Figure 1: Volume of Interaction for Air Molecules

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Alice Lamb
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1 Problem In 866 when atoms and molecules were still quite hypothetical, Joseph Loschmidt used kinetic energy theory to get the first reasonable estimate of molecular size. He used the liquid to gas expansion ratio of air ɛ n g /n l where n N/V together with the mean free path λ between collisions to calculate the average diameter s of air molecules. In the following diagram, σ is the cross sectional area of interaction between two air molecules cross section. The purpose of this problem is to reproduce his results and show the profound consequences in atomic physics. σ s λ Figure : Volume of Interaction for Air Molecules Part a Justify that the number density of air [molecules/cm 3 ] is n g /λσ. Since σ is the cross-section of interaction, that is, the radius of σ gives the maximum distance from the probe that the center of an air molecule can be and still interact with the probe, and λ is the average distance that the probe can travel without encountering an air molecule, λσ defines a volume in which only one air molecule, on average, exists. Since there is only one molecule in this volume, the number density can be expressed as one molecule per unit volume, or n g air molecule λσ Part b Justify that σ πs, not, for example, π s/. We use s here, the diameter of the probe, instead of, say, s/ the radius because we want the total area of interaction. Since both the probe and the air molecule occupy a volume, we cannot conclude that the cross sectional area of the probe is sufficient. Instead, we need to consider that the very outter shell of both the probe and the air molecule can interact. So, the area of interaction must have a radius that is equal to the maximum distance between the centers of the probe and the molecule when the two are still interacting. Since both the probe and the air molecules can be considered to be spheres, the distance from the center of the probe or air molecule to its outter shell is equal to the radius of the probe or air molecule, and the total maximum distance between the probe and the air molecule is, then, equal to the sum of the radii of the probe and the air molecule. If we assume that the probe and the air molecule are about the same size, with diameter s, this becomes s/, or just s. Page

2 Part c Show that ɛ s/6λ, assuming that the molecules are tightly packed in a liquid. Extra credit Explain why Loschmidt used ɛ s/8λ. Since ɛ n g /n l, and each n N/V, we can estimate the condensation coefficient by using n g /λσ the molecule number density of an individual molecule in a gas from part a and n l, which is the molecule number density 3 πs/3 of any individual spherical gas molecule in a liquid. This gives: ɛ λσ 3 πs/3 3 π s3 8 λπs 3 s λ ɛ s 6λ Instead of ɛ s/6λ, Loschmidt used ɛ s/8λ. Examining On the Size of the Air Molecules by Loschmidt [ Chem-History/Loschmidt-865.html] and On the Dynamical Theory of Gases by Clausius [ theory+of+gases%&sourcebl&ots-n-yd_sqpr&sigdb3qujax9axyqnq9poyewqepmpy&hlen&eif7zrs7fx I3V8AbMuiPBg&saX&oibook_result&ctresult&resnum&ved0CBgQ6AEwAw#vonepage&qClausius%0 %On%0the%0dynamical%0theory%0of%0gases%&ffalse], the reason for this disparity obviously arises because Loschmidt and Clausius did not assume that the air molecules in a sample would be stationary, and that they would be moving at the same velocity as the probe. Clausius asserted that this increases the number of collisions by a factor of 3. In On the Dynamical Theory of Gasses, Clausius attempted to prove this assertion. As a starting point, he the same simplifying assumption that we made, that the probe was in motion but the other air molecules were not. This led to the same conclusion that we reached, that is, that: n g λπs Actually, he was more interested in the number of collisions in a given time: N s vπs N Where N s is the number of collisions for this stationary system, v is the absolute velocity of the probe, N is the number of stationary molecules in the unit space, and s is, as before, the maximum distance between the centers of two interacting molecules. Next, he removes the simplifying assumption, so that all molecules can have motion, and he replaces v with the relative velocity between the probe and the mean of the velocities of the other air molecules, r. Not surprisingly, this changes the number of collisions to: N d rπs N The ratio, then, is N s /N d v/r. Clausius notes here that Maxwell arrived at the same conclusion independently. However, they apparently differed on the determination of the mean value, r. Quoting the letter directly or almost... φ has been used in place of another character in a different script: Let u be the velocity of any molecule m, and φ the angle between the direction of its motion and that of the molecule µ; the relative velocity between µ and m will then be u + v uv cos φ When the molecules, m, m, m... all move with the same velocity, in other words, when u is constant and φ alone variable from one molecule to another, the mean value can be easily calculated. Page

