www.elsolucionario.org Chapter 1: Introduction to Physics Answers to Even-Numbered Conceptual Questions The quantity T + d does not make sense physically, because it adds together variables that have different physical dimensions The quantity d/T does make sense, however; it could represent the distance d traveled by an object in the time T (a) 107 s; (b) 10,000 s; (c) s; (d) 1017 s; (e) 108 s to 109 s Solutions to Problems and Conceptual Exercises Picture the Problem: This is simply a units conversion problem Strategy: Multiply the given number by conversion factors to obtain the desired units Solution: (a) Convert the units: $114,000,000 × gigadollars = 0.114 gigadollars × 109 dollars (b) Convert the units again: $114,000,000 × teradollars = 1.14 × 10−4 teradollars × 1012 dollars Insight: The inside back cover of the textbook has a helpful chart of the metric prefixes Picture the Problem: This is simply a units conversion problem Strategy: Multiply the given number by conversion factors to obtain the desired units Solution: (a) Convert the units: 70 μ m × 1.0 × 10−6 m = 7.0 × 10−5 m μm (b) Convert the units again: 70 μ m × 1.0 × 10−6 m km × = 7.0 × 10−8 km 1000 m μm Insight: The inside back cover of the textbook has a helpful chart of the metric prefixes Picture the Problem: This is simply a units conversion problem Strategy: Multiply the given number by conversion factors to obtain the desired units Solution: Convert the units: 0.3 Gm × 109 m × = × 108 m/s s Gm Insight: The inside back cover of the textbook has a helpful chart of the metric prefixes Picture the Problem: This is simply a units conversion problem Strategy: Multiply the given number by conversion factors to obtain the desired units Solution: Convert the units: 70.72 teracalculation × 1012 calculations × 10−6 s × × μs s teracalculation = 7.072 × 107 calculations/μ s Insight: The inside back cover of the textbook has a helpful chart of the metric prefixes © Copyright 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1–1 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 1: Introduction to Physics Picture the Problem: This is a dimensional analysis question Strategy: Manipulate the dimensions in the same manner as algebraic expressions x = vt Solution: (a) Substitute dimensions the variables: for ⎛m⎞ m = ⎜ ⎟ ( s ) = m ∴ The equation is dimensionally consistent ⎝s ⎠ x = 12 at 2 (b) Substitute dimensions for the variables: ⎛m⎞ m = 12 ⎜ ⎟ ( s ) = m ∴ dimensionally consistent ⎝s ⎠ (c) Substitute dimensions for the variables: t= 2x m ⇒ s= = s = s ∴ dimensionally consistent a ms Insight: The number does not contribute any dimensions to the problem Picture the Problem: This is a dimensional analysis question Strategy: Manipulate the dimensions in the same manner as algebraic expressions Solution: (a) Substitute dimensions for the variables: (b) Substitute dimensions for the variables: (c) Substitute dimensions for the variables: ⎛m⎞ vt = ⎜ ⎟ ( s ) = m Yes ⎝s⎠ ⎛m⎞ at = 12 ⎜ ⎟ ( s ) = m Yes ⎝s ⎠ m ⎛m⎞ 2at = ⎜ ⎟ ( s ) = No s ⎝s ⎠ v2 ( m s ) = =m a m s2 (d) Substitute dimensions for the variables: Yes Insight: When squaring the velocity you must remember to square the dimensions of both the numerator (meters) and the denominator (seconds) Picture the Problem: This is a dimensional analysis question Strategy: Manipulate the dimensions in the same manner as algebraic expressions Solution: (a) Substitute dimensions for the variables: ⎛m⎞ at = 12 ⎜ ⎟ ( s ) = m No ⎝s ⎠ m ⎛m⎞ Yes at = ⎜ ⎟ ( s ) = s ⎝s ⎠ 2 (b) Substitute dimensions for the variables: (c) Substitute dimensions for the variables: 2x 2m = =s m s2 a (d) Substitute dimensions for the variables: m ⎛m⎞ 2ax = ⎜ ⎟ ( m ) = s ⎝s ⎠ No Yes Insight: When taking the square root of dimensions you need not worry about the positive and negative roots; only the positive root is physical Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1–2 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 1: Introduction to Physics Picture the Problem: This is a dimensional analysis question Strategy: Manipulate the dimensions in the same manner as algebraic expressions v = 2ax p Solution: Substitute dimensions for the variables: p ⎛m⎞ ⎛m⎞ ⎜ ⎟ = ⎜ ⎟(m) ⎝ s ⎠ ⎝s ⎠ m = m p +1 therefore p = Insight: The number does not contribute any dimensions to the problem Picture the Problem: This is a dimensional analysis question Strategy: Manipulate the dimensions in the same manner as algebraic expressions a = 2x t p Solution: Substitute dimensions for the variables: [L] = [L][T] p [T] [T]−2 = [T] p therefore p = −2 Insight: The number does not contribute any dimensions to the problem 10 Picture the Problem: This is a dimensional analysis question Strategy: Manipulate the dimensions in the same manner as algebraic expressions v = v0 + at Solution: Substitute dimensions for the variables on both sides of the equation: [L] [L] [L] = + [T] [T] [T] [T]2 [L] [L] = It is dimensionally consistent! [T] [T] Insight: Two numbers must have the same dimensions in order to be added or subtracted 11 Picture the Problem: This is a dimensional analysis question Strategy: Manipulate the dimensions in the same manner as algebraic expressions Solution: Substitute dimensions for the variables, where [M] represents the dimension of mass: F = ma = [M] [L] [T]2 Insight: This unit (kg m/s2) will later be given the name “Newton.” 12 Picture the Problem: This is a dimensional analysis question Strategy: Solve the formula for k and substitute the units Solution: Solve for k: T = 2π Substitute the dimensions, where [M] represents the dimension of mass: k= m m 4π m square both sides: T = 4π or k = k k T2 [M] [T]2 Insight: This unit will later be renamed “Newton/m.” The 4π2 does not contribute any dimensions Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1–3 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 1: Introduction to Physics 13 Picture the Problem: This is a significant figures question Strategy: Follow the given rules regarding the calculation and display of significant figures Solution: (a) Round to the 3rd digit: 3.14159265358979 ⇒ 3.14 (b) Round to the 5th digit: 3.14159265358979 ⇒ 3.1416 (c) Round to the 7th digit: 3.14159265358979 ⇒ 3.141593 Insight: It is important not to round numbers off too early when solving a problem because excessive rounding can cause your answer to significantly differ from the true answer 14 Picture the Problem: This is a significant figures question Strategy: Follow the given rules regarding the calculation and display of significant figures Solution: Round to the 3rd digit: 2.9979 × 108 m/s ⇒ 3.00 × 108 m/s Insight: It is important not to round numbers off too early when solving a problem because excessive rounding can cause your answer to significantly differ from the true answer 15 Picture the Problem: The parking lot is a rectangle 144.3 m Strategy: The perimeter of the parking lot is the sum of the lengths of its four sides Apply the rule for addition of numbers: the number of decimal places after addition equals the smallest number of decimal places in any of the individual terms 47.66 m 47.66 m 144.3 m Solution: Add the numbers: 144.3 + 47.66 + 144.3 + 47.66 m = 383.92 m Round to the smallest number of decimal places in any of the individual terms: 383.92 m ⇒ 383.9 m Insight: Even if you changed the problem to ( × 144.3 m ) + ( × 47.66 m ) you’d still have to report 383.9 m as the answer; the is considered an exact number so it’s the 144.3 m that limits the number of significant digits 16 Picture the Problem: The weights of the fish are added Strategy: Apply the rule for addition of numbers, which states that the number of decimal places after addition equals the smallest number of decimal places in any of the individual terms Solution: Add the numbers: 2.35 + 12.1 + 12.13 lb = 26.58 lb Round to the smallest number of decimal places in any of the individual terms: 26.58 lb ⇒ 26.6 lb Insight: The 12.1 lb rock cod is the limiting figure in this case; it is only measured to within an accuracy of 0.1 lb 17 Picture the Problem: This is a significant figures question Strategy: Follow the given rules regarding the calculation and display of significant figures Solution: (a) The leading zeros are not significant: 0.0000 has significant figures (b) The middle