Chapter 4 The Particle Nature of Matter. Home Work Solutions 4. Problem 4.0 (In the text book) A typical Rutherford scattering apparatus consists of an evacuated tube containing a polonium- 20 α source (5.2-M ev α s), collimators, a gold foil target, and a special alpha-detecting film. The detecting film simultaneously measures all the alphas scattered over a range from φ = 2.5 to 2.5. (See Figure (4.).) The total number of counts measured over a week s time falling in a specific ring (denoted by its average scattering angle) and the corresponding ring area are given in Table (4.). Figure 4.: (a) Side view of Rutherford s scattering apparatus: φ is the scattering angle. (b) End view of the Rutherford apparatus showing the film detector end cap with grid marking the angle φ. The α particles damage the film emulsion and after development show up as dots within the rings.
2 CHAPTER 4. THE PARTICLE NATURE OF MATTER. HOME WORK SOLUTIONS Table 4.: Data To Be Used in Problem 0 Angle φ Counts/Ring Ring Area (degree) (cm 2 ) 2.5 605 0.257 3.5 63 0.360 4.5 520 0.463 5.5 405 0.566 6.5 30 0.669 7.5 20 0.772 8.5 22 0.875 9.5 78 0.987 0.5 65.08.5 66.8 2.5 44.29 (a) Find the counts per area at each angle and correct these values for the angle independent background. The background correction may be found from a seven-day count taken with the beam blocked with a metal shutter in which 72 counts were measured evenly distributed over the total detector area of 8.50 cm 2. (b) Show that the corrected counts per unit area are proportional to sin 4 (φ/2) or, in terms of the Rutherford formula, Equation 4.6, n A = C sin 4 (φ/2) Notes: If a plot of ( n/a) versus φ will not fit on a single sheet of graph paper, try plotting log( n/a) versus log[/(sin φ/2) 4 ]. This plot should yield a straight line with a slope of and an intercept that gives C. Explain why this technique works. Solution (a) The back ground correction BG is: BG = Total Background Counts Area of the Detector = 72 8.5 = 8.47 count/cm2
4.. PROBLEM 4.0 (IN THE TEXT BOOK) 3 Table 4.2: Counts/Area and corrected counts using the data in Table (4.) Angle φ Counts/Ring Ring Area Counts/Area Corrected Counts/Area (degree) (cm 2 ) n/a 2.5 605 0.257 2354.09 2345.62 3.5 63 0.360 752.78 744.3 4.5 520 0.463 23. 4.64 5.5 405 0.566 75.55 707.08 6.5 30 0.669 449.93 44.46 7.5 20 0.772 260.36 25.89 8.5 22 0.875 39.43 30.96 9.5 78 0.987 79.03 70.56 0.5 65.08 60.9 5.72.5 66.8 55.93 47.46 2.5 44.29 34. 25.64 The background corrected count per unit area n/a is given by: n A = Counts/Ring Ring Area BG We can calculate the counts per unit area and the corrected counts per unit area using the the data given in Table(4.2). The results of the calculations are shown in Table (4.2). (b) To show that n/a sin 4 (φ/2) we plot the two quantities and the data should lie on a straight line that passes through origin and has a slope of the constant of proportionality. Table (4.3) shows the the calculations for sin 4 (φ/2) and n/a is taken from Table (4.2) It is obvious from the values of [sin(φ/2)] 4 that plotting n/a vs [sin(φ/2)] 4 is not practical. Instead we can convert the relationship between [sin(φ/2)] 4 and n/a from: to log n A [sin(φ/2)]4 = C [sin(φ/2)] 4 ( ) n = log C + log[sin(φ/2)] 4 A
4 CHAPTER 4. THE PARTICLE NATURE OF MATTER. HOME WORK SOLUTIONS Table 4.3: Calculation of [sin(φ/2)] 4. φ φ/2 sin(φ/2) [sin(φ/2)] 4 log[sin(φ/2)] 4 n/a log( n/a) 2.5.25 0.02 445585.37 6.64 2345.62 3.37 3.5.75 0.03 49763.30 6.06 744.3 3.24 4.5 2.25 0.04 420927.8 5.62 4.64 3.05 5.5 2.75 0.05 88723.85 5.28 707.08 2.85 6.5 3.25 0.06 96802.99 4.99 44.46 2.64 7.5 3.75 0.07 5465.98 4.74 25.89 2.40 8.5 4.25 0.07 3353.40 4.52 30.96 2.2 9.5 4.75 0.08 2267.0 4.33 70.56.85 0.5 5.25 0.09 4265.43 4.5 5.72.7.5 5.75 0.0 9925.3 4.00 47.46.68 2.5 6.25 0. 78.97 3.85 25.64.4 If the above equation is valid, then plotting log ( ) n A vs log[sin(φ/2)] 4 produces a straight line with a slope of and a y-intercept of log C. The plot is shown in Figure (4.2). Figure 4.2: log ( ) n A vs log[sin(φ/2)] 4. The red line is the best linear fit to the data. Using the two points on the line we get a slope of 0.94 and the y-intercept gives C = 0.06
4.. PROBLEM 4.0 (IN THE TEXT BOOK) 5 The straight line that best fit the data is shown by the red line in Figure (4.2). The straight line has a slope of 0.94 and a y-intercept = -.22, that gives C = 0.06, which means n/a is proportional to [sin(φ/2)] 4.
