physical chemistry ii

Download Physical Chemistry II

Post on 12-Mar-2015




10 download

Embed Size (px)


1-1 Chapter 1: From Classical to Quantum Mechanics Problem numbers in italics indicate that the solution is included in the Students Solutions Manual. Questions on Concepts Q1.1) How did Planck conclude that the discrepancy between experiments and classical theory for blackbody radiation was at high and not low frequencies? The experimental results and the classical theory agree in the limit of low frequencies and diverge at high frequencies. Q1.2) The inability of classical theory to explain the spectral density distribution of a blackbody was called the ultraviolet catastrophe. Why is this name appropriate? The divergence between the classical theory and the experimental results is very pronounced in the UV region. In particular, because the classical theory predicts that the spectral density increases as 2 , the total energy radiated by a blackbody is predicted to be infinite. This prediction was clearly wrong, and the lack of agreement with the experimental results was viewed by scientists of the time as a catastrophe. Q1.3) Why does the analysis of the photoelectric effect based on classical physics predict that the kinetic energy of electrons will increase with increasing light intensity? In the classical theory, more light intensity leads to more energy absorption by the solid and the electrons. At equilibrium, as much energy must leave the surface as is absorbed. Therefore, the electrons will have a greater energy. Classical physicists thought that the greater energy would manifest as a higher speed. In fact, it appears as a larger number of electrons leaving the surface with a frequency determined by the photon energy and the work function of the surface. Q1.4) What did Einstein postulate to explain that the kinetic energy of the emitted electrons in the photoelectric effect depends on the frequency? How does this postulate differ from the predictions of classical physics? Einstein postulated that the energy of light depends on the frequency, whereas in classical theory, the energy depends only on the intensity and is independent of the frequency. He further postulated that the energy of light could only be an integral multiple of . hv Q1.5) Which of the experimental results for the photoelectric effect suggests that light can display particle-like behavior? Chapter 1/From Classical to Quantum Mechanics 1-2 The fact that at very low intensity the incident light is still able to eject electrons, if its frequency is above the threshold frequency, suggests that the energy of the light can be concentrated in a region of atomic dimensions at the surface of the solid. This suggests particle-like behavior. Q1.6) In the diffraction of electrons by crystals, the volume sampled by the diffracting electrons is on the order of 3 to 10 atomic layers. If He atoms are incident on the surface, only the topmost atomic layer is sampled. Can you explain this difference? Whereas electrons can penetrate through the topmost layer into the solid, He atoms are too large to allow penetration at thermal energies. Therefore electrons sample a number of atomic layers below the surface, and He atoms are only sensitive to the outermost layer. Q1.7) In the double-slit experiment, researchers found that an equal amount of energy passes through each slit. Does this result allow you to distinguish between purely particle-like or purely wave-like behavior? No. This result would be expected for both waves and particles. In the case of particles, the same number would pass through each slit for a large number of incident particles. For a wave, the fraction of the intensity of the wave that passes through each slit would be the same. Q1.8) Is the intensity observed from the diffraction experiment depicted in Figure 1.6 the same for the angles shown in parts (b) and (c)? Yes. For all minima in a graph of intensity versus frequency, the intensity is zero as seen in Figure 1.5. Q1.9) What feature of the distribution depicted as Case 1 in Figure 1.7 tells you that the broad distribution arises from diffraction? The intensity goes through a minimum and increases again on both sides of the maximum. This can only occur through wave interference. Q1.10) Why were investigations at the atomic and subatomic levels required to detect the wave nature of particles? The wavelength of particles with greater mass is so short that particle diffraction could not be observed experimentally. Therefore, the wave-particle duality present even in heavier particles was not observed. Chapter 1/From Classical to Quantum Mechanics 1-3 Problems P1.1) The distribution in wavelengths of the light emitted from a radiating blackbody is a sensitive function of the temperature. This dependence is used to measure the temperature of hot objects, without making physical contact with those objects, in a technique called optical pyrometry. In the limit ( ) 1, hc k T >> the maximum in a plot of ( ) , versus T is given by max/ 5 . hc kT = At what wavelength does the maximum in ( ) ,T occur for T = 450, 1500, and 4500 K? According to Example Problem 3.1, max5h ck T = . 