Question

Consider the de-Broglie wavelength of an electron and a proton. Which wavelength is smaller if the two particles have (a) the same speed (b) the same momentum (c) the same energy?

Answer 1

de-Broglie wavelength,
$\lambda =\frac{h}{mv}$,

where h = Planck's constant
$m$ = mass of the particle
v = velocity of the particle
(a) It is given that the speed of an electron and proton are equal.
It is clear from the above equation that
$\lambda \propto \frac{1}{m}$
As mass of a proton, m_p > mass of an electron, m_e, the proton will have a smaller wavelength compared to the electron.

(b) $\lambda =\frac{h}{p}$ $\left(\because p=mv\right)$
So, when the proton and the electron have same momentum, they will have the same wavelength.
(c) de-Broglie wavelength $\left(\lambda \right)$ is also given by
$\lambda =\frac{h}{\sqrt{2mE}}$,
where E = energy of the particle.
Let the energy of the proton and electron be E.
Wavelength of the proton,
${\lambda }_p=\frac{h}{\sqrt{2m_{\mathrm{p}}E}}...\left(1\right)$
Wavelength of the electron,
${\lambda }_e=\frac{h}{\sqrt{2m_{\mathrm{e}}E}}...\left(2\right)$
Dividing (2) by (1), we get:
$$\begin{gathered} \frac{{\lambda }_e}{{\lambda }_p}=\frac{\sqrt{m_e}}{\sqrt{m_p}} \\ \Rightarrow \frac{{\lambda }_e}{{\lambda }_p}<1 \\ \Rightarrow {\lambda }_e<{\lambda }_p \end{gathered}$$
It is clear that the proton will have smaller wavelength compared to the electron.