Particle Nature of Matter. Solutions of Selected Problems

Chapter 4 Particle Nature of Matter. Solutions of Selected Problems 4. Problem 4.5 In the text book A Thomson-type experiment with relativistic electmns. One of the earliest experiments to show that p γm e v rather than p m e v was that of C. Neumann. The apparatus shown in Figure 4. is identical to Thomson s except that the source of high-speed electrons is a radioactive radium source and the magnetic field B is arranged to act on the electron over its entire trajectory from source to detector. The combined electric and magnetic fields act as a velocity selector, only passing electrons with speed v, where v V/Bd Equation 4.6, while in the region where there is only a magnetic field the electron moves in a circle of radius r, with r given by p Bre. This latter region E 0, B constant acts as a momentum selector because electrons with larger momenta have paths with larger radii. Figure 4.: The Neumann apparatus.

2 CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED PROBLEMS a Show that the radius of the circle described by the electron is given by r l 2 + y 2 /2y. b Typical values for the Neumann experiment were d 2.5 0 4 m, B 0.077 T, and l 0.0247 m. For V 060 V, y, the most critical value, was measured to be 0.0024 ± 0.0005 m. Show that these values disagree with the y value calculated from p m e v but agree with the y value calculated from p γm e v within experimental error. Hint: Find v from Equation 4.6, use m e v Bre or γm e v Bre to find r, and use r to find y. Solution Figure 4.2: Details of electron path in region where E 0 and B constant. a Figure 4.2 shows the path of the electron in the region where E 0 and B constant. The path intersect the detector photo plate at point f. Using the right-angle triangle cfd we get: r 2 l 2 + r y 2 l 2 + r 2 + y 2 2ry 2ry r 2 + l 2 + r 2 + y 2 l 2 + y 2 r l2 + y 2 2y

4.. PROBLEM 4.5 IN THE TEXT BOOK 3 b The combination of electric and magnetic fields selects a velocity v of the electron, given by v V/Bd, using the given numerical values we get: v V Bd 060 0.077 2.5 0 4 2.39 0 8 m/s using the classical expression for the momentum, we get: p m e v Bre r m ev Be 9. 0 3 2.39 0 8 0.077.60 0 9 7.69 0 2 m Using y 2 2yr + l 2 0 we can find the value of y that corresponds to the above value of r: 0 y 2 2ry + l 2 y 2 2 7.69 0 2 y + 2.47 0 2 2 y 2.54 0 y + 6.0 0 4 using the quadratic equation solution: y b ± b 2 4ac/2a, with a, b.54 0, and c 6.0 0 4, we get: y b ± b 2 4ac 2a +.54 0 ±.54 0 2 4 6.0 0 4 2 7.70 0 2 ± 7.29 0 2.499 0 m or 4.0 0 3 m 4.0 0 3 m

4 CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED PROBLEMS The first solution is rejected because it gives y > r. Using the relativistic expression of momentum p γmv Bre, where γ / v 2 /c 2, we get: p γm e v Bre r γm ev Be m e v Be v 2 /c 2 9. 0 3 2.39 0 8 0.077.60 9 2.39 0 8 /3 0 8 2.272 0 m The quadratic equation in y for this gives: 0 y 2 2.544 0 y + 6.0 0 4 y +2.544 0 ± 2.544 0 2 4 6.0 0 4 2.272 0 ±.248 0 2.520 0 m or 2.400 0 3 m 2.400 0 3 m The first solution is rejects because it gives y > r and obviously, the value of y calculated from the relativistic expression agrees well with the experimental value.

4.2. PROBLEM 4.6 IN THE TEXT BOOK 5 4.2 Problem 4.6 In the text book In a Millikan oil-drop experiment, the condenser plates are spaced 2.00 cm apart, the potential across the plates is 4000 V, the rise or fall distance is 4.00 mm, the density of the oil droplets is 0.800 g/cm 3, and the viscosity of the air is.8 0 5 kg m s. The average time of fall in the absence of an electric field is 5.9 s. The following different rise times in seconds are observed when the field is turned on: 36.0, 7.3, 24.0,.4, 7.54. a Find the radius and mass of the drop used in this experiment. b Calculate the charge on each drop, and show that charge is quantized by considering both the size of each charge and the amount of charge gained lost when the rise time changes. c Determine the electronic charge from these data. You may assume that e lies between.5 and 2.0 0 9 C. Solution a The drag force D on the oil droplet is given by: D 6πaηv where a is the radius of the droplet, η is the viscosity of the air, and v is the velocity of the droplet. During the fall the droplet moves under the gravity downward and the drag force upward. When the two forced become equal, the droplet moves with terminal speed and we get: mg 6πaηv since the droplet is spherical, then m 4πa 3 /3ρ, where ρ is the density of the droplet, ad the last equation becomes: or 4 3 πa3 ρg 6πaηv a 9ηv 2ρg

