Unit 12A Quantum PhysicsDue: 7:30am on Friday, March 27, 2015

To understand how points are awarded, read the Grading Policy for this assignment.

Prelecture Video: Photons (Chapter 28, Video 1)

Click the Play button below to start the video. You must answer the questions embedded in the video in order to proceed through it. When you have watchedthe entire video, answer the follow­up question below in Part A. Only this follow­up question will be graded.

Part A

In the video, we see that sunscreen protects the skin by

ANSWER:

Correct

Prelecture Video: Energy levels and quantum jumps (Chapter 28, Video 3)

Click the Play button below to start the video. You must answer the questions embedded in the video in order to proceed through it. When you have watchedthe entire video, answer the follow­up question below in Part A. Only this follow­up question will be graded.

Reflecting ultraviolet light.

Absorbing ultraviolet light.

Shifting the photon energy of ultraviolet light to a safer range.

Healing the damage caused by ultraviolet light.

Part A

The spectrum of light emitted by helium atoms is different than that emitted by hydrogen atoms because

ANSWER:

Correct

Prelecture Video: The wave nature of matter (Chapter 28, Video 2)

Click the Play button below to start the video. You must answer the questions embedded in the video in order to proceed through it. When you have watchedthe entire video, answer the follow­up question below in Part A. Only this follow­up question will be graded.

Helium is a noble gas, and hydrogen is not.

The force binding an electron to a helium nucleus is larger.

The different atoms have different quantized energy levels.

Helium atoms are more massive, so move more slowly.

Part A

Light has both a wave and a particle nature. Particles have a wave nature as well, and therefore

ANSWER:

Correct

Light on a Photoelectric Surface

When ultraviolet light with a wavelength of 400 falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is 1.10 .

Part A

What is the maximum kinetic energy of the photoelectrons when light of wavelength 290 falls on the same surface?

Use = 6.63×10−34 for Planck's constant and = 3.00×108 for the speed of light and express your answer in electron volts.

Hint 1. Calculate the photon energy

Calculate the energy of a photon with a wavelength of 290 .

Express your answer in electron volts.

Hint 1. Energy of the photon

Recall that the energy of a photon is given by

,

where 6.63×10−34 is Planck's constant, is the frequency, 3.00×108 is the speed of light, and is the wavelength.

ANSWER:

All particles decay with a certain characteristic lifetime.

Their mass is not well defined.

Their position can not be specified with absolute precision.

Their speeds can only have certain quantized values.

nm eV

K0 nm

h J ⋅ s c m/s

E nm

E

E = hf = hcλ

J ⋅ s f m/s λ

Hint 2. Calculate the work function

Calculate the work function for the metal surface in this problem, given the energy of the photon with a wavelength of 400 and the emittedphotoelectron with an energy of 1.10 .

Express your answer in electron volts.

Hint 1. Work function definition

The work function of a metal is the potential energy that a photoelectron must overcome before it can be emitted from the surface. It isessentially the binding energy of the electron to the metal.

Hint 2. Energy of the photon

Recall that the energy of a photon is given by

,

where 6.63×10−34 is Planck's constant, is the frequency, 3.00×108 is the speed of light, and is the wavelength.

ANSWER:

Hint 3. Converting to electron volts

To convert your answers into electron volts, use the relation .

ANSWER:

Correct

= 4.28 E eV

ϕ nmeV

E

E = hf = hcλ

J ⋅ s f m/s λ

= 2.01 ϕ eV

1 eV = 1.60 × J10−19

= 2.28 K0 eV

Photoelectric Effect

The following table lists the work functions of a few common metals, measured in electron volts.MetalCesium 1.9

Potassium 2.2

Sodium 2.3

Lithium 2.5

Calcium 3.2

Copper 4.5

Silver 4.7

Platinum 5.6

Using these data, answer the following questions about the photoelectric effect.