34 8 16max 23 16.626 10 J s 2.998 10 m s6.40 10 m5 1.381 10 J K 450 K = = at 450 K. max for 1500 K and 4500 K is 1.92106 m and 6.39107 m, respectively. P1.2) For a monatomic gas, one measure of the average speed of the atoms is the root mean square speed, 12 23v v ,rms kTm= = in which m is the molecular mass and k is the Boltzmann constant. Using this formula, calculate the de Broglie wavelength for He and Ar atoms at 100 and at 500 K. 3423 1 27 1rms106.626 10 J sv33 1.381 10 J K 100 4.003amu 1.661 10 kgamu1.26 10 mh hm k T m K = = = = for He at 100 K. =5.651011 m for He at 500 K. For Ar, = 4.00 1011 m and 1.791011 m at 100 K and 500 K, respectively. P1.3) Using the root mean square speed, 12 23v v ,rms kTm= = calculate the gas temperatures of He and Ar for which = 0.20 nm, a typical value needed to resolve diffraction from the surface of a metal crystal. On the basis of your result, explain why Ar atomic beams are not suitable for atomic diffraction experiments. For He, ( )( )22234223 1 27 1 936.626 10 J s40 K3 1.381 10 J K 4.003amu 1.661 10 kgamu 0.20 10 mhTk m == = For Ar, T = 4.0 K. Chapter 1/From Classical to Quantum Mechanics 1-4The argon temperature is well below its liquifaction temperature at 4 K. It will not be possible to make an atomic beam of Ar atoms with this wavelength with conventional means. P1.4) Electrons have been used to determine molecular structure by diffraction. Calculate the speed of an electron for which the wavelength is equal to a typical bond length, namely, 0.150 nm. 346 131 96.626 10 J sv 4.85 10 m s9.109 10 kg 0.150 10 mp hm m = = = = P1.5) Calculate the speed that a gas-phase oxygen molecule would have if it had the same energy as an infrared photon ( = 104 nm), a visible photon ( = 500 nm), an ultraviolet photon ( = 100 nm), and an X-ray photon ( = 0.1 nm). What temperature would the gas have if it had the same energy as each of these photons? Use the root mean square speed, 12 23v v ,rms kTm= = for this calculation. ( )34 8 11127 92 2 2 6.626 10 Js 2.998 10 m sv 864 ms32.0amu 1.661 10 kg amu 10000 10 mE h cm m = = = = for = 104 nm.The results for 500 nm, 100 nm and 0.1 nm are 3.87103 m s1, 8.65103 m s1, and 2.73105 m s1. We calculate the temperature using the formula ( )21 121 132.0kg mol 864m sv958K3 3 8.314 J mol KrmsMTR = = = = 104 nm. The results for 500 nm, 100 nm and 0.1 nm are 1.92104 K, 9.60104 K, and 9.56107 K. P1.6) Pulsed lasers are powerful sources of nearly monochromatic radiation. Lasers that emit photons in a pulse of 10-ns duration with a total energy in the pulse of 0.10 J at 1000 nm are commercially available. a) What is the average power (energy per unit time) in units of watts (1 W = 1 J/s) associated with such a pulse? b) How many 1000-nm photons are emitted in such a pulse? a) 7 180.10 J1.0 10 J s1.0 10 sEPt = = = b) 178 134 160.10 J5.0 102.998 10 m s6.626 10 J s1.000 10 mpulse pulsephotonE ENcE h = = = = Chapter 1/From Classical to Quantum Mechanics 1-5 P1.7) Assume that water absorbs light of wavelength 3.00 106 m with 100% efficiency. How many photons are required to heat 1.00 g of water by 1.00 K? The heat capacity of water is 75.3 J mol1 K1. ,1 1 6, 191 34 8 11.00g 75.3J K mol 1.00K 3.00 10 m6.31 1018.02g mol 6.626 10 J s 2.998 10 msp mp mhcE Nh N nC TC TmNM hc = = = = = = P1.8) A 1000-W gas discharge lamp emits 3.00 W of ultraviolet radiation in a narrow range centered near 280 nm. How many photons of this wavelength are emitted per second? 1 1 1 118 134 8 193.00W 1J s W 3.00W 1J s W4.23 10 s6.626 10 J s 2.998 10 ms280 10 mtotalphotonEnhcE = = = = P1.9) A newly developed substance that emits 225 W of photons with a wavelength of 225 nm is mounted in a small rocket such that all of the radiation is released in the same direction. Because momentum is conserved, the rocket will be accelerated in the opposite direction. If the total mass of the rocket is 5.25 kg, how fast will it be traveling at the end of 365 days in the absence of frictional forces? The number of photons is given by 1 120 134 8 191 J s 1 J s3.00W 225WW W2.547 10 s6.626 10 J s 2.998 10 ms225 10 mtotalphotonEnhcE = = = = The momentum of one photon is 3427 196.626 10 J s2.945 10 kg ms225 10 mhp = = = The force is given by the rate of change of momentum. ( )27 1 20 1 7 22.945 10 kg ms 2.547 10 s 7.501 10 kg msd n pFdt = =


View more >