Question

A proton is placed at point A, where the electric potential is 100 V . The proton is released from rest. Some time later, the proton has reached point B, where the electric potential is -120 V . What is the de Broglie wavelength of the proton when it reaches point B?

Answer 1

To develop this problem it is necessary to apply the concepts related to electromagnetic energy and Broglie's hypothesis.

By definition we know that the electrical energy of a proton can be expressed as

E = qV

Where,

q = Charge of proton

V = Voltage

Replacing with our values

E = qV

$E = (1.6*10^{-19})(220) \rightarrow$ It is necessary to add the two potentials

$E = 4.224*10^{-17}J$

From Broglie's hypothesis we know that the wavelength is given by

$\lambda = \frac{h}{P}$

Where,

h = Planck's constant

p = Momentum

The momentum of a particle can be expressed in terms of energy, that is,

$P = \sqrt{E*2m}$

Where,

m = mass

E = Energy (potential or kinetic)

Therefore replacing this value at lambda,

$\lambda = \frac{h}{\sqrt{E*2m}}$

$\lambda = \frac{6.625*10^{-34}}{\sqrt{(4.224*10^{-17})*2(1.67*10^{-27})}}$

$\lambda = 1.763*10^{-12}m$