Physics 30 program, machine-scorable open-ended questions : Units 1-5, Complete course

Ex Libris

UNIVERSITY OF XT

ALBERTA UNIVERSITATIS

Albertensis

UNIVERSITY LIBRARY

TABLE OF CONTENTS

PAGE

Preface iii
Introduction 1
Introduction to Machine-Scorable Open-Ended Items 1
Sample Instruction Sheet 4
Sample Machine-Scorable Open-Ended Items 5
Item Information Sheet 20
Solutions to Questions having Multiple Correct Answers 22
Answer Sheet with Solutions Displayed 31
Sample Answer Sheet, Blank for Student Use 33

i

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PREFACE

This document outlines the use of machine-scorable open-ended questions in the
evaluation of Physics 30. It is primarily designed for use by classroom
teachers of Physics 30, though it can be modified for use by teachers of
Physics 10 and Physics 20. The sample items provided were originally asked in
multiple-choice form on Grade 12 Diploma Examinations.

The use of machine-scorable open-ended questions is being considered for the
1991 series of Diploma Examinations in Physics 30, and such questions are
being introduced on the 1990 field tests. Students writing these field tests
should be made familiar with the use of this item format.

The document can also be used for professional inservice at the school or
jurisdiction level. In this case, supplementary materials suitable to local
needs could be added to the basic foundation provided here.

To obtain further copies, or to make comments about this document, you are
invited to contact:

Jack Edwards

Examination Manager, Physics 30

Student Evaluation and Records Branch

Alberta Education

Box 43, Devonian Building

11160 Jasper Avenue

Edmonton, Alberta

T5K 0L2

427-2948 Fax: 422-4200

SEPTEMBER 1989

IV

INTRODUCTION TO MACHINE-SCORABLE OPEN-ENDED ITEMS

It is now possible to have students derive the answer to an item and record
the answer in a machine-readable form. Final answers can be expressed as a
single number or as a sequence of numbers. This item format can be used as an
alternative to multiple-choice, to sequencing, or to matching formats.

Alberta Education uses the following guidelines for the construction of
machine-scorable open-ended items:

• Solutions to machine-scorable open-ended Items should be short and
involve no more than one or two steps.

• All legitimate methods of solving a problem should lead to numerical
answers that are marked correct.

• In general, each item should have the same weighting as a
multiple-choice item.

Compared with multiple-choice items there are the following benefits:

• The chances of a student guessing an answer correctly is essentially
reduced to zero.

• Answer verification methods are eliminated as students cannot work
backwards from the alternatives provided.

• Students cannot determine the correct answer by eliminating the other
distractors .

• Students cannot obtain additional clues from the list of possible
responses presented.

• Machine-scorable open-ended items individually discriminate very well
between strong and weak students at all difficulty levels.

• When these items are included on a test, the test as a whole will be
more reliable in ranking the students from strongest to weakest.

However, the following limitations still apply:

• The effectiveness of this type of item has not yet been tested for
subjects other than mathematics.

• Answer fields suitable for this type of Physics 30 item may be too
complex for use.

Alberta Education has used machine-scorable open-ended questions successfully
since January 1989 in its Mathematics 30 Diploma Examinations. The
mathematics questions used a four space answer field, accommodating numbers
between 0.1 and 999.9. All input numbers were presumed exact, and all
numerical answers were required to be rounded to the nearest tenth. A
standardized rounding instruction had to be given. Students were told to keep
their calculators running for the whole item. When this instruction was
given, multiple answers were eliminated.

1

In introducing machine-scorable open-ended questions to Physics 30 there are
four main problems:

• Physics questions do not have answers that limit to the numerical
range from 0.1 to 999.9. Allowance must be made for the use of
scientific notation and metric prefixes.