Part A

Light with a wavelength of 190 is incident on a metal surface. The most energetic electrons emitted from the surface are measured to have 4.0 ofkinetic energy. Which of the metals in the table is the surface most likely to be made of?

Hint 1. Find the energy of the incident photon

How much energy does a 190­ photon have?

Enter your answer numerically in electron volts to two significant figures.

Hint 1. Energy of a photon

The energy of a photon of frequency is . Frequency and wavelength are related as .

ANSWER:

Φ (eV)

nm eV

E nm

ν E = hν λν = c

= 6.5 E eV

Hint 2. Photoelectric effect

The work function is the energy required to remove an electron from the metal. Thus, by conservation of energy, the work function plus the kineticenergy of the emitted electron will be equal to the energy of the incident photon.

ANSWER:

Correct

Part B

Of the eight metals listed in the table, how many will eject electrons when a green laser ( ) is shined on them?

Hint 1. Find the energy of the incident photon

How much energy does a 510­ photon have?

Enter your answer numerically in electron volts to two significant figures.

Hint 1. Energy of a photon

The energy of a photon of frequency is . Frequency and wavelength are related as .

ANSWER:

cesium

potassium

sodium

lithium

calcium

copper

silver

platinum

= 510 nmλg

E nm

ν E = hν λν = c

Hint 2. Photoelectric effect

The work function is the energy required to remove an electron from the metal. When a photon is absorbed by a metal, the metal will only emit anelectron if the energy of the incident photon is greater than the work function of the metal.

ANSWER:

Correct

Part C

Light with some unknown wavelength is incident on a piece of copper. The most energetic electrons emitted from the copper have 2.7 of kineticenergy. If the copper is replaced with a piece of sodium, what will be the maximum possible kinetic energy of the electrons emitted from this newsurface?

Enter your answer numerically in electron volts to two significant figures.

Hint 1. How to approach the problem

Recall that the work function represents the energy required to remove an electron from a metal. If the work function is smaller, less energy isrequired to remove an electron.

= 2.4 E eV

0

1

2

3

4

5

6

7

8

eVK

ANSWER:

Correct

The de Broglie Wavelength

Part A

How does the de Broglie wavelength of an electron change if its momentum increases?

Hint 1. The de Broglie wavelength

The de Broglie wavelength of an electron of momentum is defined as

,where is Planck's constant. That is, the de Broglie wavelength of an electron is inversely proportional to its momentum.

ANSWER:

CorrectThe larger the electron's momentum, the shorter its de Broglie wavelength will be.

= 4.9 K eV

λ p

λ = hp

h

The de Broglie wavelength of the electron increases.

The de Broglie wavelength of the electron decreases.

The de Broglie wavelength of the electron is unchanged.

Part B

How does the de Broglie wavelength of an electron change if its kinetic energy decreases?

Hint 1. How to approach the question

Recall that magnitude of an object's momentum is the product of its mass and speed, . The de Broglie wavelength of an electron, then,can be expressed as

,

where and are the mass and speed of the electron. Note that the de Broglie wavelength of an electron is inversely proportional to its speed.

Use this information along with the relation between kinetic energy and speed of an object to determine how the de Broglie wavelength of anelectron changes with its kinetic energy.

ANSWER:

Correct

The lower the electron's energy, the larger its de Broglie wavelength will be. This agrees with the result from Part A.

± Accelerating Electrons

Part A

Through what potential difference must electrons be accelerated (from rest) so that they will have the same wavelength as an x­ray of wavelength0.135 ?

Use 6.63×10−34 for Planck's constant, 9.11×10−31 for the mass of an electron, and 1.60×10−19 for the charge on an electron. Expressyour answer using three significant figures.

p = mv

λ = hmv

m v

The de Broglie wavelength of the electron increases.

The de Broglie wavelength of the electron decreases.

The de Broglie wavelength of the electron is unchanged.

