Chapter-42 - 1599 Chapter 42 1. Kinetic energy (we use the...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1599 Chapter 42 1. Kinetic energy (we use the classical formula since v is much less than c ) is converted into potential energy (see Eq. 24-43). From Appendix F or G, we find Z = 3 for Lithium and Z = 90 for Thorium; the charges on those nuclei are therefore 3 e and 90 e , respectively. We manipulate the terms so that one of the factors of e cancels the “e” in the kinetic energy unit MeV, and the other factor of e is set to be 1.6 × 10 –19 C. We note that k = 1 4 π ε can be written as 8.99 × 10 9 V·m/C. Thus, from energy conservation, we have ( 29 ( 29 ( 29 9 19 V m C 1 2 6 8.99 10 3 1.6 10 C 90 3.00 10 eV e kq q K U r K- ⋅ × × × = ⇒ = = × which yields r = 1.3 × 10 – 13 m (or about 130 fm). 2. Our calculation is similar to that shown in Sample Problem 42-1. We set ( 29 ( 29 Cu min 5.30MeV= 1/ 4 / K U q q r α ε = = π and solve for the closest separation, r min : ( 29 ( 29 ( 29 ( 29 19 9 Cu Cu min 6 14 2 29 1.60 10 C 8.99 10 V m/C 4 4 5.30 10 eV 1.58 10 m 15.8 fm. e q q kq q r K K α α ε ε-- × × ⋅ = = = π π × = × = We note that the factor of e in q α = 2 e was not set equal to 1.60 × 10 – 19 C, but was instead allowed to cancel the “e” in the non-SI energy unit, electron-volt. 3. The conservation laws of (classical kinetic) energy and (linear) momentum determine the outcome of the collision (see Chapter 9). The final speed of the α particle is v m m m m v f i α α α α =- + Au Au , and that of the recoiling gold nucleus is v m m m v f i Au, Au = + 2 α α α . (a) Therefore, the kinetic energy of the recoiling nucleus is CHAPTER 42 1600 ( 29 ( 29 ( 29( 29 ( 29 2 2 2 Au Au, Au Au, Au 2 Au Au 2 2 4 1 1 2 2 4 197u 4.00u 5.00MeV 4.00u+197u 0.390MeV. f f i i m m m K m v m v K m m m m α α α α α α = = = + + = = (b) The final kinetic energy of the alpha particle is ( 29 2 2 2 2 Au Au Au Au 2 1 1 2 2 4.00u 197u 5.00MeV 4.00u 197u 4.61MeV. f f i i m m m m K m v m v K m m m m α α α α α α α α α α -- = = = + + - = + = We note that K K K af f i + = Au, α is indeed satisfied. 4. Using Eq. 42-3 ( 1/3 r r A = ), we estimate the nuclear radii of the alpha particle and Al to be 15 1/3 15 15 1/3 15 Al (1.2 10 m)(4) 1.90 10 m (1.2 10 m)(27) 3.60 10 m. r r α---- = × = × = × = × The distance between the centers of the nuclei when their surfaces touch is 15 15 15 Al 1.90 10 m 3.60 10 m 5.50 10 m r r r α--- = + = × + × = × . From energy conservation, the amount of energy required is 9 2 2 19 19 Al 15 12 6 1 (8.99 10 N m C )(2 1.6 10 C)(13 1.6 10 C) 4 5.50 10 m 1.09 10 J 6.79 10 eV q q K r α πε---- × ⋅ × × × × = = × = × = × 5. Kinetic energy (we use the classical formula since v is much less than c ) is converted into potential energy. From Appendix F or G, we find Z = 3 for Lithium and Z = 110 for Ds; the charges on those nuclei are therefore 3 e and 110 e , respectively. From energy conservation, we have...
View Full Document

This note was uploaded on 08/09/2010 for the course PHY 202 101A2 taught by Professor Prof.yang during the Summer '10 term at National Taiwan University.

Page1 / 36

Chapter-42 - 1599 Chapter 42 1. Kinetic energy (we use the...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online