Question
The de Broglie wavelengths associated with an electron and a proton are same. What will be the ratio of
(i) their momenta
(ii) their kinetic energies?

Answer

Data: $\lambda$ (electron $)=\lambda$ (proton)
$m$ (proton) $=1836 m$ (electron)
(i) $\lambda=\frac{h}{p}$ As $\lambda$ (electron) $=\lambda$ (proton),
$\frac{p \text { ( electron ) }}{p \text { (proton) }}=1$, where $p$ denotes the magnitude of momentum.
(ii) Assuming $v \ll C$,
$KE =\frac{1}{2} m v^2=\frac{1}{2} \frac{m^2 v^2}{m}=\frac{p^2}{2 m}$
$\therefore \frac{ KE \text { (electron) }}{ KE \text { (proton) }}=\frac{m \text { (proton) }}{m \text { (electron) }}=1836$ as $p$ is the same for the electron and the proton.

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