Question

Consider the de Broglie wavelength of an electron that strikes the back face of one of the early models of a TV screen at 1/10 the speed of light. Find the electron wavelength.

Answer 1

Answer:

$2.4\cdot 10^{-11} m$

Explanation:

The De Broglie wavelength of a particle is given by

$\lambda=\frac{h}{p}$ (1)

where

h is the Planck constant

p is the momentum of the particle

For the electron in this problem, which is relativistic, the momentum is given by

$p=\gamma m_0 v$ (2)

where:

$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{\sqrt{1-(\frac{0.10 c}{c})^2}}=1.005$ is the relativistic factor

$m_0 =9.11\cdot 10^{-31} kg$ is the rest mass of the electron

$v=0.10 