6 CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED PROBLEMS The terminal speed of the droplet in the absence of the the electric field is: the radius of the droplet is then: v y t 4 0 3 5.9 2.52 0 4 m/s The mass of the droplet is given by: 9ηv a 2ρg 9.8 0 5 2.52 0 4 2 0.8 0 3 9.8.62 0 6 m m ρv ρ 4 3 πa3 b The charge on the droplet is given by: 4 0.8 03 π.62 0 6 3 3.42 0 4 kg q mg E v + v where E is the electric field E V/d 4000/0.02 2.00 0 5 V/m, v is the downward terminal velocity in the absence of the electric field v 2.52 0 4 m/s, and v is the upward terminal speed in the presence of the electric field. Using the times given, the various v s can be calculates as: v v 4 0 3 36.0 v 2 4 0 3 7.3 v 3 4 0 3 24.0. 0 4 m/s 2.3 0 4 m/s.67 0 4 m/s

4.2. PROBLEM 4.6 IN THE TEXT BOOK 7 v 4 4 0 3.4 v 5 4 0 3 7.54 Now the charges can be calculated: 3.5 0 4 m/s 5.3 0 4 m/s q mgv + v Ev q.42 0 4 9.82.52 0 4 + v 2.52 0 4 2.00 0 5 2.76 0 5 2.52 0 4 + v 2.76 0 5 2.52 0 4 +. 0 4 0.0 0 9 C q 2 2.76 0 5 2.52 0 4 + v 2 2.76 0 5 2.52 0 4 + 2.3 0 4 3.4 0 9 C q 3 2.76 0 5 2.52 0 4 +.67 0 4.6 0 9 C q 4 2.76 0 5 2.52 0 4 + 3.5 0 4 6.7 0 9 C q 5 2.76 0 5 2.52 0 4 + 5.3 0 4 2.6 0 9 C c Now, we will try to find an integer n such that.5 0 9 < q/n < 2.0 9 : The amount of charges gained or lost are: q.67 0 9 C 6 q 2.68 0 9 C 8 q 3.67 0 9 C 7 q 4 0.67 0 9 C q 5 3 66 0 9 C q q 2 3.4 0 9 C

8 CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED PROBLEMS q q 3.6 0 9 C q q 4 6.7 0 9 C q q 5.6 0 9 C q 2 q 3.8 0 9 C q 2 q 4 3.3 0 9 C q 2 q 5 8.2 0 9 C q 3 q 4 5. 0 9 C q 3 q 5 0.0 0 9 C q 4 q 5 4.9 0 9 C Once again we find an integer n 2 such that.5 0 9 < q i q j /n 2 < 2.0 0 9 : q q 2 2 q q 3 q q 4 4 q q 5 7 q 2 q 3 q 2 q 4 2 q 2 q 5 5 q 3 q 4 3 q 3 q 5 6 q 4 q 5 3.70 0 9 C.60 0 9 C.68 0 9 C.66 0 9 C.80 0 9 C.65 0 9 C.64 0 9 C.70 0 9 C.67 0 9 C.63 0 9 C Averaging all values of q i /n and q i q j /n 2, we get: q ave.67 0 9 C

4.3. PROBLEM 4.8 IN THE TEXT BOOK 9 4.3 Problem 4.8 In the text book A parallel beam of α particles with fixed kinetic energy is normally incident on a piece of gold foil. a If 00 α particles per minute are detected at 20, how many will be counted at 40, 60, 80, and 00? b If the kinetic energy of the incident α particles is doubled, how many scattered α particles will be observed at 20? c If the original α particles were incident on a copper foil of the same thickness, how many scattered α particles would be detected at 20? Note that ρ cu 8.9 g/cm 3 and ρ Au 9.3 g/cm 3. Solution a Using Equation 4.6, we get: nφ sin φ 4 2 and so, nφ 2 nφ 2 sin φ 4 2 sin φ 2 2 sin 0 n40 n20 sin 20 4 0.74 00 0.342 6.64 cpm 4

