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chapter3

# chapter3 - Useful constants etc h:= 6.63 10 34 melectron:=...

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Problem 3.2 Find the de Broglie wavelength of (a) an electron whose speed is 1x10 8 m/s and (b) an electron whose speed is 2x10 8 m/s. γ v ( ) 1 1 v 2 c 2 := v part1 1.0 10 8 := v part2 2.0 10 8 := Part a. Do we need to worry about relativity? Let's do this non-relativistically and relativistically, and see if there is a significant difference Nonrelativistic: Relativistic λ non m v , ( ) h m v := λ rel m v , ( ) h γ v ( ) m v := λ non m electron v part1 , ( ) 7.278 10 12 × = λ rel m electron v part1 , ( ) 6.861 10 12 × = wavelength units are meters Percent error P: P λ non m electron v part1 , ( ) λ rel m electron v part1 , ( ) λ rel m electron v part1 , ( ) 100 := P 6.066 = Because a 6% error is introduced by the nonrelativistic calculation, we need to use the relativistic result. Useful constants, etc.: h 6.63 10 34 := m electron 9.11 10 31 := c 3 10 8 := ===================================================================================== Physics 107 Problem 3.1 Compare the linear momenta, total energy, and kinetic energy of a photon and a particle that have the same wavelength. Because λ h p := regardless of whether the object is a photon or a particle, if the wavelengths are the same, then the momenta p are the same. Problem 3.11 shows that the particle's total energy is nearly the same as the energy of a photon of the same wavelength PROVIDED THE PARTICLE'S TOTAL ENERGY GREATLY EXCEEDS ITS REST ENERGY; i.e. when the particle's kinetic energy is very large; it is moving very fast. From chapter 1, E m 0 2 c 4 p 2 c 2 + := In this problem, both the photon and the particle have the same momemtum p, so p 2 c 2 is the same for both. Because some of the particle's total energy is tied up in its rest energy m 0 2 c 4 , and none of the photon's total energy is tied up in rest energy, the particle must have less kinetic energy than the photon has total energy. ===================================================================================== Physics 107 1

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E m electron c 2 1 v 2 c 2 := We can use equation 1.22 to calculate the velocity joules E 8.816 10 14 × = convert to joules E E 1.60 10 19 := eV E 5.51 10 5 × = E 40 10 3 0.511 10 6 + := The electron's total energy is its 40-keV kinetic energy plus its 0.511 MeV rest energy. In units of eV, we need to use relativistic corrections for this magnitude of speed v cl 1.185 10 8 × = v cl 2 KE m electron := if this gives a v near c, then we need to go back and calculate a relativistic velocity KE 1 2 m v 2 := KE 40 10 3 1.60 10 19 := Part b. Because the velocity is greater than in part a, we clearly need a relativistic calculation. λ rel m electron v part2 , ( ) 2.712 10 12 × = ===================================================================================== Physics 107 Problem 3.3 Find the de Broglie wavelength of a 1 mg grain of sand blown by the wind at 20 m/s. This is an extremely lightweight piece of matter. Maybe its wavelength will be observable?? λ sand m v , ( ) h m v := λ sand 1 10 6 20 , ( ) 3.315 10 29 × = (using the fact that one milligram is 10 -3 grams, or 10 -6 kg) This wavelength is in meters. This is many orders of magnitude smaller than a typical size of a nucleus. We'll never "see" that a grain of sand has wave properties.
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