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 know from x-rays measurements to be 3 Angstrom, the speed of the electron is most nearly
A. 1.4 × 10−4 m/s
B. 2.4 m/s
C. 5.0 × 103 m/s
D. 2.4 × 106 m/s
E. 4.5 × 109 m/s
Bragg's Law: nλ = 2d sin θ
De Broglie wavelength: λ = h/mev
h/mev = 2d sin θ / n
v = nh / (2med sin θ)
Given:
n = 1 (first-order reflection)
d = 3 Angstrom = 3 × 10−10 m
θ = 30 → sin 30 = 0.5
me = 9.11 × 10−31 kg
h = 6.63 × 10−34 Js
v = (6.63 × 10−34)/[2( 9.11 × 10−31)(3 × 10−10)(0.5)]
= [6.63/(9.11 × 3)] × 10−34 × 1041
= [2/(3 × 3)] × 107 ≈ 0.22 × 107 = 2.2 × 106 m/s
Answer: D
A. 1.4 × 10−4 m/s
B. 2.4 m/s
C. 5.0 × 103 m/s
D. 2.4 × 106 m/s
E. 4.5 × 109 m/s
(GR8677 #91)
Solution:
Bragg's Law: nλ = 2d sin θ
De Broglie wavelength: λ = h/mev
h/mev = 2d sin θ / n
v = nh / (2med sin θ)
Given:
n = 1 (first-order reflection)
d = 3 Angstrom = 3 × 10−10 m
θ = 30 → sin 30 = 0.5
me = 9.11 × 10−31 kg
h = 6.63 × 10−34 Js
v = (6.63 × 10−34)/[2( 9.11 × 10−31)(3 × 10−10)(0.5)]
= [6.63/(9.11 × 3)] × 10−34 × 1041
= [2/(3 × 3)] × 107 ≈ 0.22 × 107 = 2.2 × 106 m/s
Answer: D
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