Physics Week 14(Sem. 2)

Ms. N. May ABC Math Student Copy

Physics Week 14(Sem. 2) Name____________________________

Particles & Waves Chapter Summary

The Wave‐Particle Duality

The ability to exhibit interference effects is an essential characteristic of waves. For example, Young's double slit experiment demonstrates how light from two closely spaced slits produces bright and dark fringes. The fringe pattern is a direct result of interference between light waves coming from each of the two slits.

One of the most important discoveries of the twentieth century is that particles can also behave like waves and produce interference patterns. For example, if Young's experimental setup were to be used and a beam of electrons were incident on a screen (with a double slit in front) the result would not be that of each electron passing through one slit or the other, demonstrating particle behavior. The actual pattern would consist of bright and dark fringes, as witnessed by the light wave experiment of Youngs'. The fringe pattern indicates that the electrons are exhibiting interference effects that are associated with waves.

The concept of how electrons can exhibit both wave and particle properties will be discussed later, it must just be accepted for now. Also, since particles can demonstrate wave properties then the question of can waves demonstrate particle properties. The answer is yes and it will be addressed within the next few sections. Actually before the wave properties of particles were discovered scientists had already discovered the particle properties of waves. Scientists now accept wave‐particle duality as part of nature. The theory states that waves can exhibit particle‐like characteristics, and particles can exhibit wave‐like characteristics.

Blackbody Radiation & Planck's Constant

All bodies, no matter their temperature, will continuously emit electromagnetic waves. The reason

we can see the sun is because it emits electromagnetic radiation in the visible region. The sun appears yellow because of its surface temperature of about 6000K, while the cooler star of Betelgeuse has a red‐orange appearance resulting from its temperature of 2900K. However, for relatively lower temperature objects emit weaker visible light waves. Thus these cooler objects don't appear to be glowing, as the hotter objects like the sun (with its intensity). Human bodies (temp. ~310K) emit electromagnetic waves in the infrared zone, therefore humans can be detected using infrared devices.

At a given temperature, the intensity of the electromagnetic radiation varies from wavelength to wavelength. As seen below in figure 29.2, the intensity per unit wavelength depends on wavelength for a perfect blackbody emitter.

A perfect blackbody at a constant temperature absorbs and reemits all of the electromagnetic radiation that falls on it. In order to account for the shape Maxwell Planck took the first steps towards the current understanding of wave‐particle duality.

In 1900 Planck used an atomic oscillator to represent a blackbody, each oscillator emitted and absorbed

Page 1 of 14

N. May

ABC M

ath S

tuden

t Cop

y


electromagnetic waves. In order to get the theoretical and experimental data to match Planck said that the energy (E) of the atomic oscillators could only have discrete values of E= 0, hf, 2hf, 3hf, and so on. In equation form

Where E in the energy, n is a positive integer, h is a constant called Planck's constant (6.6260755x10‐34J∙s), and f is the frequency of vibration (in Hertz).

Whenever the energy of a system can have only certain definite values, and nothing in between, the energy is said to be quantized. Planck's equation assumed no energy in between the discrete values of hf therefore, making the assumption that the energy was quantized. This assumption was unexpected based on traditional physics, yet has become applicable in a wide range of applications today.

Applying conservation of energy to Plank's experiment it would require that the energy carried off by the radiated electromagnetic waves must equal the energy lost by the atomic oscillators. Suppose for example, that an oscillator with an energy of 3hf emits an electromagnetic wave. Then using Planck's equation above, the next lowest amount would be 2hf. In this case, the energy carried off by the electromagnetic wave would be hf, equaling the amount of energy lost by the oscillator. Thus, Planck's model for blackbody radiation presents the idea that energy occurs as discrete packets of energy equaling hf. Einstein made the specific proposal that light consists of such energy packets.

Photons & Photoelectric Effect

Total energy and linear momentum are important concepts that not only relate to any particle, but they relate to photons too. For a particle the total energy is the potential energy and kinetic energy combined. For a photon the total energy is calculated using the equation E=nfh. As for linear momentum(p), for a particle it is mass times velocity. When applied to a photon the momentum is p=h/λ.