0 CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED PROBLEMS sin 0 n60 n20 sin 30 4 0.74 00 0.500.45 cpm sin 0 n80 n20 sin 40 4 0.74 00 0.643 0.533 cpm sin 0 n00 n20 sin 50 4 0.74 00 0.766 0.264 cpm 4 4 4 b From Equation 4.6 we get: n E 2 α so doubling E α reduces n by a factor of 4, so n at 20 25 cpm. c From Equation 4.6 we get: n Z 2 N ncu nau Z2 Cu N Cu Z 2 Au N Au where Z is the atomic number of the atom of the foil and N is the number of nuclei per unit area in the foil. The number of nuclei per unit area in a foil is given by: N Number of nuclei per unit volume foil thickness Number of nuclei per gram foil density foil thickness Avogadro s Number foil density foil thickness Molecular weight of foil in grams

4.3. PROBLEM 4.8 IN THE TEXT BOOK we then get for each foil: N Cu 6.02 023 8.9 t Cu 63.54 8.43 0 22 t N Au 6.02 023 9.3 t AU 97.0 5.90 0 22 t The number of α particles scattered from the copper foil at 20 is: ncu nau Z2 Cu N Cu Z 2 Au N Au 00 292 8.43 0 22 79 2 5.9 0 22 9.3 cpm

2 CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED PROBLEMS 4.4 Problem 4.3 In the text book a Find the frequency of the electron s orbital motion, f e, around a fixed nucleus of charge + Ze by using Equation 4.24 and f e v/2πr to obtain f e m ek 2 Z 2 e 4 2π 3 n 3 b Show that the frequency of the photon emitted when an electron jumps from an outer to an inner orbit can be written f photon kz2 e 2 2a h m ek 2 Z 2 e 4 2π 3 n 2 i n i + n f 2n 2 i n2 f n i n f For an electronic transition between adjacent orbits, n i n f and f photon m ek 2 Z 2 e 4 2π 3 n i + n f 2n 2 i n2 f Now examine the factor and use n i > n f to argue that n 3 i < n i + n f 2n 2 i n2 f n i + n f < 2n 2 i n2 f n 3 f c What do you conclude about the frequency of emitted radiation compared with the frequencies of orbital revolution in the initial and final states? What happens as n i? Solution

4.4. PROBLEM 4.3 IN THE TEXT BOOK 3 a Given that the orbital frequency of an electron in an atomic orbit is: f e v 2πr and that the electron s angular momentum is quantized, m e vr n then the velocity of the electron in an atomic orbit is: v n m e r and since the radius of an orbit is r n 2 a where a is the orbit with n, and is given by: a 2 m e ke 2 then the orbital frequency of the electron becomes: f e v 2πr n 2πm e r 2 n 2πm e n 4 a 2 m2 ek 2 e 4 2πm e 4 n 3 me k 2 e 4 2π 3 n 3 4. b an electron in an orbit n has an energy E n given by: E n m ek 2 e 4 2 2 n 2 An photon is emitted when the electron jumps from an outer orbit n i to an inner orbit n f, the difference in electron energy is carried by the photon as hf photon, i.e.

4 CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED PROBLEMS hf photon E ni E nf m ek 2 e 4 2 2 f photon m ek 2 e 4 4π 3 n 2 i n 2 i Let us consider; n 2 i n2 i n 2 i n2 f n i + n f n i n f n 2 i n2 f 2n + n f n 2n 2 i n f i n2 f The photon frequency then becomes: f photon m ek 2 e 4 2π 3 for adjacent orbits n i n f +, so ni + n f 2n 2 i n f 2 n i n f n 3 i n f + 3 < n 3 f 4.2 and n i + n f 2n 2 i n2 f n f + + n f 2n f + 2 2n f + 2n f + 2 then we need to prove that: 2n f + 2n f + 2 < n 3 f 2n f + 2n f + 2 < n f

4.4. PROBLEM 4.3 IN THE TEXT BOOK 5 or 2 + n f < 2 + 4n f + 2 Using Equations 4.2 and 4.3 we can see that: n f < 4n f + 2 4.3 n 3 i < n i + n f 2n 2 i n2 f < n 3 f 4.4 c Comparing Equations 4. and 4.4 we find, for n i n f, that: f ei < f photon < f ef Since n i n f +, as n i gets smaller, n f also gets smaller and f ei and f ef get closer together and eventually we get f ei f ef f photon

