Reading Quiz

Question 1:

Answer:

1. The alpha particle that Rutherford Geiger and Mardsen used was doubly ionized helium. It was produced during the radioactive decay of uranium, and it's charge over mass was half of the protons, because it includes the mass of two neutrons.
2. The alpha-particle was one of the two types of radiation from uranium, which had a charge per mass ratio that was half that of the proton. At the time Rutherford, Geiger, and Marsden believed these particles were doubly ionized helium.
3. An alpha particle is two neutrons and two protons
4. They alpha-particles were suspected to be doubly ionized helium, they are composed to 2 protons and 2 neutrons.
5. The alpha-particle used by Rutherford, Geiger and Mardsen is one type of radiations from uranium. Its q/m ratio is half of the proton's, which leads the scientists to suspect that it is double ionized helium.
6. When Rutherford and co. discovered that the q/m of the alpha-particle was half that of a proton, they suspected that the alpha-particle was doubly ionized helium. After conducting an experiment where they let an alpha-particle decay in an evacuated chamber and looked at the emission bands via spectroscopy, they discovered the spectral lines of ordinary helium gas, confirming their suspicions.

Question 2:

Answer:

1. Geiger and Mardson contained a raduoactive source in a shield and let some particles escape in the direction of a thin gold foil, where the particles were deflected at various angles depending on what they hit in the material.
2. The Geiger-Marsden experiment consisted in firing alpha-particles from a radioactive source at a thin metal foil (various types of metal were used). Then a zinc-sulfide screen, which emits flashes of visible light when struck, we placed at various angles from behind the thin sheet of foil. In this manner, Geiger and Marsden were able to probe the number of alpha-particles (number of flashes) as a function of the angle from behind the foil, theta.
3. Alpha Particles were shot at extremely thin gold foil. The amount of deflection was then measured.
4. A beam of alpha-particles was directed at a gold foil, and the angle at which the particles were deflected was observed through their interaction with the scintillation screen.
5. In Geiger-Marsden experiment, a narrow beam of alpha particles were used to strike various thin metal foils that were placed between it and the source, and the distribution of these alpha particles after struck can be observed on zinc sulfide screen.
6. The Geiger-Marsden experiment was designed to probe the interiors of atoms. They creates a beam of alpha-particles from a chunk of Bi 214, which they sent to an ultra-thin (~2000 atoms thick) Au foil, measuring the scattering of the alpha particles by looking at a piece of zinc sulfide through a microscope (as zinc sulfide emitted visible light when struck), which could be rotated around the Au foil.

Question 3:

Answer:

1. A much higher number of particles were deflected at large angles than the model predicted for a solid ball of positive paste with electrons scattered inside. This ultimately led to a change in the plum pudding theory of the atom because that theory could not explain this behavior.
2. The 'unexpected' result observed by Geiger and Marsden was that a few alpha-particles were deflected through angles up to and above 90 degrees. If the structure of the atom was as predicted by JJ Thomson, only very small deflections would occur, thus these results would not be possible.
3. There was some scattering at 90 degrees and greater
4. Some of the alpha particles were deflected at angles greater than 90 degrees.
5. Geiger and Marsden found that most of the alpha particles were either undeflected or deflected through very small angles of the order of 1 degree. Unexpectedly, a few particles were deflected through angles as large as 90 degrees or more. The large angle scattering is not what Thomson atomic model could account for.
6. The Thompson model of the atom could account for approx. 1˚ of deflection. However, Geiger and Marsden discovered some alpha-particles that were deflected through angles of around 90˚ or more. Rutherford famously described the results as "the most incredible event that ever happened to me in my life. It was as incredible as if you fired a 15-inch shell at a tissue paper and it came back and hit you.

Question 4:

Answer:

1. Rutherford did a classical calculation with the plum pudding model, and found that the angle corresponded to distance of closest approach of the alpha particle to the atom as a whole, using the force described by Coulombs law.
2. As the impact parameter grows smaller, the scattering angle gets larger. The impact parameter is directly proportional to the cotangent of the scattering angle divided by two.
3. the impact parameter is proportional to cot(theta/2)
4. The impact parameter and the scattering angle are related by b=(kqQ)/(mv^2)cot(theta/2).
5. According to equation 4-3, the smaller the impact parameter, the larger the scattering angle.
6. As given by equation 4-3 in the text, the impact parameter, b = kqQ/(mv^2) * cot(theta/2). Where theta is the impact parameter, k is coulomb's constant, q is the alpha-particle charge, Q is the charge of the nucleus, m is the alpha-particle mass, and v is the velocity of the alpha-particle.