Experimental evidence indicating that light consists of photons comes from a phenomenon called the photoelectric effect. The emission of electrons from a metal surface when light (of high enough frequency) shines on it is evidence of the photoelectric effect. As demonstrated the incident light caused electrons (photoelectrons) to bounce off the surface. These electrons were then attracted to a positively charged plate as part of circuit with an ammeter. The ammeter recorded a current indicating that electrons were coming from the incident light(photoelectrons). Many features of the photoelectric effect could not be explained using classical physics. In 1905 Einstein developed a theory using in part Planck's black body radiation study and he provided an explanation of the photoelectric effect. It is this contribution that primarily earned him a Nobel prize in physics. In his photoelectric effect he proposed that light travels in discrete amounts called photons and each packet contained an energy of

Where h is Planck's constant. Therefore the light energy that a light bulb emits is carried by photons. The brighter the light shining on a given area the greater the number of photons per second that strike the area. See example 1.

According to Einstein, when light strikes a metal surface the photon can give up its energy to an electron in the metal. If the photon has enough energy it can do the work to remove the electron from the metal. The amount of work required would depend on how tightly the electron was held. Consider the least strongly held electron to need a minimum amount of energy with a value called the work function (Wo). If the photon has excess energy above the amount needed to eject the electron then the remaining energy will increase the kinetic energy of the ejected electron. Thus, the least strongly held electron will by ejected with the maximum amount of kinetic energy(KEmax). Applying the conservation of energy principle Einstein proposed the description of the photoelectric effect to be

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

Page 2 of 14


According to the equation the maximum amount of energy applied to an ejected electron would be

If KE max were to be plotted on the y axis and f on the x axis, the result would be a straight line that crosses the x axis at f=fo. At the frequency (fo), the electron departs from the metal with no kinetic energy (KEmax=0). According to the equation, when the KEmax equals zero hfo equals the work function (Wo).[hfo = Wo]

The photon provides a number of features explaining the photoelectric effect that would not be possible without the photon. It is also important to note that if the frequency of the light is below the value of fo then no electrons are ejected, regardless of how intense the light is. See Example 2.

Another important feature of the photoelectric effect is that the maximum kinetic energy of the ejected electron remains unchanged even if the light intensity is increased, provided the frequency remains the same. With an increasing light intensity the number of photons striking the surface per second increases thus the number of electrons that are ejected per second increases. Thus the ejected electrons all have the same maximum kinetic energy for the same frequency.

When analyzing light and applying electromagnetic wave theory, it does not reason that when increasing the light intensity the kinetic energy of the ejected electrons does not increase. If the light intensity increased one would expect the ejected electrons kinetic energy to increase when the amplitude of oscillation increased enough. Moreover, a relatively long time exposure of low intensity light would be needed to eventually build up a sufficient amount of energy to eject an electron. This is found to be untrue, even the weakest of light intensities causes electrons to be ejected almost instantaneously, provided that the energy is above the minimum value of fo. Thus concluding that the wave theory of light does not explain all of its facets and the photon theory explains some key attributes. The photon theory also helps explain how light interacts with matter.

Photons carry energy with them and can then interact with an electron to eject it from its surface. However, photons are different than traditional particles because traditional particles (normal part.) have mass and can only travel at speeds up to but not equal to the speed of light. A photon, on the other hand, does travel at the speed of light in a vacuum and does not exist as an object at rest. The energy of a photon is purely kinetic energy and it has no rest energy and no mass. If the equation representing the total energy of an object were used (explained in next chapter)

1

Since the photon travels at the speed of light c, the left side of the equation becomes E times 0 (or zero). Thus, the right side of the equation must also be zero leading to the mass (m) must be equal to zero since the photon has a finite speed (c). Therefore the photon has no mass.

Momentum of a Photon & Compton

Although Einstein presented the photon idea in 1905 it was not very accepted until 1923, when Compton used it to explain his research. Compton's research involved the scattering of X‐rays by the electrons in graphite, X‐rays are high frequency electromagnetic waves and like light are composed of photons.

The figure above demonstrates what happens to an electron when struck by an incident electron. The result of the collision is similar to a billiard ball collision, the electron recoiled in one direction while the photon scattered in another direction. Compton observed that the scattered photon had a frequency (f'), a smaller frequency than that of the incident photon indicating that the photon lost energy. He also discovered that

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

Page 3 of 14


the difference in the frequencies depended on the angle θ at which the scattered photon leaves the collision. The phenomenon in which the scattered photon leaves with a smaller frequency than the incident photon is called the Compton Effect.