6 CHAPTER 4. THE PARTICLE NATURE OF MATTER. HOME WORK SOLUTIONS 4.2 Problem 4.5 (In the text book) (a) Construct an energy-level diagram for the He + ion, for which Z = 2. (b) What is the ionization energy for He +? Solution (a) Normal He atom is neutral and has two electrons and the charge on the its nucleus is + 2e. He + atoms lost one electron and their nuclei still have + 2e each. One can treat He + has a hydrogen atom with nuclear charge of + 2e or Z = 2. The atomic energy levels of He + are, then: So the various energy levels are: E n = 3.6 Z2 n 2 E = 54.4 ev, E 2 = 3.6 ev, E 3 = 6.04 ev, E 4 = 3.4 ev. The energy level diagram is shown in Figure (4.3). Figure 4.3: Energy level diagram for He +
4.2. PROBLEM 4.5 (IN THE TEXT BOOK) 7 (b) The single electron in the He + atom would be in the n = state. To ionize the He + is to take the remaining electron away from the nucleus, i.e. to fully ionize the He atom. To do that one has to provide the electron with 54.4 ev. So, the ionization potential for a He + atom is 54.4 ev
8 CHAPTER 4. THE PARTICLE NATURE OF MATTER. HOME WORK SOLUTIONS 4.3 Problem 4.20 (In the text book) What is the energy of the photon that could cause (a) an electronic transition from the n = 4 state to the n = 5 state of hydrogen and (b) an electronic transition from the n = 5 state to the n = 6 state? Solution The energy of transition E from n i to n f in hydrogen is: ( ) E = 3.6 n 2 i n 2 f (a) ( E = 3.6 6 ) = 0.306 ev 25 moving from n = 4 to n = 5 requires external energy, a photon with energy of 0.306 ev can initiate such transition. (b) ( E = 3.6 25 ) = 0.66 ev 36
4.4. PROBLEM 4.27 (IN THE TEXT BOOK) 9 4.4 Problem 4.27 (In the text book) Show that Balmer s formula, reduces to the Rydberg formula, ( ) n 2 λ = C 2 n 2 2 2 ( λ = R 2 ) 2 n 2 provided that (2 2 /C 2 ) = R. Check that (2 2 /C 2 ) has the same numerical value as R. Solution Since: we get: λ = C 2n 2 n 2 2 4 λ = ( ) n 2 2 2 C 2 n 2 = ( ) 22 C 2 = 22 C 2 = R n 2 ( 2 ) 2 n ( 2 2 ) 2 n 2 where R = 2 2 /C 2 since C 2 = 3.65456 0 4 cm, we get: R = 2 2 = 0972 cm 3.6456 0 5 This value of R is very close to the currently accepted value of 09737 cm.
0CHAPTER 4. THE PARTICLE NATURE OF MATTER. HOME WORK SOLUTIONS 4.5 Problem 4.39 (In the text book) 39. An electron collides inelastically, and head-on with a mercury atom at rest. (a) If the separation of the first excited state and the ground state of the atom is exactly 4.9 ev, what is the minimum initial electron kinetic energy needed to raise the atom to its first excited state and also conserve momentum? Assume that the collision is completely inelastic. (b) What is the initial speed of the electron in this case? (c) What is the speed of the electron and atom after the collision? (d) What is the kinetic energy (in electron volts) of the electron after collision? Is the approximation that the electron loses all of its kinetic energy in an inelastic collision justified? Solution (a) Energy and momentum conservation gives: 2 m ev 2 = 2 (m e + M)V 2 + 4.9 (4.) m e v = (m e + M)V (4.2) where m e is the mass of the electron, M is the mass of the Hg atom, v is the velocity of the electron before collision and V is the velocity of the combined electron and Hg after collision. We find V from Equation (4.2) as: V = m ev m e + M Using Equation (4.3) in Equation (4.) we get: (4.3) 2 m ev 2 = ( 2 (m me v e + M) m e + M = m 2 ev 2 2 m e + M + 4.9 ) 2 + 4.9
4.5. PROBLEM 4.39 (IN THE TEXT BOOK) 2 m ev 2 = ( ) 2 m ev 2 me + 4.9 m e + M m ) e = 4.9 m e + M ( ) me + M m e = 4.9 m e + M ( ) M 2 m ev 2 = 4.9 m e + M ( ) 2 m ev 2 me + M = 4.9 M ( ) 5.486 0 4 (u) + 20.97(u) = 4.9(eV ) 20.97(u) = 4.90003 ev (4.4) ( 2 m ev 2 2 m ev 2 (b) The velocity of the electron before collision can be calculated from its kinetic energy before collision. Using Equation (4.4) we get: v = = 4.90003 m e 4.90003(eV ) 5 0 3 ev/c 2 = 4.379 0 3 c = 4.379 0 3 3 0 8 =.34 0 6 m/s (4.5) (c) The velocity of the combined electron and Hg after collision is V which is also the velocity of of the electron, using Equations (4.2) and (4.5) we get: V = m e m e + M v = 5.486 0 4 (u).34 5.486 0 4 06 (u) + 20.97(u) = 3.570 m/s (4.6)
2CHAPTER 4. THE PARTICLE NATURE OF MATTER. HOME WORK SOLUTIONS (d) The kinetic energy of the electron K e after collision is : K e = 2 m ev 2 = 2 9.09 0 3 (3.570) 2 = 5.805 0 30 J = 5.805 0 30 ev.602 0 9 = 3.623 0 ev (4.7) In inelastic scattering all the incoming energy is given to the combined masses. So, it is justified to assume that the electron loses all it incoming kinetic energy, as can be seen from Equation (4.7) that the kinetic energy of the electron after collision is extremely small compared to it incoming energy.