Question 5:

Answer:

1. Geiger and Mardsen conducted their experiment with aluminum rather than gold because it had a smaller atomic number, and thus was a smaller atom, which helped them to isolate variables. They also put varying numbers of mica in front of the aluminum to slow the incident alpha particles.
2. In order to further test Rutherford's construction of the atom, Geiger and Marsden performed tests to determine the number of alpha-particles found at any scattering angle theta. Rutherford had made an equation predicting these numbers and, according to Geiger and Marsden's tests, all of Rutherford's predictions agreed with experimental values. They tested the number of alpha-particles for over four orders of magnitude.
3. they shot alpha particles at other materials and notice different amounts of scattering mostly based on nuclear.
4. They fired alpha particles at the different foils, such as aluminum in order to determine the size of the nucleus.
5. Geiger and Marsden conducted the gold foil experiment that uses a beam of alpha particles to strike an ultrathin gold foil and these alpha particles scattered through various angles. Those scattering at the angle theta is shown with a scintillator and can be viewed by the observer through a small microscope.The experiment can finally derive the method of counting the number of scintillations as a function of theta.
6. Geiger and Marsden performed a series of experiments to garner data to compare with Rutherford's theoretical expression for the number of particles scattered at any angle theta, which matched up extremely well. Since their data worked considerably well with Au, they decided to lower the atomic number of the element they were shooting at in an attempt to penetrate the atom, eventually settling for Al. By calculating the distance of closest approach r from the energy of the alpha-particle, they realized that their distance of closest approach approximated the radius of the nucleus, since this is where their experimental results deviated from the theory.

Question 6:

Answer:

1. Distance of closest approach can tell you the initial kinetic energy of the alpha particle, and for distances of the same magnitude, the particles are scattered at large angles.
2. The distance of closest approach provides an experimental upper limit on the size of the target nucleus, so long as the number of observed alpha-particles at a given scattering angle remained consistent with Rutherford's predictions.
3. how close the alpha particle gets to the center of the atom
4. An experimental upper limit for the size of the target nucleus.
5. The distance of closest approach tells us the at which distance the potential energy equals the original kinetic energy. The force law would hold when the alpha particles are not able to penetrate the nuclear charge. And with the distance of closest approach, we can estimate the radius of certain atoms.
6. As stated above the distance of closest approach approximates the radius of the nucleus that you are approaching.

Question 7:

Answer:

1. Alpha particles with lower kinetic energies and larger distances of closest aproach, the number of particles observed matched the number of particles predicted. However, this broke down at a distance of about 1.2*10^-14m, where the number of predicted particles started to become bigger than the number they observed. This indicates that the approximate size of the aluminum nucleus is 1.2*10^-14m
2. When energetic alpha particles were fired onto Aluminum to study their distance of closest approach, the number of alpha-particles scattered at large angles was no longer consistent with Rutherford's predictions. Instead, according to Figure 4-14, the observed number of alpha-particles became less than the predicted number. However, they also found that if the alpha-particles lost energy by being sent through thin mica sheets then the observed numbers once again agreed with Rutherford's predictions. Thus, the value at which the data began to deviate from Rutherford's predictions represented the approximate radius of the nucleus.
3. by calculating the distance of closest approach and then looking at whether or not your predicted scattering actually happens you can figure out the radius of the atom. in figure 4-14 when the ratio of observed and expected deviates from 1 it means that the alpha particle is interacting with the center.
4. The alpha particles were able to penetrate the nucleus, and thus they were able to determine its size.
5. When the energetic alpha particles were fired onto Aluminum, the number of them that scattered at large angles did not follow the predictions. But when their kinetic energy were reduced, the data again began to agree with the prediction. The value of distance of closest approach at which the data started to deviate from the prediction can be thought of as the surface of nucleus. From these information, Rutherford was able to estimate the radius of Aluminum atom.
6. As you can observe from figure 4-14, the value for the distance of closest approach started to deviate from the predicted value at around 1.2*10^-14. This showed that the nucleus clearly was NOT a point-like particle like Rutherford assumed in his calculations, and as such his theory did not work when the alpha particle penetrated the nucleus.