Analyzing the collision between the incident photon and the electron initially at rest should consider both total energy and linear momentum. The electrons initial kinetic energy is zero and not bound to any material. Therefore

Where the energy of the photons are taken to be E=hf. The incident photon has an energy of hf and the scattered photon has an energy of hf', KE is the energy of recoil of the electron. Since λ=c/f, it follows that λ'=c/f', and therefore the wavelength of the scattered x‐rays are larger than the incident x‐rays.

For an electron initially at rest the linear momentum is zero, thus the conservation of momentum equation becomes

Taking the momentum of a photon to be

Thus the difference in the wavelength of the scattered photon and the wavelength of the incident photon would be

1

In the above equation the h/mc term is referred to as the Compton wavelength of an electron and has the value of h/mc=2.43x10‐12m. The wavelength difference can only vary between zero and 2h/mc, because of the cos θ term.

Between the photoelectric effect and Compton's experiment there is quite convincing evidence for the particle nature of light. However, there are also other convincing theories/experiments that prove that light has wave characteristics. It is commonly accepted now

that light has particle‐wave duality and is far more complex than just a stream of particles or an electromagnetic wave.

De Broglie Wavelength

De Broglie a college student in 1923 made a suggestion that since light waves exhibit particle like behaviors that particles of matter should exhibit wave‐like behaviors. He proposed that all moving matter has a wavelength associated with it, just as waves do. He proposed that the wavelength of a particle has the same notation as that of a photon, thus λ=h/p (from the momentum eq.). Today λ is known as the de Broglie wavelength of the particle.

De Broglie's suggestion was confirmed in 1927 by a couple of two American physicists(Davisson & Germer) and independently Thomson an English scientist. Davisson & Germer directed a beam of electrons onto a crystal of nickel and observed that the diffraction pattern was identical to that seen when x‐rays were diffracted from a crystal. The wavelength of the electrons revealed by the diffraction pattern matched that predicted by deBroglie's hypothesis, λ=h/p. More recently, Young's double slit experiment, performed with electrons, reveals the effects of wave interference.

Particles other than electrons can also exhibit wave‐like properties. For instance, neutrons are sometimes used in diffraction studies to crystal structure. Although all moving particles have a deBroglie wavelength, the effects of this wavelength are observable only for particles with masses similar to that of an electron or neutron (very small). See Example 4.

Particle waves are waves of probability whose magnitude provides the probability that the particle will be found at a point. This arises from analyzing Young's double slit experiment performed with electrons. When only 100 electrons were used, no distinct pattern appeared. As the number of electrons were increased to 3000, a pattern was still hard to find but it was becoming more evident. When the number of electrons were increased to approximately 70,000, the distinct bright and dark fringes appeared. This indicated that the area of highest probability would be in the

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

Page 4 of 14


bright fringe areas, and the dark fringes corresponded to low probability areas. Since the light intensity is proportional to the square of the electric field strength (or magnetic field strength). In an analogous way, the probability of particle waves are found to be proportional to the square of the magnitude of the wave (ψ) . ψ is referred to as the wave function of a particle.

In 1925 Schrodinger & Heisenberg independently developed theoretical frameworks for determining the wave function. In studying this they established a whole new branch of physics called quantum mechanics. This name was derived from the quanta of energy that particles are considered to have.

Heisenberg Uncertainty Principle

Considering that each electron can only strike the screen at one point and that the areas of high probability only increase the likely hood of an electron hitting the screen, one can not exactly predict the exact location on the screen of a particular electron. All that can be discussed is the probability of the electron ending up in different places on the screen. It is not possible to use Newton's laws to say that the electron will be found in a direct straight line from the firing point. As the electron travels through small slits, it isn't as simply since the electron is such a small particle that demonstrates wave properties. Therefore, the 100% certainty of the exact location of a single particle is lost. Only the average behavior of a large number of particles is predictable and the behavior of any individual particle is uncertain.

While the diffracted electrons start with only an x component momentum, some are found a distance Δy

from the a single slit of width W. The momentum in the y direction (py) according to deBroglie's equation would be

This equation indicates that the smaller the slit width, the larger the uncertainty in the y component of the electron's momentum. The figure below indicates the uncertainty of the momentum and energy of a particle, it has wide applicability beyond the single slit experiment.