zeros are significant: 3.0 1×105 has significant figures Insight: Zeros are the hardest part of determining significant figures Scientific notation can remove the ambiguity of whether a zero is significant because any trailing zero to the right of the decimal point is significant Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1–4 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 1: Introduction to Physics 18 Picture the Problem: This is a significant figures question Strategy: Apply the rule for multiplication of numbers, which states that the number of significant figures after multiplication equals the number of significant figures in the least accurately known quantity Solution: (a) Calculate the area and round to four significant figures: A = π r = π (14.37 m ) = 648.729144 m ⇒ 648.7 m 2 (b) Calculate the area and round to two significant figures: A = π r = π ( 3.8 m ) = 45.3645979 m ⇒ 45 m 2 Insight: The number π is considered exact so it will never limit the number of significant digits you report in an answer 19 Picture the Problem: This is a units conversion problem Strategy: Multiply the known quantity by appropriate conversion factors to change the units Solution: (a) Convert to feet per second: m ⎞⎛ 3.28 ft ⎞ ft ⎛ ⎜ 23 ⎟⎜ ⎟ = 75 s ⎠⎝ m ⎠ s ⎝ (b) Convert to miles per hour: mi ⎛ m ⎞⎛ mi ⎞⎛ 3600 s ⎞ ⎜ 23 ⎟⎜ ⎟⎜ ⎟ = 51 h ⎝ s ⎠⎝ 1609 m ⎠⎝ hr ⎠ Insight: Conversion factors are conceptually equal to one, even though numerically they often equal something other than one They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion 20 Picture the Problem: This is a units conversion problem Strategy: Multiply the known quantity by appropriate conversion factors to change the units Solution: (a) Find the length in feet: ( 631 m ) ⎛⎜ 3.28 ft ⎞ ⎟ = 2069 ft ⎝ 1m ⎠ Find the width in feet: ( 707 yd ) ⎜ Find the volume in cubic feet: V = LWH = ( 2069 ft )( 2121 ft )(110 ft ) = 4.83 ×108 ft (b) Convert to cubic meters: ( 4.83 ×10 ⎛ ft ⎞ ⎟ = 2121 ft ⎝ yd ⎠ ⎛ 1m ⎞ ft ) ⎜ ⎟ = 1.37 × 10 m ⎝ 3.28 ft ⎠ Insight: Conversion factors are conceptually equal to one, even though numerically they often equal something other than one They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion 21 Picture the Problem: This is a units conversion problem Strategy: Multiply the known quantity by appropriate conversion factors to change the units Solution: Find the length in feet: ( 2.5 cubit ) ⎛⎜ 17.7 in ⎞⎛ ft ⎞ ⎟⎜ ⎟ = 3.68 ft ⎝ cubit ⎠⎝ 12 in ⎠ Find the width and height in feet: (1.5 cubit ) ⎛⎜ Find the volume in cubic feet: V = LWH = ( 3.68 ft )( 2.21 ft )( 2.21 ft ) = 18 ft 17.7 in ⎞⎛ ft ⎞ ⎟⎜ ⎟ = 2.21 ft ⎝ cubit ⎠⎝ 12 in ⎠ Insight: Conversion factors are conceptually equal to one, even though numerically they often equal something other than one They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1–5 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 1: Introduction to Physics 22 Picture the Problem: This is a units conversion problem Strategy: Convert the frequency of cesium-133 given on page to units of microseconds per megacycle, then multiply by the number of megacycles to find the elapsed time Solution: Convert to micro seconds per megacycle and multiply by 1.5 megacycles: ⎛ ⎞ ⎛ 1× 106 cycles ⎞ ⎛ μ s ⎞ 1s μs ⎟×⎜ ⎜ ⎟⎜ ⎟ = 108.7827757 −6 Mcycle ⎝ 9,192, 631, 770 cycles ⎠ ⎝ Mcycle ⎠ ⎝ 1× 10 s ⎠ μs 108.7827757 × 1.5 Mcycle = 160 μ s = 1.6 × 10− s Mcycle Insight: Only two significant figures remain in the answer because of the 1.5 Mcycle figure given in the problem statement The metric prefix conversions are considered exact and have an unlimited number of significant figures, but most other conversion factors have a limited number of significant figures 23 Picture the Problem: This is a units conversion problem Strategy: Multiply the known quantity by appropriate conversion factors to change the units ( 3212 ft ) ⎛⎜ mi ⎞⎛ 1.609 km ⎞ ⎟⎜ ⎟ = 0.9788 km 5280 ft ⎠⎝ mi ⎠ ⎝ Solution: Convert feet to kilometers: Insight: Conversion factors are conceptually equal to one, even though numerically they often equal something other than one They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion 24 Picture the Problem: This is a units conversion problem Strategy: Multiply the known quantity by appropriate conversion factors to change the units ⎛ msg ⎞⎛ 3600 s ⎞⎛ 24 h ⎞⎛ d ⎞ msg ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ = ×10 s h d wk wk ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ Solution: Convert seconds to weeks: Insight: In this problem there is only one significant figure associated with the phrase, “7 seconds.” 25 Picture the Problem: This is a units conversion problem Strategy: Multiply the known quantity by appropriate conversion factors to change the units (108 ft ) ⎛⎜ 1m ⎞ ⎟ = 32.9 m ⎝ 3.28 ft ⎠ Solution: Convert feet to meters: Insight: Conversion factors are conceptually equal to one, even though numerically they often equal something other than one They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion 26 Picture the Problem: This is a units conversion problem Strategy: Multiply the known quantity by appropriate conversion factors to change the units 0.20 g ⎞ ⎛ kg ⎞⎛ 2.21 lb ⎞ ⎟⎜ ⎟ = 0.23 lb ⎟⎜ ⎝ ct ⎠ ⎝ 1000 g ⎠⎝ kg ⎠ Insight: Conversion factors are conceptually equal to one, even though numerically they often equal something other than one They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion Solution: Convert carats to pounds: ( 530.2 ct ) ⎛⎜ Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1–6 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 1: Introduction to Physics 27 Picture the Problem: This is a units conversion problem Strategy: Multiply the known quantity by appropriate conversion factors to change the units Solution: (a) The speed must be greater than 55 km/h because mi/h = 1.609 km/h mi ⎞⎛ 1.609 km ⎞ km ⎛ ⎜ 55 ⎟⎜ ⎟ = 88 h ⎠⎝ mi h ⎝ ⎠ (b) Convert the miles to kilometers: Insight: Conversion factors are conceptually equal to one, even though numerically they often equal something other than one They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion 28 Picture the Problem: This is a units conversion problem Strategy: Multiply the known quantity by appropriate conversion factors to change the units ⎛ m ⎞⎛ mi ⎞⎛ 3600 s ⎞ mi ⎜ 3.00 ×10 ⎟⎜ ⎟⎜ ⎟ = 6.71×10 s ⎠⎝ 1609 m ⎠⎝ h ⎠ h ⎝ Solution: Convert m/s to miles per hour: Insight: Conversion factors are conceptually equal to one, even though numerically they often equal something other than one They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion 29 Picture the Problem: This is a units conversion problem Strategy: Multiply the known quantity by appropriate conversion factors to change the units m ⎞⎛ 3.28 ft ⎞ ft ⎛ ⎜ 98.1 ⎟⎜ ⎟ = 322 s ⎠⎝ m ⎠ s ⎝ Solution: Convert to ft per second per second: Insight: Conversion factors are conceptually equal to one, even though numerically they often equal something other than one They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion 30 Picture the Problem: This is a units conversion problem Strategy: Multiply the known quantity by appropriate conversion factors to change the units In this problem, one “jiffy” corresponds to the time in seconds that it takes light to travel one centimeter 1s s jiffy ⎛ ⎞⎛ m ⎞ −11 =1 ⎜ ⎟⎜ ⎟ = 3.3357 × 10 × 2.9979 10 m 100 cm cm cm ⎝ ⎠⎝ ⎠ Solution: (a): Determine the magnitude of a jiffy: jiffy = 3.3357 × 10−11 s (1 minute ) ⎛⎜ 60 s ⎞⎛ jiffy ⎞ 12 ⎟⎜ ⎟ = 1.7987 × 10 jiffy −11 ⎝ ⎠⎝ 3.3357 ×10 s ⎠ (b) Convert minutes to jiffys: Insight: Conversion factors are conceptually equal to one, even though numerically they often equal something other than one They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion 31 Picture the Problem: This is a units conversion problem Strategy: Multiply the known quantity by appropriate conversion factors to change the units Solution: (a) Convert cubic feet to mutchkins: L ⎞⎛ mutchkin ⎞ (1 ft ) ⎜⎝⎛ 28.3 ⎟⎜ ⎟ = 67 mutchkin ft ⎠⎝ 0.42 L ⎠ (b) Convert noggins to gallons: (1 noggin ) ⎜ 3 ⎛ 0.28 mutchkin ⎞ ⎛ 0.42 L ⎞ ⎛ gal ⎞ ⎟⎜ ⎟⎜ ⎟ = 0.031 gal noggin ⎝ ⎠ ⎝ mutchkin ⎠ ⎝ 3.785 L ⎠ Insight: To convert noggins to gallons, multiply the number of noggins by 0.031 gal/noggin Conversely, there are noggin/0.031 gal = 32 noggins/gallon That means a noggin is about half a cup Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1–7 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 1: Introduction to Physics 32 Picture the Problem: The volume of the oil is spread out into a slick that is one molecule thick Strategy: The volume of the slick equals its area times its thickness Use this fact to find the area Solution: Calculate the area for the known volume and thickness: A= V 1.0 m3 ⎛ μ m ⎞ = ⎜ ⎟ = 2.0 × 10 m h 0.50 μ m ⎝ 1× 10−6 m ⎠ Insight: Two million square meters is about 772 square miles! 