# Full text of "Physics 30 program, machine-scorable open-ended questions : Units 1-5, Complete course"

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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. 1 5 5 <D •.©© < 2 ) ©.© © © ©.© © © ( 2 ).© © ©©.•• ©<D.©<D < 2 ) < 2 >.® < 2 ) (D<D.(D<D 2 7 <5> ©.© © ©©.©© © •.© © © ©.© © © ©.© © © ©.© © ® ©.©© <Z><2>.#<2> © ®.® ® ® ®.® ® 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. 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