This question is about group/phase velocities and also De Brogilie wavelength.

What I would like to know is how to derive ratio $\lambda_e/\lambda_p$ ($\lambda_e$ and $\lambda_p$ are De Broglie wavelength for electron and proton) if we know that electron and proton have same velocities?

I know that when we say "velocity" $v$, this velocity is the same as "group velocity" $v_g$. So for start I can calculate ratio for group velocities:

\begin{align} \boxed{v_e = v_p} \longrightarrow \frac{v_{ge}}{v_{ge}} = \frac{v_e}{v_p} = 1 \end{align}

and similarly I can do for phase velocities $v_p$:

\begin{align} \frac{v_{pe}}{v_{pp}} = \dfrac{\tfrac{c^2}{v_{ge}}}{\tfrac{c^2}{v_{gp}}} = \frac{v_{gp}}{v_{ge}} = 1 \end{align}

But when i try to calculate the ratio for wavelengths I got stuck:

\begin{align} \frac{\lambda_{e}}{\lambda_{p}} = \frac{\tfrac{h}{p_e}}{\tfrac{h}{p_p}} = \frac{p_p}{p_e} = ~\longleftarrow \substack{\text{I got stuck here where i don't know}\\\text{how to use relation $\boxed{v_p = v_e}$}} \end{align}

Can anyone give me a hint? I was trying to use relation $p = \frac{\sqrt{{E_k}^2 + 2E_0E_k}}{c}$ but I got lost...

  • $\begingroup$ I already figured this out. $\endgroup$ – 71GA Jul 23 '13 at 13:28
  • $\begingroup$ Because velocity is explicitly part of the question you're going to be stuck with some $\gamma's$. $\endgroup$ – dmckee --- ex-moderator kitten Jul 23 '13 at 13:48

Let's try to express the de Broglie wavelength $\lambda_j$ of a particle in terms of its mass $m_j$ and velocity $v_j$. $$ \left(\frac{2 \pi}{\lambda_j}\right)^2 = \left(\frac{p_j}{\hbar}\right)^2 = \frac{1}{\hbar c^2}\left(E_j^2 - m_j^2 c^4\right) $$

Where we've used the relation $E^2 = p^2 c^2 + m^2 c^4$.

From here we note that $E_j=\gamma(v_j) m_j c^2$, where $\gamma(v) = \frac{1}{\sqrt{1 - \frac{v_j^2}{c^2}}}$ to see that: $$ \left(\frac{2 \pi}{\lambda_j}\right)^2 = \frac{\gamma(v_j) - 1}{\hbar c^2} m_j^2 c^4 $$

Since $v_e=v_p$, we see the only term not common between the two particles is their rest mass and so we find that: $$ \frac{\lambda_e}{\lambda_p} = \frac{m_p}{m_e} $$

Which to me is a surprising result.


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