Chapter 14 Quantum Physics

stopping voltage for the photoelectric effect

-eVstop = KEmax (Vstop is a necessarily negative voltage, so a negative sign is required on the left side to match the necessarily positive kinetic energy)

What was expected of the photoelectric effect?

1. A brighter light would yield a stronger current 2. There would be a delay between the time the light shines and when current is detected 3. If the polarity of the battery was reversed, increasing the brightness would increase the potential difference of the ejected electrons

What occurred during the photoelectric effect?

1. Elecromagnetic radiation of a certain frequency is made up of discrete bundles 2. Photons of light are absorbed or emitted as single instantaneous interactions 3. When a proton light strikes the surface of a metal, it interacts with a single electron, imparting all of its energy in the formed of increased kinetic energy for the electron.

The photon model of light is necessary to explain each of the following experimentally observed effects except: A. The ejection of electrons from a metal surface with light shining upon it. B. The stopping voltage as a function of light frequency C. The instantaneous detection of a current following the application of light to the metal surface. D. The fact that an intense light may result in no measured current where a dim UV light results in a measured current.

A; Choice A is correct because the ejection of electrons by the application of light is perfectly predictable according to the classical wave theory.

All of the following statements about the photoelectric effect are true EXCEPT: A. the intensity of the light beam does not affect the photocurrent. B. the kinetic energies of the emitted electrons do not depend on the light intensity. C. a weak beam of light of frequency greater than the threshold frequency yields more current than an intense beam of light of frequency lower than the threshold frequency. D. for light of a given frequency, the kinetic energy of emitted electrons increases as the value of the work function decreases.

A; The greater the intensity, the greater the number of incident photons and, therefore, the greater the number of photoelectrons that will be ejected from the metal surface (provided that the frequency of the light remains above the threshold). This means a larger current. Remember that the frequency of the light (assuming it is above the threshold frequency) will determine the kinetic energy of the ejected electrons; the intensity of the light determines the number of electrons ejected per time (the current).

A single photoelectron ejected from a metal target will have a greater chance of reaching the detector plate if: A. it is ejected by a dim ultraviolet light and the detector voltage is higher than the target voltage. B. it is ejected by a dim ultraviolet light and the detector voltage is lower than the target voltage. C. it is ejected by a bright infrared light and the detector voltage is higher than the target voltage. D. it is ejected by a bright infrared light and the detector voltage is lower than the target voltage.

A; This is a classic two-by-two question, where the two degrees of freedom are the nature of the light ejecting the photoelectron and the voltage between the plates. A higher initial kinetic energy will make reaching the detector plate more likely. Note that the initial kinetic energy of an ejected photoelectron depends upon the energy of the ejecting photon, hf, which is a function of frequency and not brightness (which would determine the number of photons). Since ultraviolet light has a higher frequency than infrared light, the ultraviolet light will correspond to higher KE photoelectrons that are more likely to strike the detector (eliminating choices C and D). The other factor that will increase the likelihood of striking the detector plate is if photoelectrons are attracted to it rather than repelled. By ΔPE = qV, a decreasing potential energy (indicating attraction) corresponds to a positive voltage for a negative charge like a photoelectron. Thus a higher detector voltage will attract electrons (eliminating choice B).

What measures current?

Ammeter

• Question: Suppose you use the photoelectric experiment apparatus to determine the work function of an unknown metal. Which independent variables would you want to change and which output would you want to measure? A. Vary light intensity and battery voltage, measure current B. Vary light frequency and battery voltage, measure current C. Vary light intensity and current, measure voltage D. Vary light frequency and current, measure voltage

B

The Heisenberg uncertainty principle explains: A. the statistical uncertainty associated with measurements using multiple photons. B. the physical uncertainty associated with the coupling of position and momentum. C. the uncertainty of whether a charge in a magnetic field is positive or negative. D. the uncertainty of whether to apply the rules of classical or quantum physics.

