# Question d42ce

Feb 2, 2014

de Broglie suggested that particles can behave like waves.

de Broglie postulated the wave nature of electrons and suggested that all matter has wave properties. This concept is known as wave-particle duality.

Every particle of matter with mass m and velocity v has a momentum $p = m v$. de Broglie assumed that any particle had a wavelength λ that was equal to Planck's constant $h$ divided by its momentum.

λ = h/p = h/(mv)

EXAMPLE

What is the de Broglie wavelength of an electron ("mass" = 9.11 xx 10^-31"kg") moving at a speed of $6.00 \times {10}^{6} \text{m/s}$?

Solution

λ = h/(mv) = (6.626 × 10^-34 cancel("kg")·"m"^cancel("2")cancel("s⁻¹"))/( 9.11 × 10^-31 cancel("kg") × 6.00 × 10^6 cancel("m·s⁻¹")) 
= 1.21×10^-10"m"#

Note: This is just a little bigger than the diameter of a hydrogen atom.