ΔVnm

J ⋅ s kg C

Hint 1. How to approach the problem

When an electron is accelerated through a potential difference, the change in kinetic energy of the electron equals the work done on the electron bythe electric force. Thus, you can write an expression that links the accelerating potential to the kinetic energy of the electron. Then, you need toexpress the kinetic energy of the electrons in terms of their wavelength.

Hint 2. Find the work done on an electron by the electric force

An electron, initially at point a, is accelerated to point b through a potential difference . What is the work done on the electron bythe electric force?

Let be the magnitude of the charge of an electron.

Hint 1. Potential difference

From the definition of electric potential as electric potential energy per unit charge, it follows that the potential difference ,the difference between the potential at point A and the potential at point B, equals the work done by the electric force when a unit chargemoves from A to B.

ANSWER:

Hint 3. Find the kinetic energy of the electrons

What is the kinetic energy of an electron that has the same wavelength as an x­ray of wavelength 0.135 ?

Use 6.63×10−34 for Planck's constant and 9.11×10−31 for the mass of an electron. Express your answer using three significantfigures.

Hint 1. Kinetic energy as a function of momentum

ΔV = −Vb Va

qe

= −VAB VA VB

ΔVqe

− ΔVqe

ΔVqe

− ΔVqe

K nm

J ⋅ s kg

= 1 2

The kinetic energy of a particle with mass and velocity is defined as . However, given the momentum of theparticle, the kinetic energy of the particle can also be expressed as

.

Hint 2. Find the momentum of the electrons

What is the momentum of an electron that has the same wavelength as an x­ray of wavelength 0.135 ?

Use 6.63×10−34 for Planck's constant. Express your answer using three significant figures.

Hint 1. The de Broglie wavelength

The de Broglie wavelength of a particle with momentum is defined as

,

where = 6.63×10−34 is Planck's constant.

ANSWER:

ANSWER:

ANSWER:

Correct

m v K = m12

v2 p = mv

K = p2

2m

p nm

J ⋅ s

p

λ = hp

h J ⋅ s

= 4.91×10−24 p kg ⋅ m/s

= 1.32×10−17 K J

= 82.5 ΔV V

Part B

Through what potential difference must electrons be accelerated so they will have the same energy as the x­ray in Part A?

Use 6.63×10−34 for Planck's constant, 3.00×108 for the speed of light in a vacuum, and 1.60×10−19 for the charge on an electron.Express your answer using three significant figures.

Hint 1. How to approach the problem

As in the previous part, it is useful to write an expression that links the accelerating potential to the kinetic energy of the electrons. However, thistime you need to express the energy of the electrons in terms of the energy of x­rays of given wavelength.

Hint 2. Find the energy of the x­ray

What is the energy of an x­ray of wavelength 0.135 ? Recall that x­rays are electromagnetic waves and are subject to the same quantumrelations as those of photons of light.

Use 6.63×10−34 for Planck's constant and 3.00×108 for the speed of light in a vacuum. Express your answer using threesignificant figures.

Hint 1. The energy of a photon

Given the wavelength of a photon, its energy is

,

where = 3.00×108 is the speed of light and = 6.63×10−34 is Planck's constant.

ANSWER:

ANSWER:

ΔV

J ⋅ s m/s C

E nm

J ⋅ s m/s

λ

E = h cλ

c m/s h J ⋅ s

= 1.47×10−15 E J

= 9190 ΔV V

Correct

Multiple Choice Question 28.28

In a photoelectric effect experiment, the intensity of the light is increased while the frequency is held constant.

Part A

As a result,

ANSWER:

Correct

Problem 28.14

Image intensifiers used in night­vision devices create a bright image from dim light by letting the light first fall on a photocathode. Electrons emitted by thephotoelectric effect are accelerated and then strike a phosphorescent screen, causing it to glow more brightly than the original scene. Recent devices aresensitive to wavelengths as long as 900 , in the infrared:

Part A

If the threshold wavelength is 900 , what is the work function of the photocathode?

Express your answer with the appropriate units.

