Question

Through what potential difference must an electron be accelerated from rest to have a de broglie wavelength of 400 nm ?

Answer 1

So the potential difference is 9.4229μV

Given,

mass of the electron = 9.1*10^{-31}kg

charge of electron = 1.6*10^{-19}C

From Broglie's relation we can write

mv=h/λ

So, v=h/mλ=$\frac{6.626*10^{-34} }{(9.1*10^{-31}*(400*10^{-9} ) )}$=1820.32m/s

Kinetic Energy of electron = $\frac{1}{2} mv^{2}$=eV

V=$\frac{1*m*v^{2} }{2*e}$=$\frac{1*9.1*10^{-31}*1820.32^{2} }{2*1.6*10^{-19} }$=9.4229μV

Electron

Unattached or attached to an atom, an electron is a negatively charged subatomic particle (not bound). One of the three main types of particles inside an atom, along with protons and neutrons, is an electron that is linked to the atom.

An atom's nucleus is made up of electrons, protons, and neutrons. The negative charge of the electron is balanced by the positive charge of the proton. An atom is in a neutral state if its protons and electrons are equal in number.

The differences between electrons and the other particles are numerous. They have a mass that is considerably lower, exist outside of the nucleus, and display both wave- and particle-like properties.

Learn more about Electron here:

brainly.com/question/1255220

#SPJ4