B;

When you breathe in, oxygen molecules from the atmosphere enter small sacs in your lungs called alveoli. These sacs have an average diameter of 250 micrometers when inflated. If an oxygen molecules has a mass of about 5.3 x 10^-26 kg, what is the minimum uncertainty in the speed of an O2 molecule confined in an alveolus A. 8 x 10^-9 m/s B. 8 x 10^-6 m/s C. 8 x 10^-3 m/s D. 8 m/s

B; ΔxΔp —> Δp = h/2piΔx —> Δv = h/2pi Δx x m = 6.63 x 10^-34 / 2pi (250 x 10^-6)(5.3 x 10^-26) = 1/1300 x 10^-2 = 8 x 10^-6 m/s

A red photon and a blue photon both strike a piece of unknown material at the same acute angle of incidence. Any of the following could happen EXCEPT: A. The blue photon passes through he material at a faster speed than does the red photon B. The blue photon passes through the material at a faster speed than does the red photon. C. The blue photon is reflected while the red photon is transmitted D. Both photons eject electrons from the material

B; According to the photoelectric effect, an electron can be ejected from a material by a photon whenever the photon energy hf is greater than the work function of the material. Because blue light is higher frequency than red light, the blue photon is more likely to eject an electron. This is consistent with choices A and D, eliminating them. Choice B is correct because a higher index of refraction for blue light means that the blue proton will travel slower than the red one.

An electron is confined in an electrical potential well with a diameter of the Bohr radius, a0. What is the minimum uncertainty in its velocity (take m to be the mass of the electron)? A. h / 2πa0 B. h / 2πa0m C. hm / 2πa0 D. ha0 / 2πm

B; Heisenberg's uncertainty principle states that the product of the uncertainties of position and momentum is given by ΔxΔp ≥ h / 2π, where h is Planck's constant. The size of the potential well gives the uncertainty in the position of the electron, Δx = a0. Thus Δp ≥ h / 2πa0. Since p = mv, Δp = mΔv. This leads to Δv ≥ h / 2πa0m.

A light source emits photons that hit a photodetector. All of the following would increase the TOTAL energy hitting the photodetector EXCEPT: A. increasing the area of the photodetector. B. increasing the wavelength of light being used. C. increasing the number of photons emitted per second. D. increasing the time that the photodetector collects photons.

B; If more photons (with the same energy) hit the photodetector, then more total energy will be detected. Increasing the area, the time and rate at which photons are created would all do this. Choices A, C and D can be eliminated. The energy of each photon is given by E = hf, where h is Planck's constant and f is the frequency of light. The relationship between wavelength, λ, and frequency is given by c = fλ, where c is the speed of light. Increasing the wavelength would decrease the frequency which would then decrease the energy of the photons. This would therefore decrease the total energy hitting the photodetector.

What is the wavelength of a photon that causes an electron to be emitted from a metal with a kinetic energy of 50 J? (Note: The work function of the metal is 16 J, and h = 6.626 × 10-34 J·s) A. 1.0 × 10-34 m B. 3.0 × 10-27 m C. 3.0 × 10-26 m D. 1.0 × 1035 m

B; To determine the wavelength of the light ray, first calculate its frequency from the photoelectric effect equation: Next, determine the wavelength of the incident ray of light by relating the frequency to the speed of light:

When a hydrogen atom electron falls to the ground state from the n = 2 state, 10.2 eV of energy is emitted. What is the wavelength of this radiation? (Note: 1 eV = 1.60 × 10-19 J, and h = 6.626 × 10-34 J·s.) A. 5.76 × 10-9 m B. 1.22 × 10-7 m C. 3.45 × 10-7 m D. 2.5 × 1015 m

B; To solve this question correctly, one must be careful with the units. First, convert 10.2 eV to joules: Next, to determine the wavelength of the radiation, first find the frequency: Lastly, from the wave equation c = f λ, we can calculate the wavelength of the radiation:

A beam of electrons is incident on a metal plate with a narrow slit (Δy = .1 nm). Those that pass through are subsequently detected by a scintillation detector screen. If the width of the slit is changed, what would you expect to observe on the screen? A. There will be a single split on the screen the same diameter as the slit in all cases B. The width of the spot on the screen will decrease as the slit gets wider because it's easier for the electrons to pass through a wider slit C. The width of the spot on the screen will increase as the slit gets narrower because a decrease uncertainty in the vertical position of the electron at the slit entails an increased uncertainty in its vertical momentum leaving the lit and moving to the screen. D. The width of the spot on the screen will decrease as the slit gets narrower because an increase uncertainty in the vertical momentum of the electron at the slit entails a decreased uncertainty in its vertical position.