ANSWER:

there are more photoelectrons.

the photoelectrons are faster.

Both above­mentioned effects occur.

Neither first effect nor second one take place.

nm

nm

Correct

Part B

If light of wavelength 720 strikes such a photocathode, what will be the maximum kinetic energy, in , of the emitted electrons?

Express your answer using two decimal places and include the appropriate units.

ANSWER:

Correct

Problem 28.30

Part A

What is the speed of an electron with a de Broglie wavelength of 0.14 ?

Express your answer using two significant figures.

ANSWER:

Correct

Part B

= 1.38 E eV

nm eV

= 0.35 K eV

nm

= 5.2×106 v m/s

What is the speed of a proton with a de Broglie wavelength of 0.14 ?

Express your answer using two significant figures.

ANSWER:

Correct

Problem 28.42

The allowed energies of a quantum system are 0.0 , 6.0 , and 9.0 .

Part A

What wavelengths appear in the system's emission spectrum?

Enter your answers in ascending order separated by commas.

ANSWER:

Correct

Problem 28.44

The allowed energies of a quantum system are 0.0 , 1.5 , 3.0 , and 6.0 .

Part A

How many different wavelengths appear in the emission spectrum?

ANSWER:

nm

= 2800 v m/s

eV eV eV

= 140,210,410 λi nm

eV eV eV eV

Correct

Problem 28.54

Light of constant intensity but varying wavelength was used to illuminate the cathode in a photoelectric­effect experiment. The graph of the figure shows howthe stopping potential depended on the frequency of the light.

Part A

What is the work function, in , of the cathode?

Express your answer using one significant figure.

ANSWER:

= 4N

eV

= 4 E0 eV

Correct

Problem 28.68

Electrons with a speed of 2.9×106 pass through a double­slit apparatus. Interference fringes are detected with a fringe spacing of 1.0 .

Part A

What will the fringe spacing be if the electrons are replaced by neutrons with the same speed?

Express your answer using two significant figures.

ANSWER:

Correct

Part B

What speed must neutrons have to produce interference fringes with a fringe spacing of 1.0 ?

Express your answer using two significant figures.

ANSWER:

Correct

Problem 28.85­88

m/s mm

= 0.55 Δyn μm

mm

= 1600 vn m/s

Further support for the photon model of electromagnetic waves comes from Compton scattering, in which x­rays scatter from electrons, changing direction andfrequency in the process. Classical electromagnetic wave theory cannot explain the change in frequency of the x­rays on scattering, but the photon modelcan.

Suppose an x­ray photon is moving to the right. It has a collision with a slow­moving electron, as in the figure. The photon transfers energy and momentum tothe electron, which recoils at a high speed. The x­ray photon loses energy, and the photon energy formula tells us that its frequency mustdecrease. The collision looks very much like the collision between two particles.

Part A

What happens when the x­ray photon scatters from the electron?

ANSWER:

Correct

Part B

E = hf

Its speed increases

Its speed decreases

Its speed stays the same

What happens when the x­ray photon scatters from the electron?

ANSWER:

Correct

Part C

What happens when the electron is struck by the x­ray photon?

ANSWER:

Correct

Part D

X­ray diffraction can also change the direction of a beam of x­rays. Which statement offers the best comparison between Compton scattering and x­raydiffraction?

ANSWER:

Its wavelength increases

Its wavelength decreases

Its wavelength stays the same

Its de Broglie wavelength increases

Its de Broglie wavelength decreases

Its de Broglie wavelength stays the same

Correct

Problem 28.15

Light with a wavelength of 390 shines on a metal surface, which emits electrons. The stopping potential is measured to be 0.888 .

Part A

What is the maximum speed of emitted electrons?

ANSWER:

Correct

Part B

Calculate the work function.

