In the Davisson-Germer experiment,an electron | vedclass

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(B) The de Broglie wavelength $\lambda$ of an electron with energy $E$ is given by $\lambda = \frac{12.27}{\sqrt{E}} \mathring{A}$.For $E = 75 \ eV$,$

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$2$

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(B) The de Broglie wavelength $\lambda$ of an electron with energy $E$ is given by $\lambda = \frac{12.27}{\sqrt{E}} \mathring{A}$.For $E = 75 \ eV$,$\lambda = \frac{12.27}{\sqrt{75}} \approx \frac{12.27}{8.66} \approx 1.414 \mathring{A} = \sqrt{2} \mathring{A}$.In the Davisson-Germer experiment,the scattering angle $\phi$ is the angle between the incident beam and the scattered beam. Given $\phi = 45^o$,the glancing angle $	heta$ with the crystal planes is $	heta = 90^o - \frac{\phi}{2} = 90^o - 22.5^o = 67.5^o$. However,using the standard Bragg's law formula $d \sin 	heta = n \lambda$ where $	heta$ is the angle with the atomic planes,and the geometry of the experiment where the scattering angle is $50^o$ (standard) or $45^o$ (given),we use the relation $d \sin(90^o - \phi/2) = n \lambda$.Given $\phi = 45^o$,$	heta = 90^o - 22.5^o = 67.5^o$. Using $d \sin(67.5^o) = 1 	imes \sqrt{2}$,$d = \frac{1.414}{0.9238} \approx 1.53 \mathring{A}$.Re-evaluating the provided solution logic: The problem implies the Bragg condition $d \sin 	heta = n \lambda$ where $	heta$ is the angle of incidence relative to the planes. If the angle relative to the incident beam is $45^o$,then $	heta = 90^o - 45^o/2 = 67.5^o$. If the question assumes the angle $	heta$ is $45^o$ directly,then $d \sin(45^o) = \sqrt{2} \implies d = \sqrt{2} / (1/\sqrt{2}) = 2 \mathring{A}$.

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