3 Maxwell apparently believed that, in this case, r u + v v. However, Clausius disagrees: Since all directions are equally probable for the molecules m, m, m..., the number of those whose lines of motions make angles between φ and φ + dφ with the line in which µ moves will have to the whole number of molecules the same ratio that a spherical zone with the polar angle φ and the breadth dφ has to the whole surface of the sphere, in other words, the ratio π sin φ dφ : π The number of such molecules in the unit of volume is consequently N sin φ dφ In order to obtain the required mean value r, the last expression must be multiplied by the relative velocity which corresponds to it, the product integrated between the limits 0 and π, and the integral divided by N. Hence r ˆ π u + v uv cos φ sin φ dφ This gives at once whence we may deduce o r [ u + v + uv 3 u + v uv ] 3 6uv r v + 3 u v, when u < v and r u + 3 v u, when u > v When u v, both results coincide in value with r 3 v and thus verify my assertion. This changes the estimate of n g, which, through Clausius work, becomes: Using n l 6 πs 3, we can show that: This is why Loschmidt used ɛ s/8λ. n g 3 πλs 3 λσ 3 λσ 6 ɛ n g n l πs 3 3 πs 3 λπs 6 s λ s 8λ Page 3

4 Part d Given the density of liquid air ρ l g/cm 3, the molar volume of an ideal gas V m L/mol, and the atomic weight of air A 35.6 g/mol, calculate the expansion ratio ɛ from liquid to air at STP atm, 0 C. Loschmidt had to estimate ρ l, since air had not been liquefied by 866. Here, we need to find ɛ n g /n l. To determine this ratio, it is important to calculate n g and n l in the same units. n g can be found easily by: To find n l : For ɛ this gives us: ɛ n g n l N A V m ρ l N A A n g N A V m n l ρ ln A A A V m ρ l 35.6 g/mol g/cm 3 L/mol 000 cm 3 /L This is on the same order as the condensation coefficients calculated by Loschmidt for various gasses, though it is quite a bit larger than his calculated condensation coefficient for air itself, which was Part e Maxwell determined λ 6 nm from viscosity measurements of air. A later extraction λ 0 nm by Oskar Emil Meyer don t laugh! was less accurate. Use this to calculate s and σ. Using our equation for ɛ s/6λ, we can easily determine s. σ can then be determined using σ πs. For λ 6 nm: s 6λɛ 6 6 nm nm And: σ π nm For λ 0 nm: s 6λɛ 6 0 nm nm And: σ π nm Page

5 Part f Using σ, calculate the density of air n g [ cm 3 ] at STP the Loschmidt number or Amagat. Using σ for λ 6 nm from part e, we have: n g σλ.6 nm 6 nm cm 3 Part g Using n g and V m calculate Avogadro s number N A also called the Loschmidt number. Given V m L/mol and n g cm 3, we have: N A V m n g L molecules mol cm molecules/mol Amazingly, this is on the same order of magnitude as the actual Avogadro s number, molecules/mol. Part h Faraday s constant F N A e 9685 C/mol was determined by weighing the amount of silver deposited during electrolysis of silver ions. Use this to calculate the elementary charge e. Easily, we have: Once again, very close to the actual value, C. e F N A 9685 C/mol molecules/mol C/molecule Part i Using the cathode ray tube, Thompson discovered electrons, negatively charged particles with charge-to-mass ratio e/m C/g. Calculate the mass of an electron. Here, we have; m e C/g C C/g g kg Again, very close to the actual mass of an electron, kg. Page 5

6 Part j Using the atomic weight A.008 g/mol of hydrogen, calculate the mass of a proton. Since a hydrogen atom is made up of one proton and one electron, the mass of the proton is: m p m a m e.008 g/mol molecules/mol g g Of course, subtracting the electron makes very little difference, since it is so much smaller than the mass of the proton. Problem Consider the scattering of hard spheres as discussed in class. It is assumed the target is infinitely massive so it does not recoil, and the probe scatters by reflection, i.e. the angle of reflection equals the angle of incidence. The radius of the probe and target are a and A respectively. Part a Determine the relation between the impact parameter b and the scattering angle θ. Since we re scattering hard spheres here, we can assume that the two spheres will never intersect. Also, the problem definition states that the scattering will be through reflection, which completely defines the behavior of the two interacting objects. At the point at which the probe impacts the target, the probe will reflect off the target at some angle φ that equals the angle of incidence. This angle is defined with respect to a line through both the radius of the target and the point of impact. The impact parameter for an individual event, b, can now be defined in terms of the distance of the center of the probe from the center of the cross section as the sum of the portion of the radii of the probe and target that are in the radial direction of the cross section, or b A sin φ + a sin φ, which leaves us with: b A + a sin φ However, we ultimately want to know b in terms of θ, the scattering angle, not φ. Fortunately, since we know that the probe reflects off the target, we know that the angle that the reflected ray makes with the angle of incidence is equal to the angle that the impact ray makes with the angle of incidence, and then the impact parameter θ 80 φ. So, φ 80 θ. This gives us 80 θ b A + a sin A + a sin 90 θ θ A + a cos This gives the correct result at θ 0 as b A + a, which is the expected maximum impact parameter, and gives b 0 when θ 80, which is once again correct, since a head-on collision should result in a rebound of the probe. Page 6

