# What is the velocity of de Broglie wave?

Here is my calculation: note: $$ν$$ – frequency, $$V$$ – velocity of wave, $$v'$$– velocity of object. \begin{align} λ &= h/mv' \tag{1}\\ mc^2/h &= ν \tag{2}\\ λν &= V \quad \text{(wave equation)} \tag{3} \end{align}

Substituting 1 and 2 in 3 we get:

$$h/mv' \cdot mc^2/h = V$$

Simplifying we get:

$$c^2/v' = V$$

However, according to the book, the answer is $$hv/mc$$. What is wrong with my answer?

• What is equation (2)? This appears to be a mass up of mass-energy equivalence and computing the energy of a photon. – Zhe Mar 8 '19 at 16:04
• Some de Broglie waves are not (in any meaningful sense) moving: they are standing waves (like electron orbitals in an atom) where it doesn't make sense to talk about how fast the wave moves. – matt_black Mar 8 '19 at 23:50

I think the problem arises partially from the confusing notations. "Velocity of wave" $$V$$ is essentially $$c$$, "velocity of object" $$v'$$ is usually denoted with the speed of particle $$v$$ and the energy of photon $$E$$ is for some reason expressed via mass–energy equivalence $$E = mc^2$$ when it should be just kinetic energy of the particle $$E = mv^2/2$$ (but the latter is anyways irrelevant here).

So, your equation turns into this:

$$c^2/v' = V$$ $$c^2/v = c$$ $$c = v$$

which just states that the velocities of both wave and particle are equivalent. I suggest you just combine the de Broglie equation

$$λ = \frac{h}{mv}$$

and the relation between wavelength $$λ$$ and frequency $$ν$$

$$λ = \frac{c}{ν}$$

so that

$$λ = \frac{h}{mv} = \frac{c}{ν} \quad \implies \quad v =\frac{hν}{mc}$$