In the Davisson-Germer experiment, an electro | vedclass

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(B) The de Broglie wavelength $\lambda$ of an electron with energy $E = 75 \ eV$ is given by $\lambda = \frac{12.27}{\sqrt{E}} 	ext{ \AA} = \frac{12.

Vedclass · Accepted Answer

$2$

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(B) The de Broglie wavelength $\lambda$ of an electron with energy $E = 75 \ eV$ is given by $\lambda = \frac{12.27}{\sqrt{E}} 	ext{ \AA} = \frac{12.27}{\sqrt{75}} \approx 1.417 	ext{ \AA}$.According to the Bragg's law for diffraction, $2d \sin 	heta = n \lambda$, where $	heta$ is the glancing angle.The angle of diffraction $\phi$ is given as $45^{\circ}$ with respect to the incident beam. The glancing angle $	heta$ is related to the diffraction angle $\phi$ by $	heta = 90^{\circ} - \frac{\phi}{2} = 90^{\circ} - 22.5^{\circ} = 67.5^{\circ}$.However, in many textbook contexts for this specific problem, the formula used is $d \sin \phi = n \lambda$ where $\phi$ is the scattering angle.Using $d \sin 45^{\circ} = 1 	imes 1.417 	ext{ \AA}$, we get $d = \frac{1.417}{0.707} \approx 2 	ext{ \AA}$.

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