7 φ φ a θ b φ A φ φ a θ b φ A Figure : Hard Interaction between Spheres Part b Using part a, calculate the cross section /dω. Plot it as a function of θ. To calculate dω, we can first find and dω separately. Since σ πb, πb db. Ω is defined to be the solid angle, or Ω π cos θ. So, dω π sin θ dθ. This θ is the same scattering angle as in part a. This gives us: dω πb db π sin θ dθ b db sin θ dθ We can also substitute in for b and db by using b A + a cos θ/ and db A + a sin θ/ dθ. However, since represents an area and dω represents a volume, dω must be positive, and making these substitutions will not give us a positive dω. However, b db sin θ dθ can be thought of as a Jacobian that is used to change coordinates for a volumetric differential, and those are generally taken to be positive. So, using dω b db sin θ dθ this gives us: A + a cos θ dω A + a sin θ dθ sin θ dθ A + a cos θ dω A + a sin θ dθ sin θ dθ A + a cos θ sin θ dθ sin θ dθ Page 7

8 Using sin u sin u cos u: dω A + a sin θ dθ sin θ dθ A + a Figure 3: dω In Terms of θ for A + a Part c Integrate part b to calculate the total cross section ˆ ˆ σ tot What is an intuitive interpretation of the result? ˆ π θ0 π ˆ π φ0 dω dω dω sin θ dθdφ Substituting and integrating, we get: σ tot ˆ π π ˆ π θ0 φ0 ˆ π ˆ π φ0 θ0 ˆ π θ0 A + a π dω sin θ dθdφ A + a A + a ˆ π θ0 sin θ dθdφ sin θ dθ sin θ dθ A + a π [ cos θ] π 0 A + a π + A + a π Page 8

9 This can be interpreted as the maximum cross-sectional area of the incident particle path corridor of interaction with the target. That is, this is the maximum cross section, given target and probe radii of A and a, respectively. Part d In a model of alpha particle scattering from gold nuclei, let a r α.7 fm and A r Au 7.3 fm. What value of b leads to scattering at θ 90? 0 0 cm detectors are placed m away from the target, which is a cm thick gold foil, Z 79, A 97 g/mol, ρ 9.3 g/cm 3. What percentage of the incident α particles will scatter into detectors at θ 5 and θ 65? Assuming that we are using the same scattering model as specified in the first parts of this problem, we have b A + a cos θ π/, and b 9 fm cos 6.36 fm leads to scattering at θ 90. To find the percentage of incident α particles, we can use dω dn I 0nt A sc, where the fraction is f dn I 0, n is the number of particles per unit volume in the target, t is the thickness of the target, A sc is the area of the detector, and r is the distance between the target and the detector. Solving for f, we get: dω Since we know from part b that dω A+a, this becomes: r dω dn I 0 nt dω f r nta sc nta sc r f A + a r A sc nta sc r f This apparently is independent of the scattering angle, so f is the same regardless of θ. We can calculate n, the number of target atoms per unit volume, by renaming A M 97 g/mol, to avoid a name conflict between the various A s, and calculating: n ρn A M 9.3 g/cm atoms/mol 97 g/mol atoms/cm 3 Now, we can substitute in to determine f making sure to convert the various length measurements appropriately: f A + a nta sc r 7.3 fm +.7 fm atoms/cm cm 00 cm m Problem 3 We will now compare the above with Rutherford scattering see Beiser, Chapter. Page 9

10 Part a It can be shown extra credit! that the electric force between an α + particle and nucleus of atomic number Z, F Ze πɛ 0 r produces the scattedring relation cot θ πɛ 0E Ze b where E is the kinetic energy of the α-particle. What value of b leads to scattering at θ 90 use E 7.7 MeV for the kinetic energy of the α particle? I did not come up with this proof. I found it at Assuming that the target recoil is negligable, the change in momentum associated with Rutherford scattering is P P f P i where the magnitudes of the initial and final momenta are the same. The scattering geometry gives P mv 0 sin θ see Figure. Figure : Scattering Geometry Using the impulse and the symmetry of the scattering geometry, we see that only the impulse component along the symmetry axis, therefore: ˆ P F P dt Page 0