C

Which of the following factors determine the initial kinetic energy of a photoelectron ejected from a metal plate? I. The frequency of the ejecting photon II. The potential difference between the metal plate and the detector plate. III. The binding energy of the metal A. I only Your Answer B. I and II only C. I and III only D. I, II, and III

C; The kinetic energy of a photoelectron is given by the difference between the energy imparted by the ejecting photon and the work function of the metal from which the electron is ejected: KE = hf - φ. The photon energy depends upon its frequency, so I is true (which was known already), and the work function of the metal is a measure of its binding energy, so III is true (eliminating choices A and B). Though the potential difference between the target plate from which photoelectrons are emitted and the detector plate will affect the kinetic energy of the photoelectron after it leaves the plate, it does not influence its initial kinetic energy, so II is false (eliminating choice D).

Which of the following changes to the photoelectric experiment could change the value of the stopping voltage? I. Increasing the duration of the experiment II. Changing the wavelength of the incident light III. Changing the material used for the metal A. I only B. II only C. I and III only D. II and III only

C; The photoelectric effect is essentially instantaneous, so increasing the duration of the experiment should not effect the determination of the stopping voltage, which depends on the maximum kinetic energy of the ejected photoelectrons. Statement I is thus false Statement II is true by default Changing the metal would have an affect on the stoping voltage, meaning III is true and the correct answer is D

Suppose a photoelectric experiment is conducted and the applied voltage is varied over a range of values to generate a a graph of current versus voltage similar to that shown above. If the intensity of the light is increased but all other aspects of the experiment are kept the same, which of the following aspects of the generated graph would change? I. The slope of the line from V stop to V = 0 II. The location of V stop III. The maximum value of the current A. I only B. II only C. I and III only D. I, II, and III

C; The stopping voltage depends on the maximum KE of the photoelectrons, which in turn depends upon the frequency of the incident light and the work function of the metal target. Therefore, increasing the intensity of the incident light will not alter the stopping voltage, so item II is false. Increasing intensity, will, however, increase the number of photons striking the target, and therefore will increase the number of photoelectrons and therefore the maximum current.

If the work function of a metal is 6.622 × 10-20 J and a ray of electromagnetic radiation with a frequency of 1.0 × 1014 Hz is incident on the metal, what will be the speed of the electrons ejected from the metal? (Note: h = 6.626 × 10-34 J·s and me- = 9.1 × 10-31 kg) A. 2.62 x 10^-6 m/s B. 1.07 x 10^-4 m/s C. 9.38 x 10^3 m/s D. 3.81 x 10^5 m/s

C; To determine the speed of the electrons ejected, we must first calculate their kinetic energy: Using the formula for the kinetic energy, we can now calculate the speed of the ejected electrons: Notice the wide range in the exponents for the answer choices. While the math in this question may seem complex, this allows us to round significantly.

An electromagnetic beam of wavelength y0 is incident upon an unknown metal. If electrons are ejected with maximum kinetic energy, K, what is the work function of the metal? A. hc/y0 B. hy0/c C. hc/y0 - K D. hc0/c - K

C; hc/y0 = Φ + K Φ = hc/y0 - K

A singly ionized helium atom behaves the same way as a hydrogen atom, the important difference being that there are two protons in the nucleus instead of one. What would be the energy of the photon released by an He+ ion as it transitioned from the n = 2 to the n = 1 state? A. 10.2 eV B. 13.6 eV C. 20.4 eV D. 40.8 eV

D; E(n) = 4(-13.6 ev/n^2) = -54.4 ev/n^2 Delta E = hf = |54.4 eV (1/1 - 1/2^2)| Delta E = hf = |54.4 eV (1 - 1/4)| Delta E = hf = 54.4 eV(3/4) = 40.8 eV There is a shortcut. If you remember that 10.2 eV is the energy difference for the hydrogen atom transitioning between n=2 and n=1, multiplying that factor by four for the reasons explained above yields the answer.