33 Picture the Problem: This is a units conversion problem Strategy: Multiply the known quantity by appropriate conversion factors to change the units Then use a ratio to find the factor change in part (b) Solution: (a) Convert square inches to square meters: ⎛ m2 ⎞ A = ( 8.5 in × 11 in ) ⎜ = 0.060 m 2 ⎟ ⎝ 1550 in ⎠ (b) Calculate a ratio to find the new area: Anew LnewWnew ( 12 Lold )( 12 Wold ) = = = Aold LoldWold LoldWold Anew = Aold Insight: If you learn to use ratios you can often make calculations like these very easily Always put the new quantity in the numerator and the old quantity in the denominator to make the new quantity easier to calculate at the end 34 Picture the Problem: This is a units conversion problem Strategy: Multiply the known quantity by appropriate conversion factors to change the units Solution: Convert m/s to ft/s: m ⎞⎛ 3.28 ft ⎞ ⎛ ⎜ 20.0 ⎟⎜ ⎟ = 65.6 ft/s s ⎠⎝ m ⎠ ⎝ (b) Convert m/s to mi/h: m ⎞⎛ mi ⎞⎛ 3600 s ⎞ ⎛ ⎜ 20.0 ⎟⎜ ⎟⎜ ⎟ = 44.7 mi/h s ⎠⎝ 1609 m ⎠⎝ h ⎠ ⎝ Insight: Conversion factors are conceptually equal to one, even though numerically they often equal something other than one They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion 35 Picture the Problem: This is a units conversion problem Strategy: Multiply the known quantity by appropriate conversion factors to change the units m ⎞⎛ 3.28 ft ⎞ ⎛ ⎜ 9.81 ⎟⎜ ⎟ = 32.2 ft/s s m ⎝ ⎠⎝ ⎠ Solution: Convert meters to feet: Insight: Conversion factors are conceptually equal to one, even though numerically they often equal something other than one They are often helpful in displaying a number in a convenient, useful, or easy-to-comprehend fashion 36 Picture the Problem: The rows of seats are arranged into roughly a circle Strategy: Estimate that a baseball field is a circle around 300 ft in diameter, with 100 rows of seats around outside of the field, arranged in circles that have perhaps an average diameter of 500 feet The length of each row is then the circumference of the circle, or πd = π(500 ft) Suppose there is a seat every feet Solution: Multiply the quantities to make an estimate: ft ⎞⎛ seat ⎞ ⎛ N = (100 rows ) ⎜ π 500 ⎟⎜ ⎟ = 52, 400 seats ≅ 10 seats row ft ⎝ ⎠⎝ ⎠ Insight: Some college football stadiums can hold as many as 100,000 spectators, but most less than that Still, for an order of magnitude we round to the nearest factor of ten, in this case it’s 105 Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1–8 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 1: Introduction to Physics 37 Picture the Problem: Suppose all milk is purchased by the gallon in plastic containers Strategy: There are about 300 million people in the United States, and if each of these were to drink a half gallon of milk every week, that’s about 25 gallons per person per year Each plastic container is estimated to weigh about an ounce Solution: (a) Multiply the quantities to make an estimate: ( 300 ×10 (b) Multiply the gallons by the weight of the plastic: (1×10 10 people ) ( 25 gal/y/person ) = 7.5 × 109 gal/y ≅ 1010 gal/y ⎛ lb ⎞ gal/y ) (1 oz/gal ) ⎜ ⎟ = 6.25 × 10 lb/y ≅ 10 lb/y 16 oz ⎝ ⎠ Insight: About half a billion pounds of plastic! Concerted recycling can prevent much of these containers from clogging up our landfills 38 Picture the Problem: The Earth is roughly a sphere rotating about its axis Strategy: Use the fact the Earth spins once about its axis every 24 hours to find the estimated quantities d 3000 mi = = 1000 mi/h ≅ 103 mi/h t 3h Solution: (a) Divide distance by time: v= (b) Multiply speed by 24 hours: circumference = vt = ( 3000 mi/h )( 24 h ) = 24, 000 mi ≅ 104 mi (c) Circumference equals 2πr: r= circumference 24, 000 mi = = 3800 mi = 103 mi 2π 2π Insight: These estimates are “in the ballpark.” The speed of a point on the equator is 1038 mi/h, the circumference of the equator is 24,900 mi, and the equatorial radius of the Earth is 3963 mi 39 Picture the Problem: The lottery winnings are represented either by quarters or paper dollars Strategy: There are about quarters and about 30 dollar bills per ounce Solution: (a) Multiply by conversion factors: ( ×12 ×10 (b) Repeat for the dollar bills: (12 ×10 6 ⎛ oz ⎞ ⎛ lb ⎞ quarters ) ⎜ ⎟⎜ ⎟ = 600, 000 lb ≅ 10 lb quarters 16 oz ⎝ ⎠ ⎝ ⎠ ⎛ oz ⎞⎛ lb ⎞ dollars ) ⎜ ⎟⎜ ⎟ = 25, 000 lb ≅ 10 lb 30 dollars 16 oz ⎝ ⎠⎝ ⎠ Insight: Better go with large denominations or perhaps a single check when you collect your lottery winnings! Even the dollar bills weigh over ten tons! Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1–9 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 32: Nuclear Physics and Nuclear Radiation + 235 = 133 + A + (1) Balance the atomic mass: A = + 235 − 133 − = 98 98 41 Nb Determine the missing isotope: The missing daughter product is Calculate the initial mass: mi = 1.008665 u + 235.043925 u = 236.052590 u Calculate the final mass: mf = 132.915237 u + 97.910331 u + (1.008665 u ) Calculate the mass difference: = 235.868893 u Δm = 235.868893 u − 236.052590 u = − 0.183697 u Convert the difference into the energy released: ⎛ 931.5 MeV/c ⎞ E = Δm c = 0.183697 u ⎜ ⎟ c = 171.1 MeV 1u ⎝ ⎠ Insight: The energy released by the fission is carried away primarily by the neutrons as kinetic energy 235 88 136 49 Picture the Problem: For the fission reaction n + 92 U → 38Sr + 54 Xe + neutrons , we are asked to calculate the number of neutrons produced and the energy released Strategy: Balance the atomic number and atomic mass on both sides of the reaction equation to identify the missing product Calculate the difference in mass between the parent products 01 n + 235 92 U and the daughter 136 products 88 38 Sr + 54 Xe + neutrons Convert this mass difference to energy to determine the amount of energy released in the reaction + 235 = 88 + 136 + n (1) Solution: Balance the atomic mass to determine the number of neutrons: n = + 235 − 88 − 136 = 12 n+ 235 92 136 U → 88 38 Sr + 54 Xe + 12 n Write out the complete reaction: Calculate the initial mass: mi = 1.008665 u + 235.043925 u = 236.052590 u Calculate the final mass: mf = 87.905625 u + 135.90722 u + 12(1.008665 u) = 235.91683 u Calculate the mass difference: Δm = 235.91683 u − 236.052590 u = − 0.13576 u Convert the difference into energy released: ⎛ 931.5 MeV/c ⎞ E = Δm c = 0.13576 u ⎜ ⎟ c = 126.5 MeV 1u ⎝ ⎠ Insight: This released energy is primarily distributed as kinetic energy of the neutrons 50 Picture the Problem: We are asked to calculate the amount of gasoline that would have to be burned in order to release the same energy as that released when 1.0-lb of 235 92 U undergoes nuclear fission Strategy: Calculate the amount of energy released when 1-lb of 235 92 U undergoes fission by multiplying the energy released in each fission event by the number of nuclei To calculate the number of nuclei, divide the total mass by the mass of one nuclei, 235.0 u Convert both the pound and u to kilograms Divide the released energy by the energy released by one gallon of gas in order to find the equivalent number of gallons of gas Solution: Calculate the energy released by the uranium: E= Divide by the energy released in one gallon of gas: N= mU ⎞ 1.0 lb ⎛ kg ⎞ ⎛ 1u EU = ⎟ (173 × 10 eV ) ⎜ ⎟⎜ −27 MU 235.0 u ⎝ 2.2 lb ⎠ ⎝ 1.66 × 10 kg ⎠ = 2.015 × 1032 eV (1.60 × 10−19 J/eV ) = 3.2 × 1013 J E Egasoline = 3.2 × 1013 J = 1.6 × 105 gallons of gas 2.0 × 108 J/gal Insight: The energy released by each nuclear fission event is on the order of a million times greater than the energy released by each chemical reaction when gasoline is burned Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32 – 17 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 32: Nuclear Physics and Nuclear Radiation 51 Picture the Problem: Each fission event of 235 92 U nuclei releases 173 MeV of energy Strategy: Divide the total energy by the energy per fission to calculate the number of nuclei needed Multiply the mass of one nucleus by the number of nuclei to calculate the total mass E 8.4 × 1019 J ⎛ MeV ⎞ 30 = ⎜ ⎟ = 3.035 × 10 EU 173 MeV ⎝ 1.6 × 10−13 J ⎠ Solution: Calculate the number of 235 92 U nuclei: N= Multiply by the mass of one nuclei: m = Nmu = 3.035 × 1030 ( 235.0 u ) (1.66 × 10 −27 kg/u ) = 1.2 × 106 kg Insight: This mass would occupy a volume of only 62 m3, the size of a small classroom However, it would not be wise to store it all in one room because of the danger of reaching the critical mass! 52 Picture the Problem: Each fission event of 235 92 U nuclei releases 173 MeV of energy Strategy: Divide the power output by the energy per reaction to calculate the required reaction rate R= Solution: Calculate the reaction rate: P 150 × 106 W ⎛ eV ⎞ 18 −1 = ⎜ ⎟ = 5.4 × 10 s EU 173 × 106 eV ⎝ 1.60 × 10−19 J ⎠ Insight: This rate corresponds to the fission of 2.1 mg of 235 92 U per second 53 Picture the Problem: Energy is released by the nuclear reaction: 12 H + 12 H → 13 H + 11H Strategy: Calculate the mass difference between the initial two deuterium atoms and the resulting tritium and hydrogen, using the masses in Appendix F Multiply the mass difference by c2 and convert the resulting energy to MeV Solution: Calculate the mass of 12 H + 12 H: mi = 2(2.014102 u) = 4.028204 u Calculate the mass of 13 H + 11H: mf = 3.016049 u + 1.007825 u = 4.023874 u Find the difference in mass: Δm = 4.023874 − 4.028204 u = − 0.004330 u Convert the difference into the energy released: ⎛ 931.5 MeV/c ⎞ E = Δm c = 0.004330 u ⎜ ⎟ c = 4.033 MeV 1u ⎝ ⎠ Insight: The deuterium is stable, but the tritium produced in this reaction has a half-life of 12.33 y 54 Picture the Problem: Energy is released by the nuclear reaction: 11 H + 10 n → 12 H Strategy: Calculate the mass difference between the initial two atoms and the resulting deuterium, using the masses in Appendix F Multiply the mass difference by c2 and convert the resulting energy to MeV Solution: Calculate the mass of 11 H + 10 n: Look up the mass of H: mi = 1.007825 u + 1.008665 u = 2.016490 u mf = 2.014102 u Find the difference in mass: Δm = 2.014102 u − 2.016490 u = − 0.002388 u Convert the difference into the energy released: ⎛ 931.5 MeV/c E = Δm c = 0.002388 u ⎜ 1u ⎝ ⎞ ⎟ c = 2.224 MeV ⎠ Insight: Once the deuterium is created it can be fused with another deuterium atom to release more energy in the helium fusion cycle Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32 – 18 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 32: Nuclear Physics and Nuclear Radiation 55 Picture the Problem: Energy is released by the nuclear reaction: 11 H + 12 H → 32 He + γ Strategy: Calculate the mass difference between the initial two atoms and the resulting helium-3, using the masses in Appendix F Multiply the mass difference by c2 and convert the resulting energy to MeV Solution: Calculate the mass of 11 H + 12 H: Look up the mass of mi = 1.007825 u + 2.014102 u = 3.021927 u mf = 3.016029 u He: Find the difference in mass: Δm = 3.016029 u − 3.021927 u = −0.005898 u Convert the difference into the energy released: ⎛ 931.5 MeV/c E = Δm c = 0.005898 u ⎜ 1u ⎝ ⎞ ⎟ c = 5.494 MeV ⎠ Insight: When helium-3 is produced from the two atoms, a gamma ray must carry away some of the energy in order to conserve both momentum and energy 56 Picture the Problem: We are asked to find the missing daughter product and the energy released for the fusion reaction, 12 H + 13 H → ? + 10 n Strategy: Balance the atomic number and atomic mass on both sides of the reaction equation to identify the missing product Calculate the difference in mass between the parent products (12 H + 13 H) and the daughter products (? + 10 n) Convert this mass difference to energy to determine the amount of energy released in the reaction Divide the power output by the energy per reaction to calculate the required number of reactions per second Solution: (a) Balance the atomic number: 1+1 = Z + Z = 2, which is helium, or He Balance the atomic mass: + = A +1 ⇒ A = Write the fusion reaction: Calculate the initial mass: mi = 2.014102 u + 3.016049 u = 5.030151 u Calculate the final mass: mf = 4.002603 u + 1.008665 u = 5.011268 u Calculate the mass difference: Δm = 5.011268 u − 5.030151 u = − 0.018883 u Convert the difference into the energy released: ⎛ 931.5 MeV/c E = Δm c = 0.018883 u ⎜ 1u ⎝ (b) Divide the power by the energy per reaction: R= H + 31 H → 42 He + 01 n ⎞ ⎟ c = 17.589 MeV ⎠ P 25 × 106 W ⎛ eV ⎞ 18 −1 = ⎜ ⎟ = 8.9 × 10 s −19 E 17.589 × 10 eV ⎝ 1.60 × 10 J ⎠ Insight: At this rate, it would take 19 hours to produce one mole (4 grams) of helium-4 Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32 – 19 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 32: Nuclear Physics and Nuclear Radiation 57 Picture the Problem: The Sun converts a portion of its mass into energy by means of nuclear fusion Strategy: Divide the power output of the Sun by the speed of light squared to calculate the mass per second converted into energy Calculate the total mass converted to energy by multiplying the mass conversion rate by the age of the Sun Finally, divide the converted mass by the original mass of the Sun to find the percentage that was converted to energy Solution: (a) Calculate the rate of mass conversion: 3.90 × 10 26 W Δm PSun = = = 4.33 × 109 kg/s Δt c 3.00 10 m/s × ( ) (b) Multiply the mass conversion rate by the age of the Sun: ΔM = Divide the converted mass by the original mass: ΔM 6.16 × 1026 kg = × 100 = 0.0307% M + ΔM 2.00 × 1030 kg + 6.16 × 1026 kg ⎛ 3.16 × 107 s ⎞ Δm T = ( 4.33 × 109 kg/s )( 4.50 × 109 y ) ⎜ ⎟ Δt 1y ⎝ ⎠ 26 = 6.16 × 10 kg Insight: Over its entire life span, the Sun converts a tiny amount of its mass into energy, even though it converts mass at the astonishing rate of 4.33 million metric tons per second! 58 Picture the Problem: The amount of tissue damage caused by a dose of radiation depends upon the type of particle that is absorbed by the tissue Strategy: Use equation 32-18 to set the dose in rem of the protons equal to the dose in rem of the alpha particles Solve for the dose in rad of the protons Solution: Set the doses in rem equal: ( Dose in rem )α = ( Dose in rem )p ( Dose in rad )α RBEα = ( Dose in rad )p RBEp Solve for ( Dose in rad )p : ( Dose in rad )p = ( Dose in rad )α RBEα 20 = ( 55 rad ) = 110 rad RBEp 10 Insight: The body can be exposed to twice the radiation from protons as from alpha particles because the absorption of proton radiation is half that of the alpha radiation 59 Picture the Problem: The amount of tissue damage caused by a dose of radiation depends upon the type of radiation that is absorbed by the tissue Strategy: Use equation 32-18 to set the dose in rem of the X-rays equal to the dose in rem of the heavy ion particles Solve for the dose in rad of the X-rays The RBEs are listed in Table 32-3 Solution: Set the doses in rem equal: ( Dose in rem )HI = ( Dose in rem )X-ray ( Dose in rad )HI RBEHI = ( Dose in rad )X-ray RBEX-ray Solve for ( Dose in rad )X-ray : ( Dose in rad )X-ray = ( Dose in rad )HI RBEHI RBEHI 20 = 1.0 × 103 rad Insight: The body can be exposed to twenty times the radiation from X-rays as from heavy ions because the absorption of X-ray radiation is one-twentieth that of the heavy ion radiation = 50 rad Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32 – 20 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 32: Nuclear Physics and Nuclear Radiation 60 Picture the Problem: A 78-kg person absorbs the energy from 52 mrem of particles that have an RBE of 15 Strategy: Divide the dose in rem by the RBE in order to convert it into a dose in rad Use equation 32-16 to convert the rad dosage to units of J/kg and then multiply by the person’s mass in order to find the total energy absorbed 52 × 10−3 rem = 3.5 × 10 −3 rad 15 Solution: (a) Convert the dose to rad: Dose in rad = Multiply by the mass and convert to joules: ⎛ 0.01 J/kg ⎞ E = ( Dose in rad ) m = ( 3.5 × 10−3 rad ) ⎜ ⎟ 78 kg = 2.7 mJ ⎝ rad ⎠ (b) Because the energy absorbed is inversely proportional to the RBE as long as the dose in rem remains constant, the energy absorbed will decrease if the RBE is increased Insight: If the RBE were increased to 20, the