where Lis the perpendicular distance between the slits and the screen. Therefore, the angle between the center of the pattern and the second minimum is given by 23tan2yLdλθ==. Since d±, tan≈, and we obtain 1686933(1.8910m)7.07 10 rad(4.0 10 )22(4.0010m)d−−−−×≈==×=×°×. 44. Using Eq. 37-8, we find the Lorentz factor to be 117.08881 ( / )1 (0.9900)vcγ=. With pmv=(Eq. 37-41), the de Broglie wavelength of the protons is
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