Which of the following statements is inconsistent with the Bohr model of the atom? A. Energy levels of the electron are stable and discrete. B. An electron emits or absorbs radiation only when making a transition from one energy level to another. C. To jump from a lower energy to a higher energy, an electron must absorb a photon of precisely the right frequency such that the photon's energy equals the energy difference between the two orbits. D. To jump from a higher energy to a lower energy, an electron absorbs a photon of a frequency such that the photon's energy is exactly the energy difference between the two orbits.

D; The Bohr model is based on a set of postulates originally put forward to discuss the behavior of electrons in hydrogen. In summary, these postulates state that the energy levels of the electron are stable and discrete, and they correspond to specific orbits, eliminating choice (A). They also state that an electron emits or absorbs radiation only when making a transition from one energy level to another, eliminating choice (B). Specifically, when an electron jumps from a lower-energy orbit to a higher-energy one, it must absorb a photon of light of precisely the right frequency such that the photon's energy equals the energy difference between the two orbits, eliminating choice (C). When falling from a higher-energy orbit to a lower-energy one, an electron emits a photon of light with a frequency that corresponds to the energy difference between the two orbits, This is the opposite of choice (D), which makes it the right answer.

Ultraviolet light is more likely to induce a current in a metal than visible light. This is because photons of ultraviolet light: A. have a longer wavelength. B. have a higher velocity. C. are not visible. D. have a higher energy.

D; The photoelectric effect occurs when a photon of sufficiently high energy strikes an atom with a sufficiently low work function. This means that a photon with higher energy is more likely to produce the effect. Because ultraviolet light has a higher frequency and lower wavelength than visible light, it also carries more energy according to the equation E = hf. All light travels at the speed of light, eliminating choice (B). As mentioned earlier, ultraviolet light has a shorter wavelength than visible light, eliminating choice (A). The visibility of a wave plays no role in its ability to cause the photoelectric effect, eliminating choice (C).

One light source emits bright red light while another source emits dim blue light. According to the photon theory of light: A. each red photon must have more energy than each blue photon. B. the red photons must be traveling slower than the blue photons. C. the red photons must be experiencing less diffraction than the blue photons. D. there must be more red photons being emitted per second than blue photons.

D; Whether light is thought of as a wave or as photons, all frequencies travel at 3 × 10^8 m/s in a vacuum and close to that speed in air. This eliminates choice B. (Note that even if the dispersion of light were considered, red light would travel faster than blue light). The energy of a photon is given by E = hf, where h is Planck's constant and f is the frequency of the light. Since the frequency of blue light is greater than that of red, the energy of blue photons is greater than the energy of red photons. This eliminates choice A. There is nothing in the question relating to diffraction, so we can eliminate choice C. (Note that even if diffraction were taken into account, red light diffracts more than blue light when passing through a small opening). The answer is D. Brightness corresponds to intensity, which is total energy per time per area. For red light to have higher intensity than blue light, there must be more red photons being emitted per second.

Total orbital energy equation

E = KE + PE = 1/2mv^2 - k (e^2/r

The wavelength of a red Ha spectral line characteristic of the visible hydrogen spectrum is 656 nm. How much energy does a hydrogen atom lose when it emits an Ha photon? Use h = 6.63 x 10^-34 Js

E = hc/y E = (6.63 x 10^-34 Js)(3 x 10^8) / (656 x 10^-9m) E = 20 x 10^-26 / 6.56 x 10^-7 E = 3 x 10^-19 J

If blue light of frequency 6.00 X 10^14 Hz is incident on rubidium (W = 2.26 eV), will there be photoejection of electrons? If so, what is the maximum kinetic energy an ejected electron will carry away? (Note: h = 6.626 X 10^-34 J = 4.14 X 10^-15 eV s)

E = hf E = (4.14 x 10^-15 Ev s)(6.00 x 10^14Hz) E = 2.5 eV K = hf - Φ K = 2.5 eV - 2.26 eV K = .22eV

Energy of a photon equation

E = hf = hc/lambda

Energy of a photon emitted or absorbed under the Bohr model equation

E(photon) = | ΔE atom |

Energy level of a hydrogen atom

En = -13.6 eV/n^2 where n = 1,2,3...

What determines the absorption spectrum of a single atom?

