How Is the Angle of Electron Reflection Determined in Quantum Experiments?

AI Thread Summary
The discussion focuses on determining the angle of electron reflection in quantum experiments, specifically after electrons are accelerated by a voltage of 78.0 V. The calculated speed of the electrons is 5,230,000 m/s, with a linear momentum of 4.77×10^-24 and a De Broglie wavelength of 0.139 nm. Participants suggest using the formula 2d sin(θ) = mλ for calculating the reflection angle, but there is uncertainty about the appropriate value for the order number m. The correct reflection angle, derived from the calculations, is 70.3°. This highlights the importance of understanding quantum mechanics principles in solving such problems.
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Homework Statement



For this question, h= 6.63×10–34 J.s, c = 3×108 m/s, me= 9.11×10–31 kg, e=1.6×10–19 C. Electrons are accelerated from rest by a voltage of 78.0 V. After that acceleration their speed is 5230000 m/s, their linear momentum is 4.77×10-24 and their De Broglie wavelength is 0.139 nm.

The accelerated electrons are then reflected by a crystal with crystal spacing d of 0.085 nm. Through which angle are they reflected?

(correct answer: 70.3°)

Homework Equations





The Attempt at a Solution



What formula can I use here? I tried using d sin \theta = m \lambda, where m is the order number. However I don't know what value to substitute for m...
Is there a better way of appraching this problem?
 
Last edited:
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Use 2dsin(θ) = mλ.
 

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