This principle states the limits that are imposed by nature. It basically states that it is impossible to specify both the precise momentum and position of a particle at the same time.

The energy and time uncertainty principle, as given in the figure above, states that the shorter the lifetime of a particle in a given state, the greater the uncertainty in the energy of that state.

Example 5 below demonstrates the uncertainty involved with an electron compared to that of a ping pong ball. Because of the relatively large mass of the Ping‐Pong ball, the uncertainty in position and speed are so small that they have virtually no effect on our ability to determine simultaneously where such an object would be and how fast it is moving. Also, note that the uncertainty of both depends on Planck's constant which is a very small number. Could you imagine what life would be like if Planck's constant were a larger number?

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

Page 5 of 14


N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

Page 6 of 14


N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

Page 7 of 14


N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

Page 8 of 14

1. The photon model of light is more appropriate than the wave model in explaining(1) interference(2) refraction(3) polarization(4) photoelectric emission

2. Which phenomenon is best explained by the particle nature of light?(1) interference(2) the Doppler effect(3) polarization(4) the photoelectric effect

3. Interference and diffraction can be explained by(1) the wave theory, only(2) the particle theory, only(3) neither the wave nor particle theory(4) both wave and particle theory

4. Wave-particle duality is most apparent in analyzing the motion of(1) a baseball (3) a galaxy(2) a space shuttle (4) an electron

5. What is the energy of a photon with a frequency of 3.00 × 1013 cycles per second?(1) 2.21 × 10–48 J (3) 6.63 × 10–34 J(2) 2.21 × 10–46 J (4) 1.99 × 10–20 J

6. The energy of a photon varies(1) directly as the wavelength(2) directly as the frequency(3) inversely as the frequency(4) inversely as the square of the frequency

7. The energy of a photon which has a frequency of 3.3 × 1014 cycles per second is approximately(1) 2.0 × 10–48 J (3) 5.0 × 10–19 J(2) 2.0 × 10–l9 J (4) 5.0 × 1048 J

8. Which graph best represents the relationship between the energy of a photon and its wavelength?

(1) (3)

(2) (4)

9. Blue light has a frequency of approximately 6.0 × 1014 hertz. A photon of blue light will have an energy of approximately(1) 1.1 × 10–48 J (3) 5.0 × 10–7 J(2) 6.0 × 10–34 J (4) 4.0 × 10–19 J

10. According to the quantum theory of light, the energy of light is carried in discrete units called(1) alpha particles (3) photons(2) protons (4) photoelectrons

11. What is the energy of a quantum of light having a frequency of 6.0 × 1014 hertz?(1) 1.6 × 10–19 J (3) 3.0 × 108 J(2) 4.0 × 10–19 J (4) 5.0 × 10–7 J

12. Compared to a photon of red light, a photon of blue light has a(1) greater energy (3) smaller momentum(2) longer wavelength (4) lower frequency

13. All photons in a vacuum have the same(1) speed (3) energy(2) wavelength (4) frequencyN. M

ay A

BC Math

Stud

eny C

opy

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

Page 9 of 14

14. Base your answer to the following question on the statement below.

The spectrum of visible light emitted during transitions in excited hydrogen atoms is composed of blue, green, red, and violet lines.

What characteristic of light determines the amount of energy carried by a photon of that light?(1) amplitude (3) phase(2) frequency (4) velocity

15. Which color of light has the greatest energy per photon?(1) red (3) blue(2) green (4) violet

16. In a photon-electron collision, there is conservation of(1) mass(2) energy, only(3) momentum, only(4) energy and momentum

17. Base your answer to the following question on the diagrams below, which show a photon and an electron before and after their collision.