6 CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED PROBLEMS 4.5 Problem 4.32 In the text book Wavelengths of spectral lines depend to some extent on the nuclear mass. This occurs because the nucleus is not an infinitely heavy stationary mass and both the electron and nucleus actually revolve around their common center of mass. It can be shown that a system of this type is entirely equivalent to a single particle of reduced mass µ that revolves around the position of the heavier particle at a distance equal to the electron-nucleus separation. See Figure 4.2. Here, µ m e M/m e + M, where m e is the electron mass and M is the nuclear mass. To take the moving nucleus into account in the Bohr theory we replace m e with µ. Thus Equation 4.30 becomes E n µke2 2m e a n 2 and Equation 4.33 becomes λ µke2 2m e a hc µ n 2 i m e R Determine the corrected values of wavelength for the first Balmer line n 3 to n 2 transition taking nuclear motion into account for a hydrogen, H, b deuterium, 2 H, and c tritium, 3 H. Deuterium, was actually discovered in 932 by Harold Urey, who measured the small wavelength difference between H and 2 H. n 2 i Figure 4.3: a Both the electron and the nucleus actually revolve around the center of mass. b To calculate the effect of nuclear motion, the nucleus can he considered to be at rest and m is replaced by the reduced mass µ.

4.5. PROBLEM 4.32 IN THE TEXT BOOK 7 The wavelength can be written as: λ Solution µ m e R R µ m e n 2 i n 2 i.673053 0 36 µ λ 5.977098 0 37 µ a For hydrogen atom the reduced mass is: and the wavelength is: µ H m e M H m e + M H 9.09390 0 3.673533 0 27 9.09390 0 3 +.673533 0 27 9.04434 0 3 kg λ H 5.977098 0 37 µ H 5.977098 0 37 9.04434 0 3 6.56504 0 7 m 656.504 nm b Similarly for deuterium 2 H, using M 2H 3.344495 0 27 kg, then reduced mass is: µ 2H m e M 2H m e + M 2H 9.09390 0 3 3.344495 0 27 9.09390 0 3 + 3.344495 0 27 9.0692 0 3 kg

8 CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED PROBLEMS and the wave length is λ 2H 5.977098 0 37 µ 2H 5.977098 0 37 9.0692 0 3 6.563254 0 7 m 656.3254 nm c Similarly for tritium 3 H, using M 3H 5.008234 0 27 kg, then reduced mass is: and the wave length is µ 3H m e M 3H m e + M 3H 9.09390 0 3 5.008234 0 27 9.09390 0 3 + 5.008234 0 27 9.07733 0 3 kg λ 3H 5.977098 0 37 µ 3H 5.977098 0 37 9.07733 0 3 6.562663 0 7 m 656.2663 nm

4.6. PROBLEM 4.44 IN THE TEXT BOOK 9 4.6 Problem 4.44 In the text book In a hot star, a multiply ionized atom with a single remaining electron produces a series of spectral lines as described by the Bohr model. The series corresponds to electronic transitions that terminate in the same final state. The longest and shortest wavelengths of the series are 63.3 nm and 22.8 nm, respectively. a What is the ion? b Find the wavelengths of the next three spectral lines nearest to the line of longest wavelength. Solution a For a multiply ionized atoms with just one electron left, the energy of the electron is given by: E n ke2 2a Z 2 where Z is the charge of the nucleus of the atom and R is Rydberg constant. The wavelength of an emitted radiation from such atom is then: n 2 λ Z2 R n 2 i 4.5 The shortest wavelength λ s is emitted when the electron energy change is maximum, i.e. when n i is very large or /n i /n f, so Z2 R λ s 4.6 The longest wave λ l, is emitted when the electron energy change is minimum, i.e. when n i n f +, so

20 CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED PROBLEMS Z 2 R λ l Z2 R n f + 2 [ ] 2 nf n f + 4.7 Dividing Equation 4.6 by Equation 4.7 we get: and 2 λ s nf λ l n f + n f n f + λ s λ l 22.8 63.3 0.800 n f 4 Substituting for n f 4 into Equation 4.6 we solve for Z. Z Rλ s 6.097 0 7 22.8 0 9 7.998 8 The ion has 8 charges on its nucleus it must an ion of O and since it lost all its electrons except one then it is O 7+. b The line with the longest wavelength comes from a transition from n 5 orbit to n 4 orbit. The nearest three lines will come form transition from n 6, 7, 8 to n 4, substituting with these values in Equation 4.5 we get:

4.6. PROBLEM 4.44 IN THE TEXT BOOK 2 λ Z 2 R n 2 i.424 0 9 n 2 i λ.424 0 9 4 2 6 2 4.0 nm λ 2.424 0 9 4 2 7 2 33.83 nm λ 3.424 0 9 4 2 8 2 30.38 nm