Version 001 – Quantum – tubman – (19112) 1

This print-out should have 12 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

Holt SF 23B 02001 (part 1 of 2) 10.0 points

Light of wavelength 350 nm falls on a potas-sium surface, and the photoelectrons have amaximum kinetic energy of 1.3 eV.What is the work function of potassium?

The speed of light is 3× 108 m/s and Planck’sconstant is 6.63× 10−34 J · s .

Correct answer: 2.25179 eV.

Explanation:

Let : λ = 350 nm = 3.5× 10−7 m ,

Kmax = 1.3 eV ,

h = 6.63× 10−34 J · s , and

c = 3.00× 108 m/s .

Kmax = h f − h ft

h ft = hc

λ−Kmax

=6.63× 10−34 J · s

1.6× 10−19 J/eV

×3× 108 m/s

3.5× 10−7 m− 1.3 eV

= 2.25179 eV .

002 (part 2 of 2) 10.0 pointsWhat is the threshold frequency for potas-sium?

Correct answer: 5.43417× 1014 Hz.

Explanation:

ft =h fth

=(2.25179 eV)(1.6× 10−19 J/eV)

6.63× 10−34 J · s

= 5.43417× 1014 Hz .

AP B 1993 MC 53003 10.0 points

In the photoelectric effect, the maximumspeed of the electrons emitted by a metal sur-face when it is illuminated by light dependson which of the following?I) Intensity of the lightII) Frequency of the lightIII) Nature of the photoelectric surface

1. I, II, and III

2. II and III only correct

3. III only

4. I and II only

5. I only

Explanation:According to Einstein, the maximum ki-

netic energy for the liberated electrons is

Kmax = h f − φ,

where φ is the work function of the metal.Therefore, Kmax is independent of the in-

tensity of light. It does depend on the fre-quency of the light, and the nature of thephotoelectric surface (which determines φ).

AP B 1994 FR 3a004 10.0 points

A series of measurements were taken of themaximum kinetic energy of photoelectronsemitted from a metallic surface when lightof various frequencies is incident on the sur-face. The table below lists the measurementsthat were taken.

Light Maximum KineticFrequency Energy

5× 1014 Hz 0.69 eV6× 1014 Hz 1.2 eV7× 1014 Hz 1.4 eV8× 1014 Hz 2.1 eV9× 1014 Hz 2.1 eV1× 1015 Hz 2.8 eV

Version 001 – Quantum – tubman – (19112) 2

On the axes below, plot the kinetic energyversus light frequency for the data, and drawthe best straight-line fit to the data.

0 1 2 3 4 5 6 7 8 9 100

3

6

9

12

15

18

21

24

27

30

Maxim

um

Kinetic

Energy(×

10−1eV

)

Frequency (× 1014 s−1)

Electron Energy vs Frequency

From this rough experiment, determine avalue for Planck’s constant h. The speedof light is 2.99792× 108 m/s , and 1 eV =1.60218× 10−19 J/eV .Your answer must be within ± 4.0%

Correct answer: 3.98571× 10−15 eV · s.

Explanation:The best fit line (using a linear least square

criterion) is shown below.

0 1 2 3 4 5 6 7 8 9 100

3

6

9

12

15

18

21

24

27

30

Maxim

um

Kinetic

Energy(×

10−1eV

)

Frequency (×1014 s−1)

Electron Energy vs Frequency

From the photoelectric equation, Kmax =

hf − W , Planck’s constant h is the slope ofthe best fit line of Kmax vs f . For calculationof the slope using two points on the line, notdata points. For any two points (1 and 2) onthe line

Slope =K2 −K1

f2 − f1.

For example, for points (10 Hz, 2.8 eV) and(5 Hz, 0.69 eV),

h ≈2.8 eV − 0.69 eV

1× 1015 Hz− 5× 1014 Hz

≈ 4.22× 10−15 eV · s , or least squares fit

= 3.98571× 10−15 eV · s

= (3.98571× 10−15 eV · s)

×(1.60218× 10−19 J/eV)

= 6.38582× 10−34 J · s ,

where the linear least squares fit to the datapoints is

Kmax = (3.98571× 10−15 eV · s) f

−1.27429 eV .

AP B 1993 MC 38005 10.0 points

Of the following phenomena, which providesthe best evidence that particles can have waveproperties?

1. The production of X-rays by electronsstriking a metal target

2. The scattering of photons by electrons atrest

3. The absorption of photons by electrons inan atom

4. The α-decay of radioactive nuclei

5. The interference pattern produced byneutrons incident on a crystal correct

Explanation:The interference pattern produced by neu-

trons hitting a crystal is very similar to theinterference patterns produced by light hit-ting a single slit. This interference patternshows the wavelike properties of matter.