Possible solution: For numerical answers between 0.01 and 99.99 SI
base units, there are no difficulties if the decimal is placed
between the second and the third columns of the answer field. For
numerical answers between 0.01 and 99.99 SI base units, there are no
difficulties. We will use the prefixes kilo, centi, and milli
freely, and will attach them to any SI base unit. The prefixes mega
and nano will be restricted to common usages such as megahertz (MHz)
and nanometre (nm) . No other prefixes will be used in
machine-scorable open-ended questions. We will express other large
and small numbers in scientific notation, and will ask for either the
numerical factor or the exponent. To allow for three digit accuracy
in the numerical factor, the decimal point is placed between the
second and the third columns of the answer field.

• Most numbers used in physics calculations are not exact, and answers
obtained must reflect the proper use of significant digits.

Possible solution: Machine-scorable open-ended questions can be used
as a rigorous test of the proper use of significant digits, but that
is not the intention here. In most questions, it will be obvious how
many digits are called for. In those questions where it is not
obvious, allowance for variations will be made in the key.

• Multiple answers cannot be completely eliminated, as the Data Booklet
is of three-digit accuracy, and the use of different constants or
equations as starting points automatically gives final answers that
can differ by two or three in the third digit.

Possible solution: We find more multiple answers in the Structure of
Matter unit than in any other. All such answers could be eliminated
if a five digit Data Booklet were used. However, the use of three
digit input data is customary in physics teaching. Consequently,
answer keys must allow for multiple answers. On the other hand, the
keys do not allow for answers generated by premature internal
rounding.

• Answers expressed in scientific notation to three-digit accuracy
really need six spaces in the answer field, three to accommodate the
numerical factor, one for the sign of the exponent, and two for the
exponent.

Possible solution: Six-space answer fields might prove too difficult
to use, and the mixing of two or three different answer fields in a
seven or twelve question section is undesirable. Therefore, for
questions whose full answers require the use of scientific notation,
we will use the following format:

The force is b x 10 w N when expressed in scientific notation.

The value of b (or of w) is .

2

The items

The 30 items were originally asked in multiple-choice form on previous
Physics 30 Diploma Examinations. The Item Information Sheet on pages 20 and
21 shows the item difficulties of the individual questions as tested. When
reading this table, bear in mind that the difficulties of easy questions are
above 0.750, of average questions between 0.550 and 0.750, and of difficult
questions are below 0.550. It is probable that all items will be more
difficult when asked in machine-scorable open-ended form.

The items illustrate answers that are numbers expressed to both two and three
significant digits. For answers such as 1.47 x 10 -17 , some questions call for
the 1.47, while others call for the 17 in the exponent.

Other items are included to illustrate the presence of two or more valid
answers. All valid answers are included on the Item Information Sheet on
pages 20 and 21, with only the most common answer included on the answer sheet
on pages 31 and 32. In working out the answers, we have not rounded any
intermediate answers. Explanations of the allowed variations in the answers
are found on pages 22 to 30.

Using the items

The items can be used as review sheets for physics content, or as practice
items for the question format. It is not appropriate to use the complete set
as a test, since the items were selected to illustrate the use of significant
digits, the presence of valid multiple answers, the use of scientific
notation, and the recording of partial answers. The set does not constitute a
selection of typical physics problems that can be used as a comprehensive test
in Physics 30.

Field Tests in Physics 30

All field tests in Physics 30, starting in January 1990, will contain
machine-scorable open-ended questions. The instruction page used in these
field tests is shown on page 4. Teachers are encouraged to share the contents
of this document with their students before the field tests are administered.
The form in which machine-scored open-ended questions appear on future Diploma
Examinations largely depends on the results of the 1990 field tests.

3

SAMPLE INSTRUCTION SHEET (FOR MACHINE~SCORABLE OPEN-ENDED QUESTIONS)

INSTRUCTIONS

There are five machine-scored open-ended questions each with a value of 1 mark
in this section of the test. All numbers used in the questions are to be
considered as the results of measurements.

