Quantum Mechanics - Free Particle

A free particle with initial kinetic energy E and de Broglie wavelength λ enters a region in which it has potential energy V. What is the particle’s new de Broglie wavelength?

A. λ(1 + E/V)
B. λ(1 − V/E)
C. λ(1− E/V)^{− 1}
D. λ(1 + V/E)^½
E. λ(1 − V/E)^{− ½}

(GR0177 #46)

Solution:

The initial kinetic energy of free particle:


De Broglie wavelength:


Energy of the particle when it enters the region:


So, its wavelength becomes:




Answer: E

Nuclear & Particle Physics - Bragg Diffraction

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

(GR8677 #91)

Solution:

Bragg's Law: nλ = 2d sin θ 

De Broglie wavelength: λ = h/m_ev

h/m_ev = 2d sin θ / n
v = nh / (2m_ed sin θ)

Given:
n = 1 (first-order reflection)
d = 3 Angstrom = 3 × 10⁻¹⁰ m
θ = 30 →  sin 30 = 0.5
m_e =  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