In the Davisson-Germer experiment,an electron | vedclass

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Question

(B) The de Broglie wavelength $\lambda$ of an electron with energy $E = 60 \ eV$ is given by $\lambda = \frac{12.27}{\sqrt{E}} \mathring A = \frac{12.

Vedclass · Accepted Answer

$0.9$

Vedclass · Answer

(B) The de Broglie wavelength $\lambda$ of an electron with energy $E = 60 \ eV$ is given by $\lambda = \frac{12.27}{\sqrt{E}} \mathring A = \frac{12.27}{\sqrt{60}} \approx 1.58 \mathring A$. In the Davisson-Germer experiment,the scattering angle $\phi$ is the angle between the incident and scattered beam. The glancing angle $	heta$ is related to $\phi$ by $	heta = 90^\circ - \frac{\phi}{2}$. Given $\phi = 50^\circ$,we have $	heta = 90^\circ - 25^\circ = 65^\circ$. Using Bragg's Law for diffraction,$2d \sin 	heta = n\lambda$. For the first order $(n=1)$,$d = \frac{\lambda}{2 \sin 	heta}$. Substituting the values: $d = \frac{1.58}{2 \sin 65^\circ} = \frac{1.58}{2 	imes 0.906} \approx 0.87 \mathring A$. Note: Based on standard textbook values for this specific experiment where $\phi = 50^\circ$ and $E = 54 \ eV$,the spacing $d$ is approximately $0.91 \mathring A$. Given the options provided,$0.9 \mathring A$ is the correct choice.

In the Davisson-Germer experiment,an electron beam of energy $60 \ eV$ is incident normally on a crystal surface. If the maximum intensity is observed at an angle of $50^\circ$ with the direction of the incident beam,what is the interatomic spacing in the crystal lattice in $\mathring A$?

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