ANSWER:

Correct

X­ray diffraction changes the wavelength of x­rays; Compton scattering does not

Compton scattering changes the speed of x­rays; x­ray diffraction does not

X­ray diffraction relies on the particle nature of the x­rays; Compton scattering relies on the wave nature

X­ray diffraction relies on the wave nature of the x­rays; Compton scattering relies on the particle nature

nm V

= 5.58×105 v m/s

= 2.30 A eV

Part C

Identify the metal.

ANSWER:

Correct

Problem 28.12

A photoelectric­effect experiment finds a stopping potential of 1.93 when light of 200 is used to illuminate the cathode.

Part A

From what metal is the cathode made?

ANSWER:

Correct

Part B

Potassium

Aluminum

Iron

Gold

V nm

gold

aluminum

silver

potassium

What is the stopping potential if the intensity of the light is doubled?

ANSWER:

Correct

Problem 28.20

Part A

What is the wavelength, in , of a photon with energy 0.30 ?

Express your answer using two significant figures.

ANSWER:

Correct

Part B

Is this wavelength visible, ultraviolet, or infrared light?

ANSWER:

= 1.93 V V

nm eV

= 4100 λ nm

visible light

ultraviolet light

infrared light

Correct

Part C

What is the wavelength, in , of a photon with energy 3.0 ?

Express your answer using two significant figures.

ANSWER:

Correct

Part D

Is this wavelength visible, ultraviolet, or infrared light?

ANSWER:

Correct

Part E

What is the wavelength, in , of a photon with energy 30 ?

Express your answer using two significant figures.

ANSWER:

nm eV

= 410 λ nm

visible light

ultraviolet light

infrared light

nm eV

Correct

Part F

Is this wavelength visible, ultraviolet, or infrared light?

ANSWER:

Correct

Multiple Choice Question 28.27

In a photoelectric effect experiment, the frequency of the light is increased while the intensity is held constant.

Part A

As a result,

ANSWER:

= 41 λ nm

visible light

ultraviolet light

infrared light

There are more electrons.

The electrons are faster.

Both above­mentioned effects.

Neither first effect nor second one.

Correct

Multiple Choice Question 28.35

Photon P in the figure moves an electron from energy level to energy level . The electron jumps to , emitting photon Q, and thenjumps to , emitting photon R. The spacing between energy levels is drawn to scale.

Part A

The spacing between energy levels is drawn to scale. What is the correct relationship among the wavelengths of the photons?

ANSWER:

n = 1 n = 3 n = 2n = 1

< <λP λQ λR

< <λR λP λQ

< <λQ λP λR

< <λP λR λQ

Correct

Multiple Choice Question 28.26

A light sensor is based on a photodiode that requires a minimum photon energy of 1.40 to create mobile electrons.

Part A

What is the longest wavelength of electromagnetic radiation that the sensor can detect?

ANSWER:

Correct

Multiple Choice Question 28.29

In the photoelectric effect, electrons are never emitted from a metal if the frequency of the incoming light is below a certain threshold value.

Part A

Why is this true?

ANSWER:

eV

1390 .

1890 .

888 .

388 .

nmnm

nmnm

Correct

Multiple Choice Question 28.34

You shoot a beam of electrons through a double slit to make an interference pattern. After noting the properties of the pattern, you then double the speed ofthe electrons.

Part A

What effect would this have?

ANSWER:

Correct

Conceptual Question 28.5

shows the typical photoelectric behavior of a metal as the anode­cathode potential difference is varied.

Photons of lower­frequency light don't have enough energy to eject an electron.

The electric field of low­frequency light does not vibrate the electrons rapidly enough to eject them.

The number of photons in low­frequency light is too small to eject electrons.

Low­frequency light does not penetrate far enough into the metal to eject electrons.

The fringes would get closer together.

The fringes would get farther apart.

The positions of the fringes would not change.

ΔV

Part A

Why do the curves become horizontal for ? Shouldn't the current increase as the potential difference increases?

ANSWER:

Correct

Part B

ΔV > 1 V

A positive cathode attracts all of the electrons emitted by the anode. Once is large enough to make a banch of electrons leave the anode,the current incareases by some value and stays the same until the next large enough increase in .