Version 001 – Quantum – tubman – (19112) 3

Ball DeBroglie Wavelength006 10.0 points

A ball has a mass of 0.275 kg just before itstrikes the Earth after being dropped from abuilding 22.6 m tall.What is its de Broglie wavelength? The ac-

celeration of gravity is 9.8 m/s2 and Planck’sconstant is 6.62607× 10−34 J · s .

Correct answer: 1.14483× 10−34 m.

Explanation:

Let : d = 22.6 m ,

m = 0.275 kg , and

h = 6.62607× 10−34 J · s .

From conservation of energy

1

2mv2 = mg d

v =√

2 g d =√

2 (9.8 m/s2) (22.6 m)

= 21.0466 m/s , so

p = mv = (0.275 kg) (21.0466 m/s)

= 5.78782 kg ·m/s and

λ =h

p=

6.62607× 10−34 J · s

5.78782 kg ·m/s

= 1.14483× 10−34 m .

AP B 1998 MC 34007 10.0 points

If the momentum of an electron doubles, bywhat factor would its de Broglie wavelengthbe multiplied?

1.λ′

λ=

1

4

2.λ′

λ=

1

2correct

3.λ′

λ= 4

4.λ′

λ= 1

5.λ′

λ= 2

Explanation:The de Broglie wavelength of a particle is

given by λ =h

p.

If the momentum of the electron doubles(p′ = 2 p), the new de Broglie wavelength willbe

λ′ =h

p′=

h

2 p=

1

2

h

p=

1

2λ .

Uncertainty Principle 08008 10.0 points

Suppose one wishes to use an electron beamto resolve a molecule whose size is 1 nm.The mass of an electron is 9.11× 10−31 kg

and Planck’s constant is 6.63× 10−34 J · s.What is the kinetic energy of electrons that

have the requisite wavelength?

Correct answer: 1.50785 eV.

Explanation:

Let : m = 9.11× 10−31 kg ,

h = 6.63× 10−34 J · s , and

λ = 1 nm = 1× 10−9 m .

p =h

λ

E =p2

2m=

h2

2mλ2

So

Emin =(6.63× 10−34 J · s)2

2 (9.11× 10−31 kg) (1× 10−9 m)2

= 2.41256× 10−19 J

= 1.50785 eV .

AP B 1998 MC 11009 10.0 points

Which of the following experiments provided

Version 001 – Quantum – tubman – (19112) 4

evidence that electrons exhibit wave proper-ties?I) Millikan oil-drop experimentII) Davisson-Germer electron-diffraction ex-

perimentIII) J. J. Thomson’s measurement of the

charge-to-mass ratio of electrons

1. II only correct

2. II and III only

3. I, II and III

4. I and III only

5. I only

Explanation:The Davisson-Germer electron-diffraction

experiment showed that electrons exhibitwave properties, such as diffraction and in-terference.On the other hand, Millikan’s oil-drop ex-

periment measured the charge of electrons,and J. J. Thomson’s measurement of thecharge-to-mass ration of electrons gave thevalue of electron mass (once the charge ofelectrons was known, for example, from Milli-gan oil-drop experiment).

Uncertainty of Momentum010 10.0 points

Assume we can localize a particle to an uncer-tainty of 0.1967 nm.What will be the resulting minimum uncer-

tainty in the particle’s momentum? Planck’sconstant is 6.62607× 10−34 J · s .

Correct answer: 2.68066× 10−25 kg ·m/s.

Explanation:

Let : h = 6.62607× 10−34 J · s and

∆x = 0.1967 nm = 1.967× 10−10 m .

∆x∆px ≥h̄

2=

h

4 π

∆px ≥h

4 π∆x=

6.62607× 10−34 J · s

4 π (1.967× 10−10 m)

= 2.68066× 10−25 kg ·m/s .

Hewitt CP9 32 R08011 10.0 points

In the Bohr model of the atom, what is trueabout the light emitted by an atom?

1. An electron accelerating around its orbitcontinuously emits radiation.

2. The emitted photon’s frequency is theclassic frequency at which an electron vi-brates.

3. None of these

4. The energy of the emitted photon is equalto the difference in energy between the twoorbits. correct

Explanation:Energy is conserved. When an electron

drops down an orbit and loses energy, thisenergy is emitted as a photon.

Conceptual 21 05012 10.0 points

If a one-electron atom can occupy any of 5different energy levels, how many lines mightappear in that atom’s spectrum?

1. six

2. seven

3. ten correct

4. eight

5. nine

Explanation:If there are 5 energy levels, the possible

jumps from higher to lower states are: 5 to 1,5 to 2, 5 to 3, 5 to 4, 4 to 1, 4 to 2, 4 to 3, 3 to1, 3 to 2 and 2 to 1.Thus there will be ten lines possible.

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