Read each question carefully.

Solve each question and write your answer to the appropriate number of
significant figures.

Transfer your answer to the appropriate box on the answer sheet provided.
Darken one circle in each column as necessary in order to record the answer
to the correct number of significant figures, as illustrated below. USE AN
HB PENCIL ONLY.

Sample Questions and Solutions Answer Sheet

1) If the angle of incidence is 47.3° and
the angle of refraction is 28.3°, the
index of refraction is .

sin 6.
n = — : — — L
sm 0 2

_ sin 47.3° _ _ cc
" sin 28.3° 1-55

RECORD 1.55

2) A microwave of wavelength 11 cm has a
frequency of b x 10 9 Hz. The value
of b is .

f = c/\

= (3.00 x 10® m/s)/ (0.11 m)
f = 2.727... x 10 9 Hz
RECORD 2.7

The answers 2.70, 2.72, and 2.73 will all be
marked as incorrect, as the data given are
to two significant figures only.

If you wish to change an answer, please erase your first mark completely.

1

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WHEN YOU HAVE COMPLETED PART B, PLEASE PROCEED DIRECTLY TO PART C

4

SAMPLE MACHINE-SCQRABLE OPEN-ENDED QUEST IONS

Use the following information to answer question 1.

A glass plate that is 17 mm thick has a refractive index of 1.5.

A ray of light passing through air is incident on the plate at an
angle of 32°.

1. Where the ray leaves the glass, the horizontal distance d between the
emerging ray and the normal is mm.

RECORD THE ANSWER ON THE ANSWER SHEET

- 5 -

Use the following information to answer question 2.

4.0 cm

2.0 cm

A ray of light travels from air into glass.
This diagram is not to scale.

2. The speed of light in the glass is i x 10 8 m/s.
is

The value of b

RECORD THE ANSWER ON THE ANSWER SHEET

3. If a diffraction grating with 3.00 x 10 5 lines/m gives a first-order
bright image at an angle of 10.0° on a screen that is 5.00 m away, then
the wavelength of the light is measured to be b x 10 -7 m. The value
of b is

RECORD THE ANSWER ON THE ANSWER SHEET

6

Use the following information to answer question 4.

Two plane mirrors XY and YZ are at right angles to each other.

4. A ray of light strikes mirror XY 5.0 cm from Y. The reflected ray strikes
mirror YZ at a distance d of cm from Y.

RECORD THE ANSWER ON THE ANSWER SHEET

5. A laser beam is transmitted to a satellite and back. If the time of
travel there and back is 1.23 x 10~ 2 s , the distance between the
satellite and the surface of the Earth, expressed in scientific notation,
is b x 10 w m. The value of b is

RECORD THE ANSWER ON THE ANSWER SHEET

7

Use the following information to answer question 6.

6. If the angle of incidence is 56° and the angle between the mirrors

is 113°, then the angle between the final reflected ray and the second
mirror is °.

RECORD THE ANSWER ON THE ANSWER SHEET

7. In glass (n = 1.53), a certain infra-red source has a wavelength
of 8.00 x 10 -7 m. Its frequency is b x 10 14 Hz. The value
of b is

RECORD THE ANSWER ON THE ANSWER SHEET

8

8. A polarizing filter is set so that the previously polarized light passing
through it has MAXIMUM brightness. In order to have the light passing
through it of MINIMUM brightness, the filter must be rotated
by degrees.

RECORD THE ANSWER ON THE ANSWER SHEET

9. Two negatively charged spheres placed 1.00 m apart repel each other with
a force of 10.2 N. If the distance between the spheres is increased
to 3.00 m and the charge on each sphere is increased by a factor of
exactly 4, the force of repulsion would be expected to become N.

RECORD THE ANSWER ON THE ANSWER SHEET

9

10. The electric field between two parallel plates that are 5.0 cm apart
is 3.0 x 10 2 N/C. The potential difference across the plates
is V.