Question 8:

Answer:

1. Bohr assumed that that the nucleus was a stationary point charge, and that the electron moved in a circular orbit around it such that the force causing circular motion is the same as the force described by Coulombs law.
2. In order to make his description of the atom feasible, Bohr postulated that electrons could move in certain orbits (called stationary states) without radiating and that the atom radiates when an electron makes a transition from one stationary state to another and that the frequency of the emitted radiation is the frequency correspondant to the difference in energy between the two stationary states according to Planck's theory.
3. bohr said some electron orbits were safe orbits meaning that they didnt radiate out energy. and photons were only emitted on changing orbits and the enrgy of the photon came from th difference in energy states.
4. 1. Electrons could move in certain orbits without radiating. 2. The atom radiates when the electron makes a transition from one stationary state to another and that the frequency of the emitted radiation is not the frequency of motion in either stable orbit but is related to the energies of the orbits by Planck's Theory.
5. 1) He firstly postulated that electrons could move in certain orbits without radiating. 2) He secondly postulated that the atom radiates when the electron makes a transition from one stationary state to another and that the frequency f of the emitted radiation is not the frequency of motion in either stable orbit but is related to the energies of the orbits by Planck's theory.
6. Bohr's model assumed that the electron in the hydrogen atom moved in a circular orbit around the positively charged nucleus, with the Coulomb force providing the angular acceleration. Classically, the electron would radiate energy continuously as it orbits, thus causing it to spiral into the nucleus. Bohr's postulates were an attempt to solve this problem: he believed that electrons could move in certain orbits without radiating, calling these orbits stationary states. His second postulate assumed that the electron radiates when it makes a transition from one stationary state to another and that the frequency f of the emitted radiation is NOT the frequency of motion of either stationary state, but rather the energy difference of the energy states in accordance to Planck's theory.

Question 9:

Answer:

1. Every modification made to classical physics to describe quantum must agree with the old, accepted classical physics when applied to macroscopic situations (so every modification has to still agree with all of the observed phenomena that classical physics before quantum described).
2. The "Correspondence Principle" means that any changes that scientists make to classical physics at the submicroscopic level must agree with the classical laws of physics when they are extended from this level into the macroscopic world. Although Bohr's model of the atom has been abandoned for the model explained by quantum theory, this principle has remained.
3. in thelimit of large orbits and large energies, quantum calculations must agree with classical calculation
4. When subatomic calculations are extended to the macroscopic level, they must agree with classical calculations.
5. The "Correspondence Principle" implied that "In the limit of large orbits and large energies, quantum calculations must agree with classical principles." It means that whatever modifications of classical physics are made to describe mater at the submicroscopic level, when the results are extended to the macroscopic world, they must agree with those from the classical laws of physics that have been so abundantly verified in everyday world.
6. The Correspondence Principle was Bohr's third assumption made in an attempt to determine the energies of the stable orbits. It states that "In the limit of large orbits and large energies, quantum calculations must agree with classical calculations. Thus, whatever modifications made to classical physics to describe matter in the atomic level, when brought to the macroscopic level, must agree with those from the classical model of physics that have been so tried and tested.

Question 10:

Answer:

