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Transcript
Quantum Mechanics Unit Review Answers AP Physics Conceptual Questions: 1. How does the idea that light energy is quantized explain the ultraviolet catastrophe (the fact that less light energy is put out at high frequencies by a blackbody than would be expected based on early theory). The idea that light energy is quantized explains the lack of high frequency light put off by a blackbody because high frequency light would come in high energy quanta. To produce a single quantum (now called a photon) of high frequency light requires a large amount of energy (on the atomic scale) and there is a limit to how much energy any individual electron or proton in a blackbody possesses to create a photon. As you look at higher and higher frequency light, there are fewer and fewer particles capable of creating that quantum and so fewer of those photons are made until you get to high enough frequency that none of that quantum can be made by that object. If you raised the object’s temperature that would give more energy to be divided among the particles inside the object, which would raise the maximum amount of energy any individual particle could have and therefore, the energy of the photon it could create and the frequency of the light it puts off 2. How many photons are absorbed by any one electron in a photoelectric material? How do(es) the absorbed photon(s)’s energy affect the electron? An electron absorbs one photon at any time. Absorbing a photon gives energy to the electron. This goes into raising the electron’s potential energy (moving it further away from the nucleus) and raising the electron’s kinetic energy. If the photon has enough energy, it will first raise the potential energy enough that the electron is ionized or separated from the nucleus (the amount of energy required to do this is called the work function) and any energy left over after that goes into kinetic energy. 3. What happens to the wavelength of a photon that undergoes Compton scattering off a free electron? Why? The wavelength of a photon that is scattered off an electron will increase. When a photon hits a free electron, the electron rebounds from the impact. The electron cannot simply absorb all of the photon’s energy because then momentum would not be conserved (an electron and a photon will not have the same amount of momentum if they have the same amount of energy). This means that the photon must “rebound” off of the electron and come out of the collision traveling in some direction. As the electron flies away from the collision, it has kinetic energy that must have come from the photon that hit it. This means that the photon that comes out of the collision has less energy than it did before the collision. For a photon, lower energy means lower frequency and lower frequency implies a longer wavelength. 4. a) How does Bohr’s model of the hydrogen atom explain why gaseous elements that are stimulated emit line spectra? Bohr’s model of the hydrogen atom says that the energy of the electrons in orbit around the nucleus is quantized – only particular values are allowed. The reason that gaseous elements only release particular lines is that the electron can only give up the specific amount of energy between one orbit and another. Because the energy released as electrons change orbits is emitted as a photon, only photons with energy matching the gap between orbits will ever be emitted – those photons, with their specific frequency and wavelength are the ones that make one particular line. Because there are many different orbits and many more different combinations of starting orbit and finishing orbit, there will be many different energy photons released from the gas, each one creating a different line. b) Why do those elements absorb the same wavelengths from a continuous spectrum passing through them that they would emit on their own? When an atom absorbs a photon, the electron jumps from a lower orbit to a higher orbit. The only photons that can be absorbed are the ones that have exactly the right amount of energy to lift the electron to a particular higher orbit. That amount of energy is the energy of the photon that is emitted as the electron falls back down to the lower level where it started. Therefore the lines in the emission spectrum and absorption spectrum have the same frequency and wavelength. 