RECORD THE ANSWER ON THE ANSWER SHEET

11. The acceleration of an electron in an electric field with a strength
of 7.04 N/C is w x 10^ m/s 2 , when expressed in scientific notation.
The value of the exponent b is .

RECORD THE ANSWER ON THE ANSWER SHEET

12. If a flashlight bulb draws 0.25 A from a 3.0 V battery, the power
supplied is W.

RECORD THE ANSWER ON THE ANSWER SHEET

10

13. If a proton is accelerated from rest through a potential difference
of 1.10 x 10 6 V, it attains a final speed, expressed in scientific
notation, of b x 10 w m/s. The value of b is .

RECORD THE ANSWER ON THE ANSWER SHEET

14. If a proton moves perpendicularly across a magnetic field of strength
2.5 x 10“ 2 T at a speed of 5.3 x 10 6 m/s, it would experience a
force, expressed in scientific notation, of b x 10 v N. The value
of b is

RECORD THE ANSWER ON THE ANSWER SHEET

15. A proton moving at 2.6 x 10 5 m/s perpendicularly to a magnetic field
of 1.1 T experiences a deflecting force that causes it to orbit in
a circle. The radius of the orbit, expressed in scientific notation,
is b x 10 w m, where b is .

RECORD THE ANSWER ON THE ANSWER SHEET

11

Use the following information to answer question 16.

16.

RECORD THE ANSWER ON THE ANSWER SHEET

12

Use the following information to answer question 17.

Microwave Double-Slit Experiment

w

S,

S 2

microwave

generator

second minimum < • N 2
first maximum o A,

first minimum

-C

N,

h » Aq centre
N,

A,

U

A two-source extension is placed over the horn of a microwave
generator that transmits radiation with a frequency of
5.0 x 10 10 Hz. This procedure converts the generator into a
double-slit apparatus. A detector moves along line UW.

This diagram is NOT drawn to scale.

17. If S x and S 2 are separated by 0.27 m and A 1 is 0.36 m from A 0 , the
distance from the horn to A 0 is m.

RECORD THE ANSWER ON THE ANSWER SHEET

- 13

18.

A radio station broadcasts on a wavelength of 5.0 x 10 2 m. The frequency
of the station's radio waves is MHz.

RECORD THE ANSWER ON THE ANSWER SHEET

Use the following information to answer question 19.

19. The time delay between the direct signal and the reflected signal
is b x 10 -6 s. The value of b is

RECORD THE ANSWER ON THE ANSWER SHEET

14

20. During electrolysis, the mass of aluminum, expressed in scientific
notation, that would be deposited from a solution of Al 3 + ions in
15.3 min by a current of 0.901 A is b x 10 w g. The value of b
is

RECORD THE ANSWER ON THE ANSWER SHEET

21. In Millikan's oil drop experiment, a charged particle having a mass, of

6.4 x 10 -16 kg is accelerating upwards at 2.2 m/s 2 under the influence of
an electric field between two horizontal plates having a separation of
5.0 cm. If the potential difference across the plates is 120 V, the
charge on the particle is elementary charges.

RECORD THE ANSWER ON THE ANSWER SHEET

15

Use the following information to answer questions 22 and 23.

When white light is shone through hydrogen gas, a dark line in the
absorption spectrum, corresponding to the red line in the emission
spectrum, is observed. This red line is produced by the transition
of an electron from the n - 3 to the n - 2 Bohr orbit.

22. The energy of the absorbed photon is

eV.

RECORD THE ANSWER ON THE ANSWER SHEET

23. The frequency of the absorbed red light is b x 10 14 Hz. The value
of b is .

RECORD THE ANSWER ON THE ANSWER SHEET

16

24. When zinc (W = 4.03 eV) in a photoelectric cell is irradiated by
radiation of frequency 4.52 x 10 15 Hz, the maximum kinetic energy
of photoelectrons leaving the cell is b x 1Q“ 18 J. The value
of b is .