Compared to the total momentum of the photon-electron system before the collision, the total momentum of the photon-electron system after the collision is(1) less (3) the same(2) greater

18. As the wavelength of a ray of light increases, the momentum of the photons of the light ray will(1) decrease (3) remain the same(2) increase

19. The momentum of a photon with a wavelength of 5.9 × 10–7 meter is(1) 8.9 × 1026 kg-m/s (3) 1.1 × 10–27 kg-m/s(2) 1.6 × 10–19 kg-m/s (4) 3.9 × 10–40 kg-m/s

20. What is the approximate matter wavelength of a 0.30-kilogram tennis ball moving at a speed of 30. meters per second?(1) 9.0 × 10–34 m (3) 1.4 × 10–34 m(2) 6.6 × 10–34 m (4) 7.3 × 10–35 m

21. Compared to the wavelength of a moving electron, the wavelength of a proton moving at the same speed is(1) shorter (3) the same(2) longer

22. Which of the following would best illustrate the wave properties of matter?(1) photoelectric effect(2) diffraction of electrons(3) alpha particle scattering(4) photon-particle collisions

23. If the wave properties of a particle are difficult to observe, it is probably due to the particle's(1) small size (3) low momentum(2) large mass (4) high charge

24. What is the wavelength of the matter wave associated with a bird of 1.0-kilogram mass flying at 2.0 meters per second?(1) 3.3 × 1034 m (3) 3.3 × 10–34m(2) 1.3 × 10–33 m (4) 8.6 × 10–34 m

25. If the momentum of a particle is 1.8 × 10–22 kilogram-meter per second, its matter wavelength is approximately(1) 1.2 × 10–55 m (3) 3.7 × 10–12 m(2) 2.7 × 1011 m (4) 5.0 × 10–7 m

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

Page 10 of 14

26. The threshold frequency for a photoemissive surface is 6.4 × 1014 hertz. Which color light, if incident upon the surface, may produce photoelectrons?(1) blue (3) yellow(2) green (4) red

Base your answers to questions 27 through 29 on the information below.

Light of constant intensity strikes a metal surface. The frequency of the light is increased from 6.0 × 1014 cycles per second to 9.0 × 1014 cycles per second. Photoelectrons are emitted by the metal surface when the frequency reaches 8.0 × 1014 cycles per second.

27. As the frequency of the incident light increases, the photons striking the metal surface increase in(1) number (3) speed(2) energy (4) wavelength

28. The work function of the metal surface is approximately(1) 6.0 × 10–19 J (3) 5.3 × 10–19 J(2) 2.0 × 10–19 J (4) 4.0 × 10–19 J

29. If the intensity of the incident light were increased while the frequency was kept constant, the maximum kinetic energy of the emitted photoelectrons would(1) decrease (3) remain the same(2) increase

30. When a source of dim orange light shines on a photosensitive metal, no photoelectrons are ejected from its surface. What could be done to increase the likelihood of producing photoelectrons?(1) Replace the orange light source with a red

light source.(2) Replace the orange light source with a

higher frequency light source.(3) Increase the brightness of the orange light

source.(4) Increase the angle at which the photons of

orange light strike the metal.

31. The threshold frequency in a photoelectric experiment is most closely related to the(1) brightness of the incident light(2) thickness of the photoemissive metal(3) area of the photoemissive metal(4) work function of the photoemissive metal

32. Which determines the number of electrons emitted by a photoelectric material?(1) intensity (3) frequency(2) color (4) wavelength

33. The threshold frequency for a photoemissive surface is 1.0 × 1014 hertz. What is the work function of the surface?(1) 1.0 × 10–14 J (3) 6.6 × 10–48 J(2) 6.6 × 10–20 J (4) 2.2 × 10–28 J

34. Photons with a frequency of 1.0 × 1020 hertz strike a metal surface. If electrons with a maximum kinetic energy of 3.0 × 10–14 joule are emitted, the work function of the metal is(1) 1.0 × 10–14 J (3) 3.6 × 10–14 J(2) 2.2 × 10–14 J (4) 6.6 × 10–14 J

35. A certain photoemissive material with a work function of 1.3 × 10–19 joule is exposed to incident photons with an energy of 3.3 × 10–19 joule. The maximum kinetic energy that an ejected photoelectron can attain is closest to(1) 1.0 × 10–39 J (3) 3.0 × 10–19 J(2) 2.0 × 10–19 J (4) 4.0 × 10–19 J

36. When 8.0-electronvolt photons strike a photoemissive surface, the maximum kinetic energy of ejected photoelectrons is 6.0 electronvolts. The work function of the photoemissive surface is(1) 0.0 eV (3) 7.0 eV(2) 2.0 eV (4) 14.0 eV

