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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 


Digitized by the Internet Archive 
in 2017 with funding from 
University of Alberta Libraries 


https://archive.org/details/physics30progra1989albe_0 


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. 



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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 


30 


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3 3286 09255985 1