Energy differences between ground state electrons and higher level electron orbits determine the frequencies of light a particular material absorbs.

centripetal force equation

Fc=mv^2/r = ke^2/r^2

What causes fluorescence?

Fluorescence is a special stepwise photon emission in which an excited electron returns to the ground state through one or more intermediate excited states. Each energy transition releases a photon of light. With smaller energy transitions than the initial energy absorbed, these materials can relate photons of light in the visible range.

Kinetic energy of a photon equation

KE(max) = hf - Φ h = Planck's constant f = frequency Φ = work function

What happens when the frequency of the incident phone is less than threshold frequency?

No electron is ejected

What did Rutherford discover?

Nucleus and protons

What is fluorescence?

Occurs when a species absorbs a frequency of light and then returns to its ground state

What is quantum physics?

Quantum physics is a branch of science that deals with discrete, indivisible units of energy called quanta as described by the Quantum Theory.

What electrical phenomenon results from the application of the photoelectric effect?

The accumulation of moving electrons creates a current during the photoelectric effect.

What is the photoelectric effect?

The emission of electrons from a metal when light shines on the metal

What is work function?

The minimum energy required to remove an electron from the surface of a metal

How does the work function relate to the energy necessary to emit an electron from a metal?

The work function describes the minimum amount of energy necessary to emit an electron. Any additional energy from a photon will be converted to excess kinetic energy during the photoelectric effect.

What did Rutherford assume about electron orbits?

They assumed arbitrary orbits

What is stopping voltage?

Voltage required for electrons to just fail to reach the electrode and therefore turning back and making the current 0 The work done by the field stopping the electron must be equal to electron's kinetic energy Therefore eV(stop) = 1/2mv^2

Work function equation

W = h f(T) f(T) = threshold frequency W: min energy required to eject e-

During which electronic transition is photon emission most common?

When electrons transition from higher energy state to lower energy state.

What is the emission spectrum?

a spectrum of the electromagnetic radiation emitted by a source.

What does the threshold frequency depend on?

a type of metal

What does threshold frequency depend on?

a type of metal

An electron can jump from a lower energy to a higher energy by _______ a photon of light

absorbing

The photoelectric effect is an __________ response

all or nothing

electrons liberated from the metal by the photoelectric effect will produce

current

What is the diameter of a hydrogen atom in the ground state according to the Bohr model? Use 9 x 10^-31 kg for the mass of an electron

d = 1 x 10^-10 m

What happens if the frequency of the incident photon is greater than threshold frequency?

electron will be ejected

What does the Pauli exclusion principle apply to

electrons, protons, and neutrons

When an electron falls from a higher to a lower energy orbit, it ________ a photon of light

emits

What is the frequency of a photon that, when absorbed, will ionize a hydrogen atom that begins in the n = 2 state? Planck's constant is h = 4.14 x 10^-15 eV(s)

f = 8 x 10^14 Hz

The energy of each photon remains the same unless the ________ of the light is changed

frequency

What is the Bohr model of an atom?

has electrons orbiting the nucleus. protons and nuetrons in the nucleus and electrons around it orbiting

The energy of a photon emitted or absorbed by a hydrogen atom equation

hf = |ΔE| = | 13.6 eV (1/n^2 (final) - 1/n^2(initial)

If an atom is ionized, what is its n value?

infinity

What is the Heisenberg Uncertainty Principle?

it is impossible to determine simultaneously both the position and velocity of an electron or any other particle

Angular momentum equation

mvr = nh/2i m = mass v = velocity r = radius n = principal quantum number h = planck's constant

The red Ha line has a wavelength of 656 nm. If the final state in the atomic transition that produces this line is n = 2, what was the initial state?

n(initial) = 3

What is the Pauli Exclusion Principle?

no two electrons in the same atom can have the same set of four quantum numbers

Bohr proposed that angular momentum was quantized. Bohr's model assumes that electrons have _______ with quantized energy levels

orbitals

Radius of nth orbit equation

r = (n^2h^2) / (4pi^2kme^2)

What is threshold frequency?

the minimum frequency of light required to produce the photoelectric effect

Heisenberg uncertainty relation

∆x∆p≥h/2π