RECORD THE ANSWER ON THE ANSWER SHEET

25.

If an object has a mass of
of 0.694c its rest mass is

4.0 x 10 17 kg when travelling at a speed
b x 10“ 17 kg. The value of b is

RECORD THE ANSWER ON THE ANSWER SHEET

26.

If a proton has a relativistic mass of 2.1 x 10 27 kg, its speed
is b x 10 8 m/s. The value of b is .

RECORD THE ANSWER ON THE ANSWER SHEET

17

27. The momentum of a photon with 5.1 x 10 -19 J of energy, expressed
in scientific notation, is w x 10 - ^ kg*m/s. The value of the
exponent b is .

RECORD THE ANSWER ON THE ANSWER SHEET

28. The speed, expressed in scientific notation, of an object that has a
mass of 2.2 x 10 _2 ° kg and an associated wavelength of 5.1 x 10“ 18 m
is b x 10 w m/s. The value of b is .

RECORD THE ANSWER ON THE ANSWER SHEET

18

29. The de Broglie wavelength, when expressed in scientific notation, of an
electron that has been accelerated from rest through a potential
difference of 5.0 x 10 1 V, is i x 10 w m. The value of b is .

RECORD THE ANSWER ON THE ANSWER SHEET

30. If the uncertainty in the speed of a 400 kg space probe is 0.5 m/s,
the theoretical minimum uncertainty in its position is in the order
of 10“^ m. The value of the exponent b is .

RECORD THE ANSWER ON THE ANSWER SHEET

19

ITEM INFORMATION SHEET

Item

Key

Multiple Choice
Difficulty

Source

1

6.4

0.768

June '87 #8

6.5

2

1.5

0.769

June '88 #5

3

5.78

0.697

June '87 #11

5.79

5.87

5.88

4

4.2

0.854

June '88 #4

5

1.85

0.517

Jan’ 86 #2

6

33.

0.457

Jan’ 84 #5

7

2.45

0.759

June '88 #10

2.46

8

90.

90.0

90.00

0.771

Jan' 86 #10

9

18.1

0.655

Jan' 89 #14

18.2

10

15.

0.834

Jan' 88 #21

11

12.

0.796

June '87 #16

12

0.75

0.960

Jan' 86 #19

13

1.45

0.505

Jan' 84 #17

14

2.2

0.929

Jan' 88 #26

15

2.5

0.591

Jan' 84 #23

- 20

Item

Key

Multiple Choice
Difficulty

Source

16

0.28

0.666

June’ 87 #28

17

16.

0.702

June' 88 #34

18

0.60

0.879

Jan' 84 #35

19

3.1

0.617

Jan’ 84 #32

3.13

3.14

3.15

20

7.71*

0.844

Jan’ 86 #38

21

20.

0.353

Jan' 84 #46

22

1.89

0.646

Jan' 86 #45

1.90

23

4.56

0.703

Jan' 86 #46

4.57

4.58

4.59

24

2.35*

0.656

Jan' 84 #37

25

2.9

0.733

Jan’ 88 #48

26

1.8

0.644

June' 87 #49

27

27.

0.830

Jan' 88 #50

28

5.9

0.791

June '87 #51

29

1.7

0.705

Jan' 88 #54

30

37.

0.645

Jan’ 89 #49

lotes:

Questions 11, 27, and 30 will be marked correct with trailing zeros included,
even through the answer is a whole number.

Questions 18 and 22 will be marked incorrect if the trailing zero (depending
on method used in 22) is not included.

For question 21, an integral answer is required to demonstrate understanding
of the quantum nature of electrical charge.

*For explanation of these single answers refer to page 27 for question 20 and
to page 30 for question 24.