5. What components are necessary to build a laser? A laser must have a material that will “lase”, that is which will undergo stimulated emission. For this to happen, the material should have a metastable high-energy state where electrons can remain for some time. This allows a population inversion to develop and be maintained. To get the electrons up to the metastable high-energy state without stimulating emission, it is useful if the material will also fluoresce, so you can lift it up to a higher energy state from which it will quick decay into the metastable state. The final piece is a set of good (but not perfect) mirrors to keep the photons bouncing back and forth, continually stimulating emission. At least one of the mirrors must let some of the photons out to create the laser beam. 6. Why do we believe DeBroglie’s idea that particles like an electron are also waves in some way? There are two major points that makes DeBroglie’s idea convincing: electron diffraction and Bohr’s orbits. It turns out that electrons shot through two tiny slits will produce an interference pattern on the screen behind the slits, exactly like light with the same wavelength. Realistically, you can’t construct slits small enough to diffract electrons, but you can use the regular spacing between atoms in a crystalline solid as a set of slits that serves admirably, producing exactly the results that the wave theory of electrons predicts. The explanation of why Bohr’s model only includes the particular orbits it does was perhaps more significant when the theory was first proposed. If electrons are waves, they can only exist in orbits where the wave reinforces itself by ending up in the same phase at the end of the orbit as it did in the beginning of the orbit so that the ends of the wave constructively interfere. This means that the only orbits allowed in Bohr’s model are the ones where an integer number of electron waves fit around the circumference of the orbit. 7. Why is the quantum mechanical model of the atom different from Bohr’s model of the hydrogen atom? Because DeBroglie had already shown that Bohr’s model fits with a wave model of electron behavior, there was no need to adjust the model simply to include that revolutionary idea. However, the quantum mechanical model of the atom does make two major changes to Bohr’s model. First it does away with Bohr’s orbits because the Heisenberg uncertainty principle won’t allow enough precision in an electron’s position and momentum to maintain a nice circular orbit. Orbits are replaced with orbitals which are regions of space where the electron is likely to be found. Second, the orbitals occupy all three dimensions of space. This opens up more variety in orbital type. Bohr’s orbits only varied in size but quantum mechanical orbitals vary in size, shape and complexity. They have three quantized numbers that describe them instead of just one: n, the principal quantum number, describes how the wave fits in the radial direction (away from the nucleus) and sets the base level for the energy of the orbital; l, the angular momentum quantum number describes how the wave fits around the circumference of the orbital [incidentally telling us how much angular momentum an electron in that orbital will have] and affects the way an electron in that orbital will respond to other electrons [which is expressed in the electron’s energy in that orbital]; and ml, the magnetic quantum number, describes the orientation of an orbital with angular momentum, which affects the energy of the orbital when an external magnetic field is applied to the atom (called the Zeeman effect). [There is one other quantum number that applies to electrons in an atom, but it a property of the electron itself, not the orbital. The electron’s spin, ms, tells the way the electron’s intrinsic angular momentum (separate from orbital angular momentum) is pointing and the way a magnetic field will affect the electron] 8. The same types of orbits exist in every atom that there is, though they become more appealing to an electron in an atom that has more protons. a) How does having more electrons change the energy of orbits in an atom? Electrons repel each other, which results in a higher potential energy for the electrons in multielectron atoms. Because the potential energy of an electron in an atom is negative, this means that other electrons makes the potential energy of the electron less negative, bringing it closer to 0 J and freedom. All orbits end up at a higher energy because of the repulsion between electrons, but some orbits experience the repulsion to a greater degree. The higher the angular momentum quantum number of the orbit that the electron is in – this is the number we represent with l, which describes the wave’s circumferential behavior – the more the electron’s energy will be raised. In an atom with only one electron, all orbits with the same principle quantum number – this is the number we represent with n, which describes the wave’s radial behavior – have the same energy. In an atom with more than one electron, the orbits in the same energy level with have different energies, based on their angular momentum quantum numbers. The l = 1 orbits will be at slightly higher energy than the l = 0 orbits for the same shell, as shown in the picture. b) How does an external magnetic field change the energy of the orbits