energy absorbed would be reduced to 2.0 mJ as long as the dose remained 52 mrem The new radiation is more effective in creating biological damage, so a smaller amount of energy (in rad) is needed to produce the same biologically equivalent dose (in rem) 61 Picture the Problem: A 0.17-kg cancerous growth receives a 225-rad dose of radiation Strategy: Multiply the radiation dose by the mass of the tumor in order to calculate the absorbed energy in joules To find the change in temperature, solve equation 16-13 for the temperature change, where C = 4186 J kg ⋅ K Solution: (a) Calculate the absorbed energy: ⎛ 0.01 J ⎞ E = ( Dose in rad ) m = ( 225 rad )( 0.17 kg ) ⎜ ⎟ = 0.38 J ⎝ rad ⎠ (b) Solve equation 16-13 for the temperature: ΔT = Q E 0.3825 J = = = 0.54 mK mc mc 0.17 kg ( 4186 J/kgK ) Insight: The radiation does not cause significant heating of the tumor 62 Picture the Problem: A 72-kg patient receives a radiation dose of 32 mrad over their entire body from alpha particles that have a relative biological effectiveness (RBE) of 13 Strategy: Use equation 32-18 to calculate the dosage in rem Use equation 32-16 to convert the rad dosage to units of J/kg and then multiply by the patient’s mass in order to find the total energy absorbed Solution: (a) Calculate the rem dosage: Dose in rem = ( Dose in rad )( RBE ) = ( 0.032 rad )(13) = 0.42 rem (b) Multiply the dosage by the patient’s mass: ⎛ 0.01 J/kg ⎞ E = ( Dose in rad ) m = ( 0.032 rad )( 72 kg ) ⎜ ⎟ = 0.023 J ⎝ rad ⎠ Insight: The received dose is about twice that received from inhaling radon over the course of a year (see Table 32-4) Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32 – 21 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 32: Nuclear Physics and Nuclear Radiation 63 Picture the Problem: A patient receives a dose of radiation by ingesting a radioactive pharmaceutical containing 32 15 P Strategy: In order to calculate the number of electrons emitted over the seven days, we must calculate the total number of nuclei that decay This will be the difference between the number of nuclei initially present and the number remaining after the seven-day period First, use the half-life of the phosphorus to calculate the decay constant Then use equation 32-11 and the decay constant to determine the initial number of nuclei Subtract from the initial number of nuclei the number present after seven days, which can be calculated using equation 32-9 Multiply the number of electrons by the energy in each electron to find the total energy absorbed Divide the energy absorbed by the mass of the tissue to determine the dosage in rad Multiply the dosage by the RBE to calculate the dosage in rem ln ln = = 0.04854 d −1 T1/ 14.28 d Solution: (a) Calculate the decay constant: λ= Calculate the initial number of nuclei: N0 = Calculate the final number of nuclei: N = N e − λ t = ( 2.39 × 1012 ) e Calculate the number of electrons: ne = N − N = 2.39 × 1012 − 1.70 × 1012 = 6.9 × 1011 electrons (b) Find the energy absorbed: E = ne Ee = ( 6.9 ×1011 ) ( 705 keV ) (1.6 × 10−16 J/keV ) = 0.078 J (c) Determine the dose in rad: ( Dose in rad ) = Calculate the dose in rem: dose in rem = ( dose in rad ) RBE = ( 62 rad )(1.50 ) = 93 rem R0 λ = 1.34 × 106 s −1 ⎛ 86, 400 s ⎞ 12 ⎜ ⎟ = 2.39 × 10 nuclei d 0.04854 d −1 ⎝ ⎠ ( ) − 0.04854 d −1 ( 7.00 d ) = 1.70 ×1012 nuclei E 0.078 J ⎛ rad ⎞ = ⎜ ⎟ = 62 rad m 0.125 kg ⎝ 0.01 J/kg ⎠ Insight: The total dosage the patient will receive if all of the original nuclei decay is 323 rem, close to the lethal dose of 500 rem As it is, 93 rem is sufficient to damage the patient’s blood-forming tissues 64 Picture the Problem: An α particle (charge +2e ) and a β particle (charge − e ) deflect in opposite directions when they pass through a magnetic field Strategy: Recall the radius of curvature of a charged particle in a uniform magnetic field, r = mv (equation 22-3) qB when answering the conceptual question Solution: In general, the amount of deflection is inversely proportional to the radius of curvature; that is, a large radius implies very little deflection Recall, however, that the radius of curvature in a magnetic field (equation 22-3) is directly proportional to a particle’s mass and inversely proportional to its charge Therefore, the α particle—which has twice the charge but roughly 8000 times the mass—has a larger radius of curvature by a factor of 4000 We conclude that the β particle deflects by the greater amount Insight: We can also answer this question in terms of acceleration: F = qvB = ma The greater the acceleration a of the particle, the greater its deflection in a certain period of time We can therefore write the ratio: aβ qβ vB mβ qβ mα qβ 7295mβ = = = = 3647 ⇒ aβ = 3647aα The β particle will deflect much more than the α aα qα vB mα qα mβ 2qβ mβ Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32 – 22 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 32: Nuclear Physics and Nuclear Radiation 65 Picture the Problem: Radioactive samples A and B have equal half-lives, but the initial activity of sample A is twice that of sample B Strategy: Use the principles involved in the description of radioactive decay to answer the conceptual question Solution: The two samples have equal half-lives, therefore their numbers decrease with time by the same factor, e − λ t It follows that the activity of sample A will always be twice that of sample B We conclude that the ratio of the activity of sample A to that of sample B after two half-lives have elapsed is 2.00 Insight: If sample A had a shorter half-life than sample B, after awhile the activity of sample A would decrease and fall below the activity of sample B 66 Picture the Problem: The initial activity of radioactive sample A is twice that of sample B After two half-lives of sample A have elapsed, the two samples have the same activity Strategy: Use the principles involved in the description of radioactive decay to answer the conceptual question Solution: In two half-lives, the activity of sample A will be reduced to one-quarter its initial value The initial activity of sample B was half that of sample A, but after two half-lives of sample A its activity is now one-quarter the initial activity of sample A—that is, the two samples have the same activity It follows that the activity of sample B decreased by a factor of two in the same time that the activity of sample A decreased by a factor of four Therefore, the ratio of the half-life of B to the half-life of A must be 2.00 (the half-life of sample B is twice as long as that of sample A) Insight: If sample A had the same half-life as sample B, the population of each would decrease with time by the same factor ( e − λ t ) and the activity of sample A would always remain twice that of sample B 67 Picture the Problem: A coal-burning power plant and a nuclear power plant each consume fuel in order to produce a given amount of electricity Strategy: Compare the energy yield per kilogram of coal and the energy yield per kilogram of uranium in order to determine the relative masses of the fuels used by each power plant Solution: The amount of coal burned in a conventional power plant is much greater than the amount of uranium consumed in a nuclear power plant for a given amount of energy production The reason is that in a coal-powered plant, energy is released as a result of chemical reactions In a nuclear-powered plant the reactions occur within the nucleus, and therefore they release millions of times more energy than comparable chemical reactions As a result, the amount of coal that is burned is much greater than the amount of uranium that is consumed Insight: For instance, if we compare mid-grade bituminous coal in a 38% efficient power plant with a 32% efficient nuclear power plant that uses 3% enriched 235U fuel, we discover that in order to produce 8760 GWh (one year of continuous operation of a 1.0 GW power plant) the coal-burning plant must consume 3.15 million tons of coal while the nuclear plant needs only 38.7 tons of uranium fuel Keep in mind that the mass of the uranium fuel decreases a little bit (about 1.10 kg) as some of its mass is converted into energy, while the “coal” mass actually increases as it is reacted with oxygen to produce ash, CO2, NOx, SOx, H2O, and various other