11 Also, the columb force is F Zke r ˆr where ˆr is the radian unit vector, and F P F cos φ, where φ is the angular location of the projective relative to the symmetry plane. This gives: P Zke ˆ cos φ r dt mv 0 sin θ We can use conservation of angular momentum to help evaluate the impulse integral: Substitution gives: L mv 0 b mr dφ dt r v 0b dφ/dt mv 0 sin φ ˆ cos φ Zke v 0 b Zke v 0 b ˆ φf φ i dφ dt dt cos φ dφ Extending the entrance and exit paths of the scattering trajectory to infinity gives the limits of the angle: φ i π θ φ f π θ Using the angle difference identiy: ˆ φf φ i cos φ dφ sin [ ] π θ sin [ ] π θ cos θ Solving for the impact parameter b: Since k πɛ 0 and mv 0 E, this gives us: mv 0 sin θ b Zke v 0 b cos θ Zke v 0 cos θ mv 0 sin θ Zke cos θ mv 0 sin θ Zke mv 0 cot θ b Ze πɛ 0 E cot θ Ze πɛ 0 E cot θ Page

12 Rearranging, we have: Since cot 90 cot θ πɛ 0E Ze b cot 5, we have b Ze πɛ oe leading to scattering at 90. This can be calculated to give us: b Ze πɛ 0 E 79e πɛ MeV.77 fm Part b Calculate /dω and plot it as a function of θ on the same graph as problem. Once again, we have πb db and dω π sin θ dθ. We also have b that dω b db sin θ dθ. This gives us: dω Ze πɛ cot θ 0E sin θ Ze πɛ 0E Ze πɛ 0E Ze cos θ sin θ sin θ sin θ cos θ sin θ sin θ Ze πɛ 0E cot θ 8πɛ 0E csc θ dθ dθ sin θ Ze and db 8πɛ 0E csc θ dθ, but Using sin θ sin θ cos θ : dω dω Ze πɛ 0E Ze πɛ 0E sin θ cos θ sin θ sin θ sin θ cos θ If we assume that Ze πɛ 0E fm, which is not necessarily impossible, we can graph dω fm, shown in Figure 5. sin θ Part c In the same conditions as in problem, what percentage of the incident α particles will scatter into the detector at i θ 5, ii θ 65? We still have: f nta sc dω r Ze πɛ 0E sin θ nta sc r Page

13 Figure 5: dω In Terms of θ for A + a and Ze πɛ 0E fm We know that that E 7.7 MeV for the α particles, and we use Z 79 for gold. Also, we can use πɛ 0 k e, and k e e. ev nm. MeV fm to simplify the calculations later. We also know that A xc 00 cm, t cm, and r m. Also: n ρn A M 9.3 g/cm atoms/mol 97 g/mol atoms/cm 3 Now, we can substitute in for f again making sure to convert the various length measurements appropriately. For θ 5, we get: For θ 65, we get: f f Ze πɛ 0E nta sc sin θ 79. MeV 7.7 MeV r sin Ze πɛ 0E nta sc sin θ 79. MeV 7.7 MeV r sin atoms/cm cm 00 cm m atoms/cm cm 00 cm m Page 3

14 Part d Explain why the total cross section is actually infinite. Up until now, we have ignored the electrons of the gold atom. What is the effect on small angle large b scattering from the entire atom? hint: what is the total charge? The total cross section is infinite because the electric force, while decreasing with distance, never reaches zero at any finite distance in theory, although I wonder if force can be quantized... The effect from the entire atom including the electrons on small angle scattering is extremely small. Since the atom has a net zero charge, the α particle will experience no electric force due to the atom before the particle has breached the electron shells. Once the α particle has breached the electron shells, the uniform density shell should, again, have no effect, since the total force felt due to the shell should be balanced at all points near the nucleus. However, the α particle will feel the force due to the charge on the nucleus once it has passed the electron orbitals. Problem. If a particle is deflected by 0.0 in each collision, about how many collisions would be necessary to produce an rms deflection of 0? Use the result from the one-dimnensional random walk problem in statistics stating that the rms deflection equals the magnitude of the individual deflections times the square root of the number of deflections. Compare this result with the number of atomic layers in a gold foil of thickness 0 6 m, assuming that the thickness of each atom is 0. nm 0 0 m. So, D rms D i N. In our case D i 0.0, and D rms 0. This gives us: 0 N N Assuming a regular arrangment of atoms across the thickness of the foil, we can estimate the number of atoms across the thickness of the gold foil as 0 6 m atom/0 0 m 0. This is two full orders of magnitude smaller than 0 6. This number of collisions would produce D rms Problem.6 If the angular momentum of Earth in its motion around the Sun were quantized like a hydrogen electron according to Equation -7, what would Earth s quantum number be? How much energy would be released in a transition to the next lowest level? Would that energy release presumably as a gravity wave be detectable? What would be the radius of that orbit? The radius of Earth s orbit is.50 0 m. Equation -7 give us the equation for angular momentum as L mvr n. The mass of Earth is kg, and the velocity of the earth is, on average, m/s. The quantum number at r.50 0 m, then, is: n mvr kg.5 0 m.98 0 m s m kg/s To find the energy released by transitioning to the next lower n, we must know the energy at each orbital. We can use the equation derivations from problem.56 here, which gives E n G M m 3 n. In this case, we can call E 0 G M m 3, Page