21

SOLUTIONS TO QUESTIONS HAVING MULTIPLE CORRECT ANSWERS

Solution to question 1

n R * sin 0 R = n ,* sin 0^

0 R = sin" 1 (sin 32°/1.5)

0 R = 20.69°
sin 0 R = d/thickness

d = (17 mm) (tan 20.69°) = 6.42 mm
d = 6.4 mm

Note: If the angle is rounded to two significant digits (0 R = 21°), an

answer d = 6.5 mm is obtained. Although acceptable, this answer
is less desirable as it involves a calculation based on a prematurely
rounded value.

Bubble 6.4 or 6.5

- 22

Solution to question 3

Method 1: diffraction grating

X = (d*sin 0 ) /n

= sin 10 . 0°/ ( 3 . 00 x 10 5 /m)
X = 5.79 x 10” 7 m

Note: If d is calculated first by taking the reciprocal of 3.00 x 10 5 lines/m

and rounding to 3 significant figures, an answer of X = 5.78 x 10 -7 m
is obtained using the same method as above. Although acceptable, this
answer is less desirable as it involves a calculation based on a
prematurely rounded value.

Bubble 5.78 or 5.79

Method 2: double slit

x = t # tan 0

= (5.00 m) (tan 10.0°)

= 0.8816 m
x = 0.882 m
X = dx/nZ

= 0.882 m/[(3.00 x 10 5 /m)(5.00 m)]
X = 5.88 x 10 -7 m

Note: If d is calculated first and rounded to 3 significant figures

(d = 3.33 x 10 -6 m) as in method 1, and answer of X = 5.87 x 10 -7 m
is obtained. Although acceptable, this answer is less desirable for
the same reason as above.

Bubble 5.87 or 5.88

Both methods are equally valid.

- 23

Solution to question 7

^air “ \jlass* n glass

= (8.00 x 10" 7 m) ( 1 . 53 )

\ air = 1.224 x 10 -6 m
f = c/X air

= (3.00 x 10 8 m/s ) / ( 1 . 224 x 10“ 6 m)
f = 2.451 x 10 14 Hz
f = 2.45 x 10 14 Hz

Note: If the wavelength is rounded to three significant digits,

\ a1r = 1.22 x 10“ 6 m, then an answer of f - 2.46 x 10 14 Hz is
obtained. Although acceptable, this answer is less desirable
as it incorporates a prematurely rounded intermediate value.

Bubble 2.45 or 2.46

24

Solution to question 9

Method 1:

= 10.2 N < 4 2 / 3 . GO 2 )
F 2 = 18.1 N

Bubble 18.1

Method 2:

g A g B = FR 2 /k = 1.1346 x 10~ 9 C 2
F 2 = k(4g A X4g B )/fl 2

= (8.99 x 10 9 N*m 2 /C 2 )(16)(1.1346 x 10 9 C 2 )/(3.00 m) 2
F 2 — 18.1 N

Note: Students who make an assumption regarding the relationship between
g A and g B (e.g. g A = g B = 3.37 x 1Q~ 5 C) , and prematurely round the
value to three significant figures, will obtain an answer of
F = 18.2 N. This procedure is barely acceptable.

Bubble 18.1 or 18.2

Both methods are equally valid for an answer of 18.1 N; however, the answer
18.2 N is barely acceptable.

- 25 -

Solution to question 19

distance by reflection d R = 2x/(4.00 km) 2 + (2.00 km) 2

d R = 8.944 km

direct distance d D = 8.00 km

path difference = d R - d D = 0.944 km
time delay = path difference/c

h 0.944 km/ (3.00 x 10 5 km/s)
= 3.147 x 10 -6 s
time delay = 3.1 x 10~ 6 s

Note: Because of the subtraction of distances to obtain the path difference,

the number of significant figures in the final answer is two.

Students who are unaware of the rule of precision that governs
addition and subtraction operations may use the input data to
determine the number of significant figures allowed in the final
answer. In this case, depending on how partial answers are rounded,
students may obtain answers of (3.13 or 3.15) x 10 -6 s. Although less
desirable, these answers will be accepted.