compounds 68 Picture the Problem: We are asked to determine the number of protons and neutrons in several isotopes Strategy: Calculate the number of neutrons by subtracting the atomic number from the atomic mass (N = A − Z) The number of protons is equal to the atomic number (Z) Solution: (a) Find the number of neutrons and protons in (b) Calculate the number of neutrons and protons in 211 82 232 90 Th: Pb: neutrons = 232 − 90 = 142 protons = 90 neutrons = 211 − 82 = 129 protons = 82 neutrons = 60 − 27 = 33 protons = 27 Insight: Note that the ratio of neutrons to protons is about 1.5 for Th-232 and Pb-211, but is smaller (about 1.2) for the smaller isotope, Co-60 (c) Calculate the number of neutrons and protons in 60 27 Co: Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32 – 23 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 32: Nuclear Physics and Nuclear Radiation 69 Picture the Problem: Three different decay processes produce three different daughter nuclei Strategy: Balance the decay equations for the atomic number and the atomic mass Determine the chemical symbol of the daughter isotope from the atomic number Solution: (a) Find the value of A A and Z in 210 82 Pb → Z X + He: A = 210 − = 206 Z = 82 − = 80, which is mercury, or Hg Identify the daughter isotope: A Z (c) Find the value of A and Z A − in 239 92 U → Z X + e + υ : A = 239 − = 239 Z = 92 − ( −1) = 93, which is neptunium (Np) Identify the daughter isotope: A Z (c) Calculate the value of A and Z in 116 C → ZA X + e + + υ : A = 11 − = 11 Z = − = 5, which is boron, or B Identify the daughter isotope: A Z X= X= 206 80 239 93 X= X= X = 115 X = 206 80 239 93 11 Hg Np B Insight: The total charge and total mass number is always conserved for any radioactive decay process 70 Picture the Problem: The dose in rad that is required to give a cancerous tumor a certain dose in rem depends upon the relative biological effectiveness (RBE) of the alpha radiation Strategy: Use equation 32-18 to calculate the dose in rad from the given dose in rem and the RBE found in Table 32-3 Solution: Solve equation 32-18 for the dose in rad: Dose in rem = Dose in rad × RBE Dose in rem 3800 rem Dose in rad = = RBE 10 to 20 = 190 to 380 rad Insight: Because the RBE for alpha particles lies in the range 10 to 20, the dose in rad cannot be fixed to a single value 71 Picture the Problem: The dose in rem that is received by a patient depends upon the dose in rad and the relative biological effectiveness (RBE) of the gamma radiation Strategy: Use equation 32-18 to calculate the dosage in rem The RBE for gamma rays is given in Table 32-3 Dose in rem = Dose in rad × RBE = ( 260 )(1) = 260 rem Solution: Apply equation 32-18 directly: Insight: The dose in rem is the same as the dose in rad because the RBE for gamma rays is one 72 Picture the Problem: The two radioactive decay series that begin 232 208 90Th and end with 82 Pb are shown in the figure at the right Strategy: Use the representations of alpha decay, A A− 4 A A − to Z X → Z − Y + He and beta minus decay, Z X → Z +1 Y + e determine the daughter nuclei of each decay process illustrated in the figure Solution: The nine intermediary nuclei on the upper series are as follows: 228 228 228 224 220 216 212 212 212 88 Ra, 89 Ac, 90Th, 88 Ra, 86 Rn, 84 Po, 82 Pb, 83 Bi, and 84 Po The tenth intermediary, on the lower (left) series, is 208 81 Tl Insight: A decay chain like this one can produce a wide variety of daughter isotopes in a rock The different half-lives of each process produce the daughter products in specific ratios The measurement of the ratios of these daughter isotopes can therefore give a reasonable estimate to the elapsed time from which the 23290Th was originally produced or inserted into the rock Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32 – 24 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 32: Nuclear Physics and Nuclear Radiation 73 Picture the Problem: The radioactive decay of potassium-40 can be used to determine the age of a Moon rock Strategy: Solve equation 32-9 for the age of the rock, where the decay constant is calculated from the half-life Then use equation 32-9, setting N = 19.5% (the amount of potassium present today) and solving for the time t when N = 10.0% of the potassium-40 remains ln ln = = 5.78 × 10−10 y −1 T1 1.20 × 109 y Solution: (a) Calculate the decay constant of K-40: λ= Solve equation 32-9 for the time: N = N e − λt t=− (b) Calculate the time until only 10% remains: t=− ⎛ N ⎞ ln ⎜ ln ( 0.195 ) = 2.83 × 109 y ⎟=− −10 −1 5.78 × 10 y λ ⎝ N0 ⎠ ⎛ N ⎞ ⎛ 0.100 ⎞ ln ⎜ ln ⎜ ⎟=− ⎟ = 1.16 × 10 y −10 −1 N 5.776 10 y 0.195 × λ ⎝ 0⎠ ⎝ ⎠ Insight: Part (b) of this problem could also have been calculated setting N = 100% and calculating the total time for 90.0% of the potassium to decay The remaining time required for 90.0% of the nuclei to decay would then be the total time minus the time that was found in part (a) and that has already elapsed for 80.5% of the nuclei to decay 74 Picture the Problem: The mantle of a gas lantern contains a small amount of radioactive 232 90 Th Strategy: Calculate the activity by multiplying the decay constant by the number of nuclei present The decay constant is calculated from equation 32-10 Determine the number of nuclei present by dividing the mass of the thorium-232 by the mass on one thorium atom (found in Appendix F) Solution: Calculate the decay constant of Th-232: λ= ln ln 1y ⎛ ⎞ −18 −1 = ⎜ ⎟ = 1.563 × 10 s 10 T1 1.405 × 10 y ⎝ 3.156 × 10 s ⎠ Calculate the number of Th-232 nuclei: N= ⎞ M ( 0.325 g ) ⎛ 1u 20 = ⎜ ⎟ = 8.43 × 10 atoms m th 232.04 u ⎝ 1.66 × 10−24 g ⎠ Find the activity: R = λ N = (1.563 × 10−18 s −1 )( 8.43 × 1020 ) = 1.32 × 103 Bq (b) The activity would be reduced by a factor of 2, because it is inversely proportional to half-life Insight: If the half-life were doubled, the decay constant would become 2.467 × 10−11 y −1 and the activity would become 658 Bq 75 Picture the Problem: An unknown isotope produces the same nucleus by β − decay that 214 84 Po produces by α-decay Strategy: Determine the daughter product in alpha decay of polonium-214 by balancing the atomic number and atomic mass in the decay equation Write an equation for β − decay that produces the same daughter product and balance that equation in order to determine the parent isotope Solution: Calculate the value of A A and Z in 214 84 Po → Z X + He: A = 214 − = 210 Z = 84 − = 82, which is lead, or Pb Identify the daughter isotope: A Z Calculate the value of A and Z − in: ZA A → 210 82 Pb + −1 e + υ : A = 210 + = 210 Z = 82 + ( −1) = 81, which is thallium, or Tl Identify the parent isotope: A Z X= A= 210 82 210 81 Pb Tl Insight: A third radioactive decay that produces lead-210 is β + -decay from 210 83 Bi Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32 – 25 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 32: Nuclear Physics and Nuclear Radiation 76 Picture the Problem: The maximum Coulomb force, the electrostatic potential energy, and the initial kinetic energy of an alpha particle can be determined from the distance of its closest approach between a stationary nickel nucleus Strategy: Use equation 19-5 to find the Coulomb force, where the charge on the alpha particle is 2e and the charge on the nickel nucleus is 28e Calculate the potential energy using equation 20-8, and then find the initial kinetic energy by setting it equal to the maximum electrostatic potential energy F= Solution: (a) Calculate the Coulomb force: kqQ k ( 2e )( 28e ) = r2 d2 (8.99 ×10 = U= (b) Find the electrostatic potential energy: Nm /C ) ( 56 ) (1.60 × 10 −19 C ) (15 ×10 −15 m) = 57 N kqQ k ( 2e )( 28e ) = r d (8.99 ×10 = Nm /C2 ) ( 56 ) (1.60 × 10−19 C ) 15 × 10−15 m ⎛ MeV ⎞ U = 0.86 pJ ⎜ ⎟ = 5.4 MeV −1 ⎝ 1.60 × 10 pJ ⎠ K i = U = 5.4 MeV (c) Determine the initial kinetic energy: Insight: Decreasing the separation distance by a factor of will quadruple the Coulomb force and double the potential energy, requiring that the alpha particle has twice as much initial kinetic energy 77 Picture the Problem: The activity in curies of a sample of pure 226 88 Ra can be determined from its mass Strategy: Calculate the decay constant from the half-life of 226 88 Ra as listed in Appendix F Determine the number of nuclei by dividing the total mass of