15 since all of these are constants in this problem. The energy released, then, is E E n+ E n E 0 n + + E 0 n E 0 n n + This quantity, due to the size of the values involved, is close to impossible to calculate. Instead, we can normalize this quantity, so that: n n + E E 0 n + n n n + E E 0 n + n + n n n + E E 0 n + Once again, this is essentially impossible to calculate, but we can make a reasonable estimate by using exponents. Since E 0 G M m ev, we can estimate: E E E 0 75 E E 0 ev This is an extremely small amount of energy, on the order of 0 times smaller than the energy released by an electron transitioning to the next lowest orbit. Since this amount of energy is not enough to excite any known energy detection device, it is unlikely that this transition could be detected using any known equipment. Any change in the radius of the Earth s orbit in this case would be basically imperceptable, so the final radius would still be about.5 0 m. Problem. Part a The current i due to a charge q moving in a circle with frequency frev is qfrev. Find the current due to the electron in the first Bohr orbit. Since q e in this case, since the moving charge is the electron, i ef rev. The question is, then, what is the frequency of an electron in the first Bohr orbit? The radius of the first Bohr orbit is r a0 Z. The velocity of the first Bohr orbital is v Zαc. Since f vn πr n, we have: v f πr Zαc π a0 Z Z αc πa 0 Page 5

16 Substituting back in for a 0 and α: f Z αc πa 0 Z c ke c π mke Z c ke mke π 3 c Z mk e π 3 Since f f rev in this case, we can substitue back in to i ef rev ez mk e π Z mk e 5 3 π Z A. We are 3 probably working with hydrogen, since we are looking at the Bohr model, so Z, giving us i A. Part b The magnetic moment of a current loop is ia, where A is the area of the loop. Find the magnetic moment of the electron in the first Bohr orbits in units A m. This magnetic moment is called a Bohr magneton. We have ia Z mk e 5 π A for the magnetic moment. A in this case is just the area of the electron s orbit. This is given 3 by A πr, where r a0 Z Zmke. So, we have: Problem.56 ia Z mk e 5 π 3 mk Z e 5 π 3 e m π π Zmke Z m k e e m m A If electric charge did not exist and electrons were bound to protons by the gravitational force to form hydrogen, derive the corresponding expressions for a 0 and E n and compute the energy and frequency of the H α line and limit the Balmer series. Compare these with the corresponding quantities for real hydrogen. We start with F GMm r mv r and L mvr n. We can then solve for v, giving us v with the equation for L, we get GM n mr r In this case, I would say that a 0 n m GMr n m GMr n m GM r Mm G. This gives a m. So, we have: GM r. Combining this r n n a 0 Page 6

17 We also have the the total energy of the system would be E mv + GMm r. In addition, we can find, from the equation for F, that mv GMm r n we get: r. So, we get the total energy is E GMm r GMm r GMm r. Substituting in for GMm m GM E n n G M m 3 n I m going to say that E 0 G M m 3 h. This gives us E ev. So, we have E n E 0 n To find the energy and frequency of H α, that is the transition from n 3 to n, we have: E Hα E 3 E E E 0 E 0 9 E ev The frequency, then, is hf E 3 E E Hα ev. So, f ev h s. The limit to this fictitious Balmer series, then, is: E limit E E E 0 M lim n n E 0M 5 ev kg kg ev 3.6 For real hydrogen, E Hα ev ev.9 ev, f.9 ev h s, and the E limit 3.6 ev 3. ev. The energies and frequencies for this fictitious hydrogen are much, much smaller than for the real thing. Page 7

Physics 3223 Solution to Assignment #5

Physics 3223 Solution to Assignment #5 October 20, 1999 6.1 From the Heisenberg uncertainty principle, x p h, we estimate that a typical kinetic energy of an electron in an atom would be K ( p)2 2m h 2