Wave or particle?

1. Under what conditions the particles having mass $m$ and speed $v$ ($v < c$) scattered on a periodic structure having a constant $d$ (distance between consecutive parallel plans), show stronger wave like properties?

2. Taking into account the wave like properties of particles, indicate the limits when the classical notions can be applied safely to an electron and a proton if both have an energy of 10 eV?

3. A bullet having a mass of $40g$ has a speed of $v =1000 m/s$.

a) What is its de Broglie wavelength?

b) Why the bullet does not display wave like properties, (for example being subject to diffraction effects)?

4. What acceleration potential is necessary for an electron to show wave like properties by diffraction on to a crystal having a lattice constant of $d =0.3 nm$?


1. The wave like properties are stronger when the value of the associated de Broglie wavelength $h/(m v)$ is close to the lattice constant $d$ of the crystal.

2. The wave like properties can be applied (considered) to a particle when the dimension of the diffraction slits or of the obstacles that limits the particles motion, are much less than or comparable to the particle de Broglie wavelength. When $d \gg \lambda$, the wave like properties are negligible.


$d \gg h/\sqrt{2mE_c}$


$d \gg 0.39 nm$ foe electron and $d \gg 9.9 pm$ for the proton.

Observation: for the same energy, the de Broglie wavelength for nucleons (proton and neutron) is about 1000 times less than the de Broglie wavelength of the electrons. This is why low energy nucleons are more inclined to undergo diffraction effects than electrons.

3. a) In the this case (bullet) case the associated de Broglie wavelength is

$\lambda = h/(mv) =1.66*10^{-35} m$

b) Comparing to the dimension of the “available” slits in nature (for example the smaller distance between crystalline plans is about $10^{-10} m$, the above wavelength is very small, so that the diffraction effects can not be observed in this case.

4. See answer to question 2.

For diffraction to take place one needs: $2d*sin\theta =k\lambda$

Since $\lambda = h/\sqrt{2meU}$ one has

$U\geq h^2/(8med^2)$   that is $U \geq 4.19 V$