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,m_el = 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 an electron is given by:$$\lambda = \frac{h}{p}$$where h is Planck's constant, p is the momentum of the electron, and m is the mass of the electron.The momentum of an electron with kinetic energy K is given by:$$p = \sqrt{2mK}$$where m is the mass of the electron and K is the kinetic energy of the electron.Substituting these equations into the equation for the de Broglie wavelength, we get:$$\lambda = \frac{h}{\sqrt{2mK}}$$For an electron with kinetic energy of 980 eV, the de Broglie wavelength is:$$\lambda = \frac{6.626 	imes 10^{-34} 	ext{ J s}}{\sqrt{2(9.11 	imes 10^{-31} 	ext{ kg})(980 	imes 1.60 	imes 10^{-19} 	ext{ J})}} = 3.87 	imes 10^{-11} 	ext{ m}$$The first minimum of the electron diffraction pattern occurs when the path length difference between the two waves is equal to one wavelength. This occurs when the angle of diffraction is given by:$$\sin 	heta = \frac{\lambda}{d}$$where d is the width of the slit.For a single slit, the width of the slit is equal to the wavelength of the light. Therefore, the first minimum of the electron diffraction pattern occurs when the angle of diffraction is:$$\sin 	heta = \frac{3.87 	imes 10^{-11} 	ext{ m}}{411 	imes 10^{-9} 	ext{ m}} = 0.00094$$$$	heta = 0.054^\circ = 0.095 	ext{ mrad}$$Therefore, the correct answer is A.

A Single Slit Is Illuminated at Normal Incidence with a Parallel