in an atom? An external magnetic field will adjust the energy of the orbits in different ways depending on how the orbit is oriented – which is indicated by the magnetic quantum number, m. The orbits with higher m values will be adjusted in energy by a greater amount, producing an energy level diagram like the one shown to the left. Notice that this means we will perceive an energy difference in the photon given off when an electron drops from or into different orbits with the same l value but different m values. This shows up on the object’s spectrum as multiple lines produced where there would only be one if there were no magnetic field. This picture shows how one spectrum might look different in a magnetic field. Practice Calculations: 1. When a monochromatic light with a wavelength of 254 nm falls onto the surface of a particular metal it causes a photocurrent to flow. A stopping voltage of 2.30 V is required to totally block the photocurrent. a) What is the work function of the material? Ephoton = hf = Kmax + 0 0 = hf - Kmax and Kmax = eV 8 3 x 10 m/s hf = 4.136 x 10-15 eV·s x 254 x 10-9 m = 4.89 eV and Kmax = e x 2.30 V = 2.30 eV 0 = hf - Kmax = 4.89 eV – 2.30 eV = 2.59 eV b) What is the cutoff wavelength for this metal? Cutoff value is when photon has just enough energy to overcome work function and can give no kinetic energy to the electron. hf0 = 0 0 2.59 eV f0 = h = 4.136 x 10-15 eV·s = 6.262 x 1014 Hz c 3 x 108 m/s = f = 6.626 x 1014 Hz = 4.79 x 10-7 m = 479 nm 0 c) Will light with a wavelength of 523 nm be able to cause a photocurrent from this material? No, it has a longer wavelength and therefore a lower frequency than 479 nm light, which means that the energy of one photon is lower than the photon energy for 479 nm light. Lower energy means that it will not be able to add enough energy to “overcome” the work function and free an electron. 2. A different metal has a cutoff wavelength of 365 nm. a) What is the work function of this metal? c hf0 = 0 and =f c 3 x 108 m/s f0 = = 365 x 10-9 m = 8.219 x 1014 Hz 0 0 = hf0 = 4.136 x 10-15 eV·s x 8.219 x 1014 Hz = 3.40 eV = 5.44 x 10-19 J b) What stopping voltage would be required to prevent a monochromatic light of 67.4 nm from creating a photocurrent? hf = Kmax + 0 and eV = Kmax Kmax Kmax = hf - 0 and V= e c 3 x 108 m/s f = = 67.4 x 10-9 m = 4.451 x 1015 Hz Kmax = hf - 0 = 4.136 x 10-15 eV·s x 4.451 x 1015 Hz – 3.40 eV = 15.01 eV Kmax 15.01 eV V= e = = 15.01 V e 3. An experiment involves scattering 0.071 nm photons off of free electrons. What is the wavelength of the scattered photons at each of the following angles? a) 30° b) 70 ° c) 130° h 6.626 x 10-34 J·s a) = m c(1 – cos) = 9.11 x 10-31 kg x 3 x 108 m/s(1 – cos30) = 3.24 x 10-13 m = 0.000324 nm 0 f = i + = 0.071 nm + 0.000324 nm = .071324 nm h 6.626 x 10-34 J·s b) = m c(1 – cos) = 9.11 x 10-31 kg x 3 x 108 m/s(1 – cos70) = 1.59 x 10-12 m = 0.00159 nm 0 f = i + = 0.071 nm + 0.00159 nm = .07259 nm h 6.626 x 10-34 J·s c) = m c(1 – cos) = 9.11 x 10-31 kg x 3 x 108 m/s(1 – cos130) = 3.98 x 10-12 m = 0.00398 nm 0 f = i + = 0.071 nm + 0.00398 nm = .07498 nm 4. Using the Bohr model of the hydrogen atom, calculate the orbital radius, orbital speed and energy for the following energy levels: a) n = 2 b) n = 3 c) n = 6 d) n = 7 n2h2 = 0.05298 nm x n2 4mke2 nh n vn = = 1.164 x 10-4 m2/s r 2mr 2mk2e4 1 1 En = - ( h2 )n2 = -13.6 eV x n2 rn = a) r2 = 0.05298 nm x 22 = 0.21192 nm 2 v2 = = 1.164 x 10-4 m2/s = 1.0985 x 106 m/s 0.21192 x 10-9 m 1 E2 = -13.6 eV x 22 = -3.4 eV b) r3 = 0.05298 nm x 32 = 0.47682 nm 3 v3 = = 1.164 x 10-4 m2/s 0.47682 x 10-9 m = 7.324 x 105 m/s 1 E3 = -13.6 eV x 22 = -1.51 eV c) r6 = 0.05298 nm x 62 = 1.9073 nm 6 v6 = = 1.164 x 10-4 m2/s 1.9073 x 10-9 m = 3.662 x 105 m/s 1 E6 = -13.6 eV x 62 = -0.38 eV d) r7 = 0.05298 nm x 72 = 2.5960 nm 7 v7 = = 1.164 x 10-4 m2/s 2.5960 x 10-9 m = 3.129 x 105 m/s 1 E7 = -13.6 eV x 72 = -0.28 eV 5. Find the photon energy and wavelength of the light emitted for each of the following quantum leaps inside a hydrogen atom: (note – check out problem 6) a) n = 6 to n = 2 b) n = 7 to n = 2 c) n = 7 to n = 6 Ephoton = Ei - Ef f= E h = c f a) Ephoton = E6 – E2 = -0.38 eV – (-3.4 eV) = 3.02 eV = 4.83 x 10-19 J E 4.83 x 10-19 J f = h = 6.626 x 10-34 J·s = 7.30 x 1014 Hz c 3 x 108 m/s = f = 7.30 x 1014 Hz = 4.10 x 10-7 m = 410 nm b) Ephoton = E7 – E2 = -0.28 eV – (-3.4 eV) = 3.12 eV = 4.99 x 10-19 J E 4.99 x 10-19 J f = h = 6.626 x 10-34 J·s = 7.54 x 1014 Hz c 3 x 108 m/s = = = 3.98 x 10-7 m = 398 nm f 7.54 x 1014 Hz c) Ephoton = E7 – E6 = -0.28 eV – (-0.38 eV) = 0.1 eV = 1.602 x 10-20 J E 1.602 x 10-20 J f = h = 6.626 x 10-34 J·s = 2.42 x 1013 Hz c 3 x 108 m/s = f = 2.42 x 1013 Hz = 1.24 x 10-5 m = 12.4 m 6. Calculate the deBroglie wavelength for an electron in the second and third energy levels of a Bohr atom. Check that an integer number of wavelengths fit around the