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conantee
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« on: January 08, 2010, 01:57:57 AM »

91. When a narrow beam of monoenergetic electrons impinges on the surface of a single metal crystal at an angle of 30 degrees with the plane of the surface, first-order reflection is observed. If the spacing of the reflecting crystal planes is known from x-ray measurements to be 3 angstroms, the speed of the electrons is most nearly
(A) 1.4 \times 10^{-4} \mbox{ m/s}
(B) 2.4  \mbox{ m/s}
(C) 5.0 \times 10^{3} \mbox{ m/s}
(D) 2.4 \times 10^{6} \mbox{ m/s}
(E) 4.5 \times 10^{9} \mbox{ m/s}
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« Reply #1 on: October 16, 2010, 10:58:42 AM »

If you don't know Bragg's condition or electron wave, at least you should eliminate 2 or 3 choices. First, (E) is wrong since it is faster than the speed of light. Next, electron in a beam for this kind of experiment should be very fast. (A) is hopelessly small, while (B) is just the speed of car, roughly speaking. Hence, you have only (C) and (D). You are expert familiar with the topic, you can even say that the answer is (D) by order of magnitude.

Suppose you don't know Bragg's condition, you can just use the relation between momentum and wavelength of matter wave. In order to probe the distance of 3 angstroms, you also need roughly wavelength of about 3 angstroms. By De Broglie relation, we have
\lambda = h/p = h/ m v
Hence, v =   h /m \lambda = 6.62 \cdot 10^{-34}/(9.11 \cdot 10^{-31} \times 3 \cdot 10^{-10} ) 
v \sim 6 \cdot 10^{-34} / (10 \cdot 10^{-31} \times 3 \cdot 10^{-10}) \sim 2 \cdot 10^{-34-1+31+10} \sim 2 \cdot 10^6

The closest answer is (D).

The correct answer is therefore (D). 30 of 100 people answered correctly.

Note: The Bragg's condition is  a \sin \phi = m \lambda, where m is integer, a is the spacing between reflecting planes, and \phi is the angle between particle incoming wave and the reflecting plane.
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