37. The maximum kinetic energy of an electron ejected from a metal by a photon depends on(1) the photon's frequency, only(2) the metal's work function, only(3) both the photon's frequency and the metal's

work function(4) neither the photon's frequency nor the

metal's work function

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

Page 11 of 14

38. As the intensity of monochromatic light on a photoemissive surface increases, the maximum kinetic energy or the photoelectrons emitted(1) decreases (3) remains the same(2) increases

39. Electromagnetic radiation of constant frequency incident on a photoemissive material causes the emission of photoelectrons. If the intensity of this radiation is increased, the rate of emission of photoelectrons will(1) decrease (3) remain the same(2) increase

40. The threshold frequency of a photoemissive surface is 7.1 × 1014 hertz. Which electromagnetic radiation, incident upon the surface, will produce the greatest amount of current?(1) low-intensity infrared radiation(2) high-intensity infrared radiation(3) low-intensity ultraviolet radiation(4) high-intensity ultraviolet radiation

41. Base your answer to the following question on the graph below which represents the maximum kinetic energy of photoelectrons for varying frequencies for three different metals.

The slope of each graph represents(1) the work function(2) Planck's constant(3) the threshold frequency(4) the kinetic energy

Base your answers to questions 42 and 43 on the graph below which represents the maximum kinetic energy of photoelectrons as a function of incident electromagnetic frequencies for two different photoemissive metals, A and B.

42. The threshold frequency for metal A is(1) 1.0 × 1014 Hz (3) 3.0 × 1014 Hz(2) 2.0 × 1014 Hz (4) 0.0 Hz

43. The work function for metal B is closest to(1) 0.0 J (3) 2.0 × 10–3 J(2) 2.0 × 10–19 J (4) 1.5 × 10–4 J

44. The graph below shows the relationship between the frequency of radiation incident on a photosensitive surface and the maximum kinetic energy (KEmax) of the emitted photoelectrons.

The point labeled A on the graph represents the(1) incident photon intensity(2) photoelectron frequency(3) threshold frequency(4) work function energy

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

Page 12 of 14

45. Base your answer to the following question on the graph below which represents the relationship between the maximum kinetic energy of emitted photoelectrons and the frequencies of the photons incident upon a photoemissive surface.

A photon whose frequency is equal to the threshold frequency strikes the photoemissive surface. What is the maximum kinetic energy of the emitted photoelectron?(1) 5.0 eV (3) –2.0 eV(2) 2.0 eV (4) 0 eV

46. The graph below shows the maximum kinetic energy of photelectrons ejected from a metal as a function of frequency of incident electromagnetic radiation

What is the work function of the metal?(1) 6.6 × 10-34 J (3) 2.0 × 1015 J(2) 1.3 × 10-18 J (4) 3.0 × 1048 J

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

Page 13 of 14

Base your answers to questions 47 and 48 on the information below.

When an electron in an excited hydrogen atom falls from a higher to a lower energy level, a photon having a wavelength of 6.58 × 10–7 meter is emitted.

47. Calculate the energy of a photon of this light wave in joules. [Show all calculations, including the equation and substitution with units.]

48. Is this photon an x-ray photon? Justify your answer.

49. Base your answer to the following question on the information and diagram below. The diagram shows the collision of an incident photon having a frequency of 2.00 × 1019 hertz with an electron initially at rest.

Calculate the initial energy of the photon. [Show all calculations, including the equation and substitution with units.]

50. Base your answer to the following question on the information below.

Louis de Broglie extended the idea of waveparticle duality to all of nature with his matterwave equation, Ø λ = hmv

hmv, where Ø is the particle’s

wavelength, m is its mass, v is its velocity, and h is Planck’s constant.

Using this equation, calculate the de Broglie wavelength of a hydrogen nucleus (mass = 6.7 × 10–27 kg) moving with a speed of 2.0 × 106 meters per second.[Show all work, including the equation and substitution with units.]

51. A photon has a wavelength of 9.00 ] 1010 meter. Calculate the energy of this photon in joules.

52. Determine the frequency of a photon whose energy is 3.00 × 10–19 joule.

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

N. May

ABC M

ath S

tuden

y Cop

y

Page 14 of 14

Documents

Physics Week 14(Sem. 2)