Solutions to Homework problems Chapters 29. Including a

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1492 PARTICLES AND WAVES Using values for melectron and mproton taken from the inside of the front cover, we find that the speed of the proton is proton - velectron m =. x m/s -27 = 2.45xlO m/S v - [melectron: proton (450 106 )(9.11X10-31 1.67 X 10 kg] kg I 3 I 25. REASONING AND SOLUTION The momentum of the photon is p = h/A and that of the electron is p = mv. Equating and solving for the wavelength of the photon, 26. REASONING The de Broglie wavelength A is related to Planck's constant h and the magnitude p of the particle's momentum. The magnitude of the momentum can be related to the particle's kinetic energy. Thus, using the given wavelength and the fact that the kinetic energy doubles, we will be able to obtain the new wavelength. SOLUTION The de Broglie wavelength is The kinetic energy and the magnitude of the momentum are 1 2 KE = - mv 2 (6.2) p=mv or (7.2) (29.8) where m and v are the mass and speed of the particle. Substituting Equation 7.2 into Equation 6.2, we can relate the kinetic energy and momentum as follows: Substituting this result for p into Equation 29.8 gives h h A = p = ~2m(KE)
and Dividing the two equations and rearranging reveals that h A, ~ = ~2m(KE)f h = /KE), (KE)f ~2m(KE\ Using the given value for Aiand the fact that KEf = 2 (KEi ) , we find or Chapter 29 Problems 1493 Applying this expression for the final and initial wavelengths Af and Ai' we obtain 27. I SSMI REASONING AND SOLUTION The de Broglie wavelength A of the woman is given by Equation 29.8 as A = h / p, where p is the magnitude of her momentum. The magnitude of the momentum is p = mv , where m is the woman's mass and v is her speed. According to Equation 3.6b of the equations of kinematics, the speed v is given by v = ~~UyY ~, since the woman jumps from rest. In this expression, a y = -9.80 m/ S2 and Y = -9.5 m. With these considerations we find that = 6.63 X 10-34 J ·s = 11.2 X 10-36 m I (41 kg)~2( -9.80 m/ S2)( -9.5 m) ---- 28. REASONING The width of the central bright fringe in the diffraction patterns will be identical when the electrons have the same de Broglie wavelength as the wavelength of the photons in the red light. The de Broglie wavelength of one electron in the beam is given by Equation 29.8, A eIect ron\ = h / p, where p = mv. SOLUTION Following the reasoning described above, we find A =A red light electron
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Page 9: Chapter 29 Problems 1495 31. REASON

Solutions to Homework problems Chapters 29. Including a

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