circumference of the orbit. h h = p = mv and based on the results from question 5 h h a) = p = mv h h b) = p = mv 6.626 x 10-34 J·s = (9.11 x 10-31 kg)(1.0985 x 106 m/s) 6.626 x 10-34 J·s = (9.11 x 10-31 kg)( 7.324 x 105) = 6.62 x 10-10 m = 0.662 nm = 9.931 x 10-10 m = 0.9931 nm C2 = 2r2 = 2(0.21192 nm) = 1.33 nm C2 = 2r3 = 2(0.47682 nm) = 2.996 nm C 1.33 nm = = 2.009 ≈ 2 0.662 nm C 2.996 nm = = 3.017 ≈ 3 0.9931 nm 7. a) You shine a red light on an electron and measure its position with an uncertainty of 750 nm. What is the minimum uncertainty about its momentum? How much uncertainty in its speed? h h 6.626 x 10-34 J·s x·p ≥ so p ≥ = = 1.41 x 10-28 kg·m/s 2 2·x 2·750 x 10-9 m b) You switch to a blue light and measure the position with an uncertainty of 450 nm. What is the minimum uncertainty in its momentum? How much uncertainty in its speed? h h 6.626 x 10-34 J·s x·p ≥ so p ≥ = = 2.34 x 10-28 kg·m/s 2 2·x 2·450 x 10-9 m 8. a) How many different emission lines can be produced by an electron jumping from the third energy level of a single-electron atom to the 2nd energy level of an atom under normal conditions? There are 9 different orbitals in the 3rd energy level, with quantum numbers (n, l, ml): (3,0,0), (3,1,1), (3,1,0), (3,1,-1), (3,2,2), (3,2,1), (3,2,0), (3,2,-1), (3,2,-2). In a single-electron atom, all orbitals in the same energy level are degenerate – an electron in them has exactly the same amount of energy. So there is only one energy value possible in the 3rd energy level in a single-electron atom. There are 4 different orbitals in the 2nd energy level, with quantum numbers (n, l, ml): (2,0,0), (2,1,1), (2,1,0), (2,1,-1). In a single-electron atom, all orbitals in the same energy level are degenerate – an electron in them has exactly the same amount of energy. So there is only one energy value possible in the 2nd energy level in a single-electron atom. This means that an electron jumping from the 3rd energy level to the 2nd energy level can only lose 1 amount of energy no matter which orbital they jump from and which they jump to so there will only be one photon energy and one emission line produced by electrons jumping between those levels b) How many different emission lines can be produced by an electron jumping from the third energy level of a multi-electron atom to the 2nd energy level of an atom under normal conditions? There are 9 different orbitals in the 3rd energy level, with quantum numbers (n, l, ml): (3,0,0), (3,1,1), (3,1,0), (3,1,-1), (3,2,2), (3,2,1), (3,2,0), (3,2,-1), (3,2,-2). In a multi-electron atom, there is a difference in energy for orbitals with different values of angular momentum (l), though orbitals with different values of the magnetic quantum number are degenerate. So there are three different energy value possible in the 3rd energy level in a single-electron atom. There are 4 different orbitals in the 2nd energy level, with quantum numbers (n, l, ml): (2,0,0), (2,1,1), (2,1,0), (2,1,-1). In a multi-electron atom, there is a difference in energy for orbitals with different values of angular momentum (l), though orbitals with different values of the magnetic quantum number are degenerate. So there is are two energy values possible in the 2nd energy level in a single-electron atom. This means that an electron jumping from the 3rd energy level to the 2nd energy level can lose 6 different amounts of energy depending on which l-value orbital they jump from and which they jump to so there will be six photon energies and six emission lines produced by electrons jumping between those levels c) How many different emission lines can be produced by an electron jumping from the third energy level of a multi-electron atom to the 2nd energy level of an atom in a strong magnetic field? There are 9 different orbitals in the 3rd energy level, with quantum numbers (n, l, ml): (3,0,0), (3,1,1), (3,1,0), (3,1,-1), (3,2,2), (3,2,1), (3,2,0), (3,2,-1), (3,2,-2). In a multi-electron atom in a magnetic field, there is a difference in energy for all different orbitals. So there are nine different energy values possible in the 3rd energy level in a single-electron atom. There are 4 different orbitals in the 2nd energy level, with quantum numbers (n, l, ml): (2,0,0), (2,1,1), (2,1,0), (2,1,-1). In a multi-electron atom, there is a difference in energy for orbitals with different values of angular momentum (l), though orbitals with different values of the magnetic quantum number are degenerate. So there is are four energy values possible in the 2nd energy level in a single-electron atom. This means that an electron jumping from the 3rd energy level to the 2nd energy level can lose 36 different amounts of energy depending on which l-value and ml-value orbital they jump from and which orbital they jump to so there will be 36 photon energies and 36 emission lines produced by electrons jumping between those levels