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anyanavicka [17]
3 years ago
15

An electron and a proton are each accelerated from rest through a potential difference of 100 V. Afterward, which particle has t

he larger de Broglie wavelength?
Physics
1 answer:
Lena [83]3 years ago
3 0

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.

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