Question

An electron and a proton are each accelerated from rest through a potential difference of 100 V. Afterward, which particle has the larger de Broglie wavelength?

Answer 1

Explanation:

The De-Broglie wavelength in terms of potential difference is given by:

$\lambda=\dfrac{h}{\sqrt{2meV} }$

Where,

h is Planck's constant

m is mass of charged particle

V is potential difference

e is the amount of charge

It means that the De-Broglie wavelength is inversely proportional to the mass.

Since, the mass of the proton is more than the mass of the electron. So, the De- Broglie wavelength of the electron is larger than proton.