Assume the Hydrogen atom consists of a sphere with radius R and with a uniform positive charge density in which a negatively charged electron moves freely. Show that the electron experiences a force- C~ r toward the center of the sphere and show that the orbits of the electron in such an atom would be ellipses. It is easiest to work in Cartesian coordinates and to use the following representation of an ellipse: ( x, y ) = ( a cos φ, b sin φ ) with φ ∈ [0 , 2 π ] . 2 A static sphere centered at r = 0 with radius R has a uniform positive charge density ρ = 3 e/ (4 πR 3 ). Due to spherical symmetry, the sphere’s static electric field has the form E = E ( r )ˆ r , where ˆ r is the radial unit vector. Since E depends only on one variable ( r ) it can be quickly determined from Gauss’ law 1 ....
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ps02soln - 1 Physics 316 Solution for homework 2 Spring...

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