Electrons emerge from an electron gun with a speed of 2.0 × 10⁶ m/s and then pass through a pair of thin parallel slits. Interference fringes with a spacing of 2.7 mm are detected on a screen far from the double slit and fairly close to the center of the pattern. What would the fringe spacing be if the electrons were replaced by neutrons with the same speed? (mel = 9.11 × 10^-31kg, mneutron = 1.67 × 10^-27kg) A) 1.5 µm B) 4.9 µm C) 0.93 nm D) 1.1 µm E) 1.5 nm

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Quizplus · Accepted Answer

The de Broglie wavelength is given by:$$\lambda = \frac{h}{p}$$where h is Planck's constant and p is the momentum of the particle. The momentum of the electrons is:$$p = mv = (9.11 	imes 10^{-31} 	ext{ kg})(2.0 	imes 10^6 	ext{ m/s}) = 1.82 	imes 10^{-24} 	ext{ kg m/s}$$The de Broglie wavelength of the electrons is:$$\lambda_e = \frac{6.63 	imes 10^{-34} 	ext{ J s}}{1.82 	imes 10^{-24} 	ext{ kg m/s}} = 3.65 	imes 10^{-10} 	ext{ m}$$The momentum of the neutrons is:$$p = mv = (1.67 	imes 10^{-27} 	ext{ kg})(2.0 	imes 10^6 	ext{ m/s}) = 3.34 	imes 10^{-21} 	ext{ kg m/s}$$The de Broglie wavelength of the neutrons is:$$\lambda_n = \frac{6.63 	imes 10^{-34} 	ext{ J s}}{3.34 	imes 10^{-21} 	ext{ kg m/s}} = 1.98 	imes 10^{-13} 	ext{ m}$$The fringe spacing is given by:$$d = \frac{\lambda L}{a}$$where L is the distance from the slits to the screen and a is the distance between the slits. The fringe spacing for the electrons is:$$d_e = \frac{(3.65 	imes 10^{-10} 	ext{ m})(1.0 	ext{ m})}{2.7 	imes 10^{-3} 	ext{ m}} = 1.35 	imes 10^{-6} 	ext{ m} = 1.35 \mu	ext{m}$$The fringe spacing for the neutrons is:$$d_n = \frac{(1.98 	imes 10^{-13} 	ext{ m})(1.0 	ext{ m})}{2.7 	imes 10^{-3} 	ext{ m}} = 7.33 	imes 10^{-11} 	ext{ m} = 73.3 	ext{ nm}$$Therefore, the fringe spacing for the neutrons is 1.5 µm.

Electrons Emerge from an Electron Gun with a Speed of 2.0