A positive cathode attracts all of the electrons emitted by the anode. Since the number of the electrones emitted by the anode is limited, onceall of the electrons reach the cathode, a further increase in does not cause any further increase in the current.

A positive anode attracts all of the electrons emitted by the cathode. Since the number of the electrones emitted by the cathode is limited, onceall of the electrons reach the anode, a further increase in does not cause any further increase in the current.

A positive anode attracts all of the electrons emitted by the cathode. Once is large enough to make a banch of electrons leave thecathode, the current incareases by some value and stays the same until the next large enough increase in .

ΔVΔV

ΔV

ΔV

ΔVΔV

Δ < 0 V Δ < 0 V

Why doesn't the current immediately drop to zero for ? Shouldn't prevent the electrons from reaching the anode?

ANSWER:

Correct

Part C

The current is zero for . Where do the electrons go? Are no electrons emitted or if they are, why is there no current?

ANSWER:

Correct

Conceptual Question 28.18

Part A

Can an electron with a de Broglie wavelength of 2 pass through a slit that is 1 wide? Select the correct answer and explanation.

ANSWER:

ΔV < 0 V ΔV < 0 V

If an electron has enough kinetic energy when it leaves the cathode it might just reach the anode even when .

If an electron has enough potential energy when it leaves the anode it might just reach the cathode even when .

If an electron has enough potential energy when it leaves the cathode it might just reach the anode even when .

If an electron has enough kinetic energy when it leaves the anode it might just reach the cathode even when .

ΔV < 0 VΔV < 0 VΔV < 0 V

ΔV < 0 V

ΔV < −2.0 V

Electrons are still emitted, but they do not have sufficient kinetic energy to reach the cathode.

Electrons are not emmited for , which is defined by the material of the cathode.

Electrons are not emmited for , which is defined by the material of the anod.

Electrons are still emitted, but they do not have sufficient kinetic energy to reach the anode.

ΔV < −Vcritical

ΔV < −Vcritical

μm μm

Correct

Multiple Choice Question 28.32

Light consisting of 6.0 photons is incident on a piece of iron, which has a work function of 4.7 .

Part A

What is the maximum kinetic energy of the ejected electrons?

ANSWER:

Correct

Reading Question 28.01

Part A

Yes. If the electron was a classic particle it would be impossible, but electron can pass through the slit due to quantum tunneling.

Yes. Electron wavelength is only a measure of how rapidly its amplitude varies as measured along its direction of motion and has nothing to dowith passing through the slit.

No. Electron can only pass through a slit that is wider than the electron wavelength.

It is impossible to say because electron can only pass through a slit that is wider than the half of the electron wavelength. More precise valuesof the wavelength and the slit width are necessary.

eV eV

1.3

4.7

10.7

6.0

eVeVeV

eV

The photoelectric effect tells us that __________.

Hint 1.

ANSWER:

Correct

Reading Question 28.02

Part A

The energy of a photon depends on __________.

ANSWER:

Correct

electrons have a wave nature

a photon can be converted into an electron

electrons are the conductors in metals

light has a particle nature

its mass

its charge

its frequency

its speed

Reading Question 28.03

Part A

Which of the following wave properties are exhibited by particles?

ANSWER:

Correct

Reading Question 28.04

Part A

When an electron in a quantum system drops from a higher energy level to a lower one, the system __________.

ANSWER:

Correct

diffraction

superposition

interference

All of the listed answers are correct.

emits an electron

emits a photon

emits a plasmon

emits a neutron

Reading Question 28.05

Part A

If you precisely measure the position of a particle, you __________.

ANSWER:

Correct

Score Summary:Your score on this assignment is 110%.You received 27.77 out of a possible total of 28 points, plus 3 points of extra credit.

destroy information about its speed

destroy its wave nature

cause it to diffract

cause the particle to be annihilated