A single slit is illuminated at normal incidence with a parallel beam of light having a wavelength of 411 nm. The entire central band of the diffraction pattern is observed at ±90°. The illumination is now replaced by a nonrelativistic beam of electrons, each having a kinetic energy of 980 eV. When this beam hits the slit at normal incidence, at what angle will the first minimum of the electron diffraction pattern occur? (h = 6.626 × 10^-34 J ∙ s, mel = 9.11 × 10^-31 kg, 1 eV = 1.60 × 10^-19 J) A) 0.095 mrad B) 0.071 mrad C) 0.046 mrad D) 0.12 mrad E) 0.14 mrad

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Quizplus · Accepted Answer

The de Broglie wavelength of the electron is:$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2melK}}$$where h is the Planck constant, me is the electron mass, l is the kinetic energy, and p is the momentum.Substituting the given values, we get:$$\lambda = \frac{6.626 	imes 10^{-34} 	ext{ J s}}{\sqrt{2(9.11 	imes 10^{-31} 	ext{ kg})(980 	imes 1.60 	imes 10^{-19} 	ext{ J})}} = 4.11 	imes 10^{-10} 	ext{ m}$$The first minimum of the electron diffraction pattern occurs at an angle θ given by:$$sinθ = \frac{\lambda}{d}$$where d is the slit width.Since the entire central band of the diffraction pattern is observed at ±90°, we have:$$d = \frac{\lambda}{2} = 2.06 	imes 10^{-10} 	ext{ m}$$Substituting this value into the equation above, we get:$$sinθ = \frac{4.11 	imes 10^{-10} 	ext{ m}}{2.06 	imes 10^{-10} 	ext{ m}} = 2$$This is impossible, since sinθ cannot be greater than 1. Therefore, there is no first minimum in the electron diffraction pattern.

A Single Slit Is Illuminated at Normal Incidence with a Parallel