Bubble 3.1

3.13,

or

3.15

Solution to question 20

Method 1:

m = ItA/(Fv)

_ (0.901 C/s ) ( 15 . 3 min) ( 60 s/min) (26 . 98 g/ mol)

(9.65 x 10 4 C/mol)

m = 7.71 x 10~ 2 g

Method 2:

Al 3 + + 3e~ -> A1
m Al = n Ar M Al
n Al = n e ^
n e = It/F

_ (0.901 C/s ) ( 15 . 3 min) ( 60 s/min)

(9.65 x !0 4 C/mol)

n e = 8.571 x 10“ 3 mol
n A1 = (8.571 x 10“ 3 mol ) /3
n A1 = 2.857 mol

m A1 = (2.857 x 10" 3 mol) (26.98 g/mol)
m A1 = 7.71 x 10 -2 g

Note: Premature internal rounding of n A1 = 2.86 x 10“ 3 mol will produce

an answer of m A1 = 7.72 x 10" 2 g. This answer is not acceptable as
it is obtained from an inappropriately rounded intermediate value.

Bubble 7.71

Solution to question 22

Method 1 :

E n = E 1 /n 2

E 3 = -13.6 eV/9 = -1.511 eV
E 2 = -13.6 eV/4 = -3.400 eV
photon energy E = -A E = -(£ f - E, )

= -(-3.400 eV + 1.511 eV)
E = 1.89 eV

Note: Solutions may be done in J rather than eV with the answer converted
to eV at the end. Such solutions also have answer of 1.89 eV.

Bubble 1.89

Method 2 :

1/X = R h [ ( l/n f 2 ) - (1/n, 2 )]

= R h [(1/4) - (1/9)]

= 5K h /36

\ = 36/(5 x 1.10 x 10 7 /m)

X = 6.545 x 10 -7 m = 6.55 x 10 -7 m
E - hc/\

= (6.63 x 10 -34 J # s ) (3 . 00 x 10 8 m/s)/(6.55 x 10“ 7 m)
E = (3.04 x 10" 19 J) / ( 1 . 60 x 10" 19 J/eV)

E - 1.90 V

Note: The trailing zero MUST be included for an answer correct
to 3 significant figures.

Bubble 1.90

Both methods are equally valid.

- 28

Solution to question 23

Method 1: following from #22

f - E/h

= (1.89 e V ) (1.6 x 10' 19 J/eV)/(6.63 x 10~ 34 J*s)
f = 4.56 x 10 14 Hz

Note: Other variations of this solution include using E = 1.90 eV

and/or h = 4.14 x 10 -15 eV*s. These variations produce answers
of 4.59 x 10 14 Hz and 4.57 x 10 14 Hz.

Bubble 4.56, 4.57, or 4.59

Method 2: using the Rydberg equation

f = c/\ and l/\ = * h [(1/4) - (1/9)]
f = (3.00 x 10 8 m/s)/(6. 55 x 10" 7 m)
f = 4.58 x 10 14 Hz

Bubble 4.58

Both methods are equally valid.

29

Solution to question 24

max “ hf W

= (6.63 x 1CT 34 J/Hz ) ( 4 . 52 x 10 15 Hz) - (4.03 eV)(1.60 x 10~ 19 J/eV)
= (2.997 x 10 -18 J) - (6.448 x 10 -19 J)
fk max = 2.35 x lO" 18 J

Note: Solutions done using h = 4.14 x 10 15 eV/Hz produce the same answer
as above.

Solutions which round intermediate answers to hf = 3.00 x 10 -18 J
and/or W = 6.45 x 10 -19 J produce an answer of max = 2.36 x 10 -18 J.
This answer is not acceptable as it is obtained from an inappropriately
rounded intermediate value.

Bubble 2.35

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