1. Moving charges radiate energy, if the electron was orbiting the nucleus, it would loose energy and crash into the nucleus (in less than a microsecond). He tried to fix this by saying that the electron could orbit in quantized energy levels, only radiating when it switched to a different energy level, described by the radiation or emission lines that we see experimentally. He also assumed that the nucleus was stationary, which according to relativity is the same as assuming that it has infinite mass. There is also a problem with the idea of circular orbits where energy jumps don't agree with the observed energies unless elliptical orbits are allowed
2. The Bohr model was problematic in that it just stated that electrons could move in certain orbits without radiating, an assertion which did not really have much evidence to support it. In addition the Bohr model could not explain the observed fine structure of the hydrogen spectral lines. Although Sommerfield attempted to solve this problem, his energy calculations were classical rather than relativistic and thus failed to explain the fine structure of the spectral lines. In addition, the Bohr model failed to extend to atoms that were more complicated than hydrogen. (To me it seemed like everything else he asserted was not problematic, but this is where I got a little bogged down in the calculations so I may have missed something.)
3. electron orbits are not circular. no real explanation of why some orbits are safe. it only really worked for hydrogen.
4. The assumption that the nucleus is fixed, is equivalent to the nucleus having infinite mass. The electrons are accelerating as the orbit the nucleus, but the frequency of their emissions should change as they lost energy and spiraled into the nucleus.
5. 1) The Bohr's model was proved difficult to apply to atoms more complicated than hydrogen. Quantitative calculations of the energy levels of atoms of more than electron could not be made from the model. 2) Bohr's theory state that the energy of an electron in the first Bohr orbit is proportional to the square of the nuclear charge, but Moseley reasoned hat the energy of a characteristic x-ray photon should vary as the square of the atomic number of the target element in the x-ray tube.
6. Bohr assumed that the nucleus is stable and unmoving, which is equivalent to approximating the nucleus as having infinite mass, and as such the momentum of the nucleus does not equal the momentum of the electron, seemingly breaking the law of conservation of momentum. Bohr's model itself does not match with the classical model for large energy levels n when ∆n = 2,3,...

Question 11:

Answer:

1. Moseley's experiments describe the two possible valence shells, the one closest to the nucleus with two electrons (the k alpha and k beta x ray emissions), and the other with 8 electrons (the l series). The Z-1 dependency rather than just z is because of shielding by the other electron in that valence shell. Thus the atomic number of an atom became more that just an organization scheme in the periodic table, it had to do with the properties of the element itself.
2. Mosely surmised that x-rays came from the transitions of the innermost electrons in an atom, which are shielded from the outermost electrons and interatomic binding forces. So when an electron was bumped completely out of the atom from one of these inner orbitals, electrons in higher orbitals would transition down to fill its place, which caused the K- and L-series spikes in X-ray spectra. From his experimentation Moseley was then able to determine values for Z of different elements, some of which were wrong at the time. In arranging the periodic table according to the values of Z that Moseley obtained, rather than by weight, the periodic table become completely in agreement with chemical properties of elements and was even able to predict the discovery of new elements. (I would appreciate going over this experiment and the implications of the experiment in class because, although I think I understood it, I am not certain I understood everything.)
3. mosely saw the spacing of emissions of photons from different atoms were regular even at different levels which means that there is shieleding form the outer levels at the inner levels and shielding from the inner levels at the outer levels.
4. There are small differences in the energies of electrons with a given n that were not predicted by Bohr's model. Moseley was able to rearrange the periodic table by their Z number in such a way that completely agreed with their chemical properties. He also showed that there were gaps in the table resulting from undiscovered elements.
5. Moseley's experiments arranges the elements by the Z number obtained from the Moseley plot, rather than weight, gave a periodic chart in complete agreement with the chemical properties. He also indicated the possibilities of undiscovered elements due to the gaps in the periodic table.
6. Moseley notes that the x-ray line spectra (the sharp spikes superimposed in the Bremsstrahlung we saw earlier) varied in a regular way from element to element, believing this to be an effect of the innermost electrons of the atoms, as they were shielded by the outermost electrons by the intermediate orbit electrons. Using the Bohr model of the atom as a guide, he reasoned that the the energy, and by extension the frequency, of a characteristic photon should vary as the square of the nuclear charge, ultimately discovering that the square root of the frequency f = An ( Z - b), where An and b were constants for each characteristic line, with b = 1 for the k series and 0.74 for the L series. This had HUDE implications for the then current structure of the periodic table,which was arranged by weight, as we now had a way to the atomic number was the method of structure for the elements, and as such could even predict elements.

Question 12:

Answer:

1. Is current dependent on the kinetic energy of electrons? I thought it was just the number of electrons.
2. I don't think I had any questions about the Franck-Hertz experiment. It seemed fairly clear to me, but I will take time to let it sink in and let you know if any questions arise.
3. fine
4. What determines whether or the not the collision will be elastic?
5. There is nothing unclear regarding Frack-Hertz experiment.
6. This experiment was ingenious. No questions for now.