the sample by the mass of one nucleus Finally, multiply the decay constant by the number of nuclei to calculate the activity in becquerels, and convert the answer to curies Solution: Calculate λ from the half-life: λ= ln ln = = 4.33 × 10−4 y −1 T1 1.60 × 103 y Determine the number of nuclei: N= ⎞ 1u M ⎛ 0.0017 kg ⎞ ⎛ 21 =⎜ ⎟ = 4.5 × 10 ⎟⎜ −27 m ⎝ 226.025406 u ⎠ ⎝ 1.66 ×10 kg ⎠ Calculate the activity: 1y ⎛ ⎞ R = λ N = ( 4.33 × 10− y −1 )( 4.5 ×1021 ) ⎜ ⎟ × 3.16 10 s ⎝ ⎠ ⎛ ⎞ Ci = 6.2 × 1010 Bq ⎜ ⎟ = 1.7 Ci 10 ⎝ 3.7 × 10 Bq ⎠ Insight: The unit of curie is based on the number of decays per second of a 1-gram sample of pure 226 88 Ra (refer to the discussion surrounding equation 32-6) Therefore, a 1.7-gram sample should have an activity of 1.7 Ci 78 Picture the Problem: We are asked to compare the half-lives of two samples Sample A starts out with four times as many nuclei, but after two days the two samples contain the same number of radioactive nuclei Strategy: Use equation 32-9 to set the number of nuclei of the two samples equal after 2.00 days have elapsed The initial number of nuclei is equal to 4N for sample A and N for sample B Solve the resulting equation for the ratio of the half-lives Solution: (a) Because nuclei of type A are decaying at a faster rate than type B, type B has the longer half-life NA = NB (b) Set N A = N B after two days and simplify: ( N0 ) e − λA t = ( N ) e − λB t = e( λA − λB ) t Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32 – 26 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 32: Nuclear Physics and Nuclear Radiation (λ Take the natural log of both sides and solve for the decay constant of sample B: A − λ B ) t = ln λB = λA − ln t ln ln ln ln − = − = 0.693 d −1 T1 2, A t 0.500 d 2.00 d Write the decay constant of sample A in terms if its half-life: λB = Solve for the half-life of sample B: T1 2, B = ln λB = ln = 1.00 d 0.693 d −1 Insight: The half-life of sample B is double the half-life of sample A, and in the two-day period sample A decayed for half-lives but sample B decayed for only half-lives 79 Picture the Problem: The range of radii of stable nuclei can be determined from the range of mass numbers for stable nuclei Strategy: Use equation 32-4 to calculate the maximum and minimum radii Set the surface area equal to the surface area of a sphere and calculate the ratio Repeat for the volume of a sphere rmin = (1.2 × 10−15 m ) (1) 1/3 Solution: (a) Calculate the maximum and minimum radii: = 1.2 × 10−15 m rmax = (1.2 × 10−15 m ) ( 209 ) 1/3 = 7.1× 10−15 m Write the range of radii: 1.2 ×10−15 m ≤ r ≤ 7.1× 10−15 m (b) Calculate the ratio of surface areas: S max 4π rmax ⎛ rmax ⎞ ⎛ 7.1× 10−15 m ⎞ = =⎜ ⎟ =⎜ ⎟ = 35.2 Smin 4π rmin ⎝ rmin ⎠ ⎝ 1.2 × 10−15 m ⎠ (c) Calculate the ratio of volumes: Vmax 43 π rmax ⎛ rmax ⎞ ⎛ 7.1× 10−15 m ⎞ = =⎜ ⎟ =⎜ ⎟ = 209 −15 Vmin ⎝ rmin ⎠ ⎝ 1.2 × 10 m ⎠ π rmin 2 3 Insight: Note that the ratio of the surface areas is the ratio of the mass numbers raised to the 2/3 power and that the ratio of the volumes is equal to the ratio of the mass numbers 80 Picture the Problem: The density of a neutron star is the same as the density of a nucleus, so that a neutron star is very much smaller than our Sun even though its mass is comparable Strategy: Set the volume of the neutron star equal to the mass divided by the density Set the volume equal to the volume of a sphere and solve for the radius = π r3 ρ m Solution: Solve equation 15-1 for V: V= Solve the expression for the radius: ⎛ 3m ⎞ r =⎜ ⎟ ⎝ 4πρ ⎠ 1/3 ⎡ ( 0.50 ) ( 2.00 × 1030 kg ) ⎤ ⎥ =⎢ ⎢⎣ 4π ( 2.3 × 1017 kg/m3 ) ⎥⎦ 1/3 = 10 km Insight: Neutrons stars are extremely small, dense objects If the entire Earth were this dense, its radius would be only 184 m Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32 – 27 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 32: Nuclear Physics and Nuclear Radiation 81 Picture the Problem: The age of a carbon-containing specimen can be determined from its mass and activity Strategy: Equation 32-14 shows that the activity of a natural sample of carbon has an activity of 0.231 Bq per gram Find the activity per gram of the mummy sample, and use it together with equation 32-13 and the initial activity per gram in order to determine the age of the sample The decay constant for carbon-14 is 1.21× 10− y −1 Solution: (a) Find the activity per gram of the carbon in the mummy: R1g = (b) Use equation 32-13 to find the age of the sample: t= λ R 1.38 Bq = = 0.1765 Bq/g m 7.82 g ln R0,1g R1g = ⎛ 0.231 Bq ⎞ ln ⎜ ⎟ = 2220 y −4 −1 1.21× 10 y ⎝ 0.1765 Bq ⎠ Insight: The activity of the one-gram sample is over 75% the initial activity, so the sample is younger than one half-life 82 Picture the Problem: We are asked to calculate the number of nuclear fissions that would operate a light bulb for 2.5 days given that the conversion of energy is 32% efficient Strategy: Multiply the power dissipated by the time elapsed to calculate the energy consumed by the light bulb Divide this energy by the energy harvested from one fission reaction to find the number of reactions required to light the bulb Finally, multiply the number of reactions by the mass of one uranium atom to calculate the total required mass Solution: (a) Calculate the energy consumed: E = Pbulb t = (120 W )( 2.5 d )( 86, 400 s/d ) = 26 MJ Divide the energy by the energy extracted from one reaction: N reaction = E 25.9 MJ MeV ⎛ ⎞ = ⎜ −19 ε Efission ( 0.32 )( 212 MeV ) ⎝ 1.60 ×10 MJ ⎟⎠ = 2.39 × 1018 reactions = 2.4 × 1018 reactions (b) Multiply the number of reactions by the mass of uranium: m = N reaction M U = ( 2.39 ×1018 ) ( 235.043925 u ) (1.66 ×10 −27 kg/u ) = 9.3 × 10 −7 kg = 0.00093 g = 0.93 mg Insight: This mass is miniscule when compared with the 2.5 kg of coal (assuming 27 MJ of energy released per 1.0 kg of bituminous coal and a 38% conversion efficiency) that would need to be burned to produce the same energy 83 Picture the Problem: Energy is released when three helium atoms fuse to create one carbon-12 nucleus Strategy: Subtract the mass of the carbon atom from the mass of three helium atoms (the nuclei of which are alpha particles) to calculate the mass difference The masses in Appendix F include the mass of six electrons with the carbon atom and two electrons each with the helium atoms, so the electrons all subtract out Multiply the mass difference by the speed of light squared to calculate the energy released Solution: (a) Some of the initial mass of the alpha particles is converted to energy Therefore, the mass of carbon-12 is less than the mass of the three alpha particles (b) Calculate the change in mass: Δm = ( 4.002603 u ) − 12.000000 u = 0.007809 u Convert the mass difference to energy : ⎛ 931.5 MeV/c E = Δm c = ( 0.007809 u ) ⎜ 1u ⎝ ⎞ ⎟ c = 7.274 MeV ⎠ Insight: Older, red-giant stars have consumed their primary fuel by fusing hydrogen into helium, and subsequently they fuse helium nuclei (alpha particles) into carbon as outlined in this problem As mentioned in Section 32-6, hydrogen fusion releases 27 MeV per cycle, so we can see that helium fusion produces less energy for the star Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32 – 28 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 32: Nuclear Physics and Nuclear Radiation 84 Picture the Problem: A heavy dose of gamma rays would be sufficient to melt a block of water Strategy: The dosage is the energy per unit mass The latent heat of fusion is also measured in energy per unit mass, so that we need only convert the latent heat of fusion for water into a dosage equivalent in rad ( Dose in rad ) = L = ( 33.5 × 104 Solution: Convert the latent heat to rad: ⎛ rad ⎞ J/kg ) ⎜ ⎟ = 33.5 × 10 rad ⎝ 0.01 J/kg ⎠ Insight: A very large dose of radiation is necessary to melt the ice However, the small doses that are used to irradiate food can kill off any bacteria present without significantly affecting the temperature of the food 85 Picture the Problem: We are asked to calculate the dosage of radiation necessary to heat water by one degree Strategy: Use equation 16-13 to calculate the heat needed to raise the temperature of water by 1.0 C° Divide this heat by the mass and convert the units into a dosage equivalent in rad Solution: (a) Calculate the heat needed: Q = mcw ΔT = (1.00 kg )( 4186 J/kg ⋅ K )(1 C° ) = 4186 J Divide the heat by the mass and convert to rad: ⎛ 4186 J ⎞ ⎛ 1.0 rad ⎞ ⎜ ⎟⎜ ⎟ = 4.2 ×10 rad ⎝ 1.0 kg ⎠ ⎝ 0.01 J/kg ⎠ (b) The dosage in rad is the energy per unit mass required to heat the water The dosage will stay the same Insight: A very large dose of radiation is necessary to heat water However, the small doses used to irradiate food can kill off any bacteria present without significantly affecting the temperature of the food 86 Picture the Problem: A patient absorbs radiation when she undergoes a chest X-ray Strategy: Use equation 32-18 to write the dose in rad and then convert the dose from rad to joules per kilogram Multiply the dose by the mass of the patient in order to calculate the energy absorbed dose in rem ⎛ 35 mrem ⎞ =⎜ ⎟ = 41.2 mrad RBE ⎝ 0.85 ⎠ Solution: Calculate the dose in rad: dose in rad = Convert to J/kg: ⎛ 0.01 J/kg ⎞ dose = ( 0.0412 rad ) ⎜ ⎟ = 0.412 mJ/kg ⎝ rad ⎠ Multiply by the mass exposed to the X-rays: E = dose ( m ) = ( 0.412 mJ/kg ) ( 14 ⋅ 72 kg ) = 7.4 mJ Insight: This energy is insufficient to cause any measurable heating of the body 87 Picture the Problem: A 226 88 Ra nucleus recoils after it emits a 0.186-MeV photon Strategy: Use the conservation of momentum to calculate the speed of the recoiling nucleus The momentum of the photon is its energy divided by the speed of light (equation 30-11) Solution: Set pRa = pγ and solve for the speed of the nucleus: E c E 1u ⎛ ⎞ ⎛ 0.186 MeV ⎞ v= = ⎜ ⎟ ⎟⎜ mc ( 226.025406 u ) ⎝ 931.5 MeV/c ⎠ ⎝ c ⎠ mv = = ( 8.83 × 10−7 ) c = 264 m/s Insight: The recoil speed of the nucleus is larger than the thermal speed (use equation 17-13 to find that vrms = 182 m/s at T = 300 K if the radium atom were in the gas phase) of the radium Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32 – 29 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 32: Nuclear Physics and Nuclear Radiation 88 Picture the Problem: All of the energy released in one hour by the alpha decay from a 50.0-g sample of to heat a 4.75-kg container of water 239 94 Pu is used Strategy: First calculate mass difference of the nuclear decay reaction in order to find the amount of energy released by each decay event Then use equation 32-9 to calculate the number of nuclei that decay in one hour Multiply the number of decays in one hour by the energy released per decay to calculate the total heat absorbed by the water Finally, solve equation 16-13 for the change in temperature of the water when it absorbs the heat from the alpha decays mi = 239.052158 u Solution: Calculate the mass difference 235 in the reaction 239 94 Pu → 92 U + He: mf = 235.043925 u + 4.002603 u = 239.046528 u Δm = 239.046528 u − 239.052158 u = − 0.005630 u Convert the mass difference to energy released per decay: ⎛ 931.5 MeV/c ⎞ E = Δm c = ( 0.005630 u ) ⎜ ⎟ c = 5.244 MeV 1u ⎝ ⎠ Calculate the decay constant: λ= Determine the initial number of nuclei: N0 = ln ln ⎛ y ⎞⎛ d ⎞ −9 −1 = ⎜ ⎟⎜ ⎟ = 3.28 ×10 h T1/ ( 2.41× 10 y ) ⎝ 365.25 d ⎠ ⎝ 24 h ⎠ ⎞ m 0.0500 kg ⎛ 1u 23 = ⎜ ⎟ = 1.26 × 10 M 239.052158 u ⎝ 1.66 × 10−27 kg ⎠ n = number of decays = N − N = N − N e − λ t = N (1 − e − λ t ) Determine the number of decays in one hour: ( n = (1.26 × 1023 ) − e ( ) − 3.28×10−9 h −1 (1.00 h ) ) = 4.13 ×10 14 decays Multiply the decays by the energy per decay: ⎛ 1.60 × 10−13 J ⎞ E = ( 4.13 × 1014 decays ) ( 5.244 MeV/decay ) ⎜ ⎟ MeV ⎝ ⎠ = 347 J Solve equation 16-13 for the temperature change: ΔT = Q 347 J = = 0.0175 K mw cw ( 4.75 kg )( 4186 J/kg ⋅ K ) Insight: The temperature increase in the water is barely measurable Alpha decay produces a tiny amount of thermal energy when compared with nuclear fission reactions 89 Picture the Problem: The energy from alpha decay in a sphere of 235 92 U heats up the sphere, until the rate at which energy is produced in the sphere equals the rate at which the energy is radiated away through blackbody radiation Strategy: Determine the mass of the sphere and the number of nuclei present from the density and radius of the sphere Calculate the decay constant from the half-life (equation 32-10) and the activity from the number of nuclei and the decay constant (equation 32-11) Use the mass difference for each reaction to find the energy released per decay Multiply the energy released by the activity to calculate the power output of the sphere Set this power output equal to the power radiated by blackbody radiation (equation 16-19) and solve for the temperature of the sphere Solution: Calculate the mass of the sphere: m = ρV = ρ ( 34 π r ) = (18.95 g/cm3 ) 43 π ( 2.25 cm ) = 0.9042 kg Determine the number of nuclei present: N= ⎞ M ⎛ 0.9042 kg ⎞ ⎛ 1u 24 =⎜ ⎟⎜ ⎟ = 2.317 × 10 nuclei −27 m ⎝ 235.04 u ⎠ ⎝ 1.6605 ×10 kg ⎠ Calculate the decay constant: λ= ⎞⎛ ln ⎛ ln 1y ⎞ −17 −1 =⎜ ⎟⎜ ⎟ = 3.121× 10 s T1/ ⎝ 7.038 × 10 y ⎠ ⎝ 3.156 × 107 s ⎠ Use equation 32-11 to write the activity: R = λ N = ( 3.121× 10−17 s −1 )( 2.317 × 1024 ) = 7.230 ì 107 s Copyright â 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32 – 30 www.elsolucionario.org James S Walker, Physics, 4th Edition Chapter 32: Nuclear Physics and Nuclear Radiation Calculate the mass difference 231 for the decay 235 92 U → 90 Th + He: mi = 235.043925 u mf = 231.036297 u + 4.002603 u = 235.0389 u Δm = 235.0389 u − 235.043925 u = − 0.005025 u Convert Δm to released energy: ⎛ 931.5 MeV/c E = Δm c = ( 0.005025 u ) ⎜ 1u ⎝ ⎞ ⎟ c = 4.681 MeV ⎠ Determine the power released: P = ER = ( 4.681 MeV ) ( 7.230 × 107 s −1 )(1.60 × 10−13 J/MeV ) = 5.42 × 10−5 J/s = 54.2 μ W Set the power equal to the power emitted by blackbody radiation: P = eσ A (T − Ts4 ) ⇒ T = T= ( 5.67 ×10 P + Ts4 eσ A 5.42 × 10 −5 W −8 W m ⋅ K ) 4π ( 0.0200 m ) + ( 293 K ) = 293.0019 K Calculate the change in temperature: ΔT = T − Ts = 293.0019 − 293.0000 K = 1.9 mK Insight: This temperature difference would be difficult to detect The alpha decay considered here does not lead to the tremendous heating that occurs during the nuclear fission of uranium-235 (see Section 32-5) 90 Picture the Problem: The activity of an iodine-131 sample depends upon the decay constant and the number of radioactive nuclei it contains Strategy: Find the decay constant from the half-life using equation 32-10 Solution: Solve equation 32-10 for the decay constant: λ= ln ln = = 9.98 × 10−7 s −1 T1/ 8.04 d × 86,400 s/d Insight: The longer the half-life, the smaller the decay constant λ 91 Picture the Problem: The activity of an iodine-131 sample depends upon the decay constant and the number of radioactive nuclei it contains Strategy: Use the decay constant of 9.98×10−7 s−1 from the previous question, together with the sample size in order to find the activity of the sample Solution Calculate the activity: ⎛ ⎞ Ci = 1.2 Ci R = λ N = ( 9.98 × 10−7 s −1 )( 4.5 × 1016 ) ⎜ 10 −1 ⎟ ⎝ 3.7 × 10 s ⎠ Insight: If the half-life were decreased to 4.02 d, more nuclei must decay per second in order that the same number of nuclei can decay in half the time The activity would increase to 2.4 Ci 92 Picture the Problem: The activity of an iodine-131 sample depends upon the decay constant and the number of radioactive nuclei it contains ⎛ ln ⎞ Strategy: Use equations 32-10 and 32-11 to find the dependence of R on T1 : R = λ N = ⎜ N Use this expression ⎜ T1 ⎟⎟ ⎝ ⎠ to answer the conceptual question Solution: If the half-life T1/ of iodine-131 were only half of its actual value, the activity R of the sample would be increased to twice its initial value because R is inversely proportional to T1/ Insight: A shorter half-life means that more nuclei decay per second, leading to a higher activity of the sample Copyright © 2010 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 32 – 31 www.elsolucionario.org ... upward, its speed decrease as it passes window would be less than the decrease in its speed as it passes window 1, again because it is traveling slower as it passes window Copyright © 2010 Pearson... dimensions in the same manner as algebraic expressions xt = ( m )( s ) = m ⋅ s Solution: (a) Substitute dimensions for the variables: No v2 m2 s2 m = = Yes x m s x m = Yes t s2 v ms m = = Yes t s. .. the dimensions in the same manner as algebraic expressions Solution: (a) Substitute dimensions for the variables: (b) Substitute dimensions for the variables: (c) Substitute dimensions for the