Reading Quiz

Question 1:

Answer:

1. Blackbody radiation curves could not be predicted by models for short wavelengths, and there is no lag time before electrons are emitted excited by incident light as one would expect from the classical wave theory based on the average energy of the light
2. In conflict with the classical theory of light, there was no intensity below which a current could not be detected. Furthermore, increasing the rate of energy falling on the cathode does not increase the maximum kinetic energy of the electrons.
3. It was contradictory to the classical theory of light that, in observations of the photoelectric effect, there was no minimum intensity of light below which the current was absent. In addition, stopping voltage, or the voltage difference between the cathode and anode at which no electrons reached the anode, was independent of the intensity of the incident light.
4. The fact that there is no time lag between turning on a light source and the appearance of photoelectrons contradicts classical theory. This occurs, even for low intensity light, because even though there are fewer photons than high intensity light, the photons that are present have enough energy to eject electrons from the metal. Similarly, counter to classical expectation, there is no minimum intensity that corresponds with a zero current below that intensity level. The reason for this is the same as for above; even at low intensities when there are a limited number of photons, those photons present do carry enough energy to eject electrons.
5. 1) There is zero time lag in between turning on a light incident to a cathode and the detection of electrons. Classically, there should be a way to calibrate the intensity of the light so that a time lag of your choosing can be observed, but this isn't the case. 2) Light intensities that are too low to provide electrons with escape energy should result in zero emission of electrons. However, there is no such minimum intensity observed.
6. Classical theory says an increase in intensit would increase electron velocity because it comes from a transfer of energy. This is not the case. Another example is a dim light would be expected to show a time lag because the energy transfer would take an amount of time to free the electrons. this is not the case.
7. 1) Classical view on light tends to think of it as wave propagation, and thus reducing the intensity of light wave will result in reduced amount of emitted electrons. However, this is not the case as in the photoelectric effect. 2) Classical physics concludes that there must be time lag between the turning on of the light source and the appearance of photoelectrons since it assumes that the intensity of incident light is power per unit area and the incident energy is distributed uniformly over the illuminated surface. But in the experiment no time lag has been discovered.
8. 1) In Lenard's experiment, he observed that the maximum current produced was proportional to the light intensity, which was classically expected, since doubling the energy per unit time incident on the cathode should double the number of electrons emitted. As such, he predicted that there would be a minimum intensity for which no current would flow. This, however, was not the case, as the current was never absent. 2) A second seemingly contradictory observation is the lack of "time lag" between between turning on a light source and the appearance/detection of photoelectrons. Classically, the time required for an atom to acquire enough energy to emit an electron can be calculated from the intensity of the incident radiation. As such you should theoretically adjust the intensity of the light so that the time lag can range from several minutes to hours, but no matter the adjusted intensity, no time lag is seen.

Question 2:

Answer:

1. Einstein said that the energy of light must be discrete quanta, localized energy, rather than spread through the entire area it shone upon.
2. Einstein suggested that the energy of light is universally quantized in photons. Thus when a photon strikes a surface the entirety of its energy can be absorbed by a single electron, and this may be enough to remove the electron from the surface.
3. Einstein resolved these contradictions by asserting that the energy quantization that Planck found in solving the blackbody radiation problem was actually a universal characteristic of light.
4. Einstein accepted Planck's energy quantization as a universal characteristic of light and used this to develop the photoelectric effect equation.
5. He proposed a quantization of light, where light energy is composed of discrete quanta each of energy hf. He built this theory off of Planck's energy quantization theory used to solve the blackbody radiation problem.
6. einstein said the the wave could be modeled as a wave packet with a certain amountof energy proportional to frequency and intensity is the number of photons.
7. Einstein resolve the apparent contradiction by assuming that energy quantization used by Planck in solving the black-body radiation was a universal characteristic of light. Light energy consisted of discrete quanta with energy hf for each, and these quanta are called photons, rather than being distributed evenly in the space through which it propagated.
8. Einstein assumed that Plank's quantized energy, or quanta, that was used in solving the black body radiation problem was a universal characteristic of light. When one of these quanta, which he called photons, penetrated the surface of the cathode, all of it's energy had the potential to be absorbed by an electron. Thus, the maximum kinetic energy of an electron is given by the energy of the photon (hf) - the energy necessary to remove the electron from the surface it's bonded to (phi). This is a consequence of energy conservation, so (hf- phi) = (1/2mv^2) max. This predicts that the slope of V0 vs. f = h/e, which Millikan eventually showed as correct.

Question 3:

Answer:

1. The kinetic energy of the released electron is based on its charge and the stopping potential (where the electric potential energy is no longer high enough for the electrons to move), rather than the intensity of the incident light.
2. The kinetic energy of a released electron is determined by the difference in energy of a photon, hf, and the work function, phi, which is a characteristic of the metal.
3. The kinetic energy of a released electron is dependent on the frequency of the incident light, the work function of the metal that composes the cathode, and the amount of energy lost by the electron in traversing the metal before it is released.
4. The (maximum) kinetic energy of a released photon is given by the energy of the incident photon minus the work function of the material.
5. The maximum kinetic energy will be hf - work function. Where the work function is a characteristic of the metal.
6. there is a flat minimum energy that is required for electrons to be released. This is the threshold energy. This is and te frequency or energy of incoming waves is what determines the kinetic energy of the eelectron
7. The maximum kinetic energy of the released electron equals hf-(phi), with f the frequency of incident light, h the Planck constant, and phi the work function.
8. As I stated before, the kinetic energy of the released electron is given by the energy of the photon that bombarded the electron minus the energy needed to release the electron from it's surface, giving a relation (hf - phi). Einstein proved that e*Vo = (1/2mv^2)max = (hf - phi). Meaning that the charge of the electron, the Vo used in the experiment can also be used as factors.

Question 4:

Answer:

1. The energy of a photon is given by multiplying Planck's constant by the frequency of the photon (or hc/lambda)
2. The energy of a photon is be given by hf, where h is Planck's constant and f is the frequency.
3. In the photon picture of light, only the frequency of the light determines the energy of the photon.
4. The energy of the photon is given by hf.
5. The stopping potential.
6. the frequency
7. The plank constant h and the incident light's frequency f, and the energy equals the product of these two quantities.
8. In the "photon" picture, you have the Vo used in the experiment and the charge of the electron, since the frequency of the photons f is equal to c/wavelength, which describes wave like properties.

Question 5:

Answer:

1. The work function phi tells the lowest frequency or biggest wavelength necessary for ejecting an electron from the metal.
2. The work function tells us the amount of energy necessary to remove an electron from the surface.
3. The work function tells you the amount of energy necessary to remove an electron from the surface of the cathode (how hard it is to release the electron).
4. The work function tells you the minimum amount of energy necessary to remove an electron from the metal. It is characteristic of the material and differs among metals.
5. The minimum energy required to remove an electron from a surface.
6. it tells you the minmium amount of energy or work requried to pull an electron off of a solid
7. The work function (phi)is a characteristic of the metal, and if it is the energy necessary to remove an electron from the surface, we can use energy conservation to solve the maximum kinetic energy of an electron leaving the surface that is equal to hf-(phi).
8. The work function phi tells you now much energy is needed to release an electron from the bonds of the surface. Its value is a characteristic of the metal that makes up said surface

Question 6:

Answer:

1. The intensity of the incident light gives the average number of photons per time, but and the maximum photoelectric current is directly related to the number of photons that excite and release electrons.
2. The intensity of the light determines how many photons are striking the surface. The more photons there are, the more electrons will be emitted to generate current.
3. The result that the maximum photoelectric current is proportional to the intensity of incident light is explained by the fact that higher intensity light corresponds to more photons being incident on the surface. Thus more electrons are also released from the surface. Therefore, the photoelectric current rises because more electrons are passing through rather than that each individual electron has more energy.
4. Maximum photoelectric current is proportional to the light intensity because increasing the number of photons incident on the metal increases the number of electrons emitted.
5. The minimum frequency for photoelectric effect and the corresponding threshold wavelength are related to the work function by taking hc/(wavelength)
6. the intensity is the number of photons released. More photons above the work threshold means more electrons knocked off. current is charge per time through a point. more electrons means more electrons and more electric charge flowing. so more current
7. In the photon model of light, electrons in the metal absorbs energy from incident photons in order to be emitted. The energy of the photon depends on the Planck constant and the incident light's frequency. And the intensity of the incident light beam depends on how many photons in unit time; however, each electron only absorb one photon for being emitted, so only the energy of the photon matters. The photoelectric current depends on the intensity: the greater the intensity is, the more photons there are. This means there will be more electrons emitted which will increase the current because current is the flow of charge.
8. By setting Vo = 0, you can show that phi = hf, where the frequency f can be redefined as c/ lambda, where lambda is the wavelength of the photon that transferred its energy to the released electron. Wavelength is, not surprisingly, only an applicable value when talking about waves. Photons with having wavelengths greater than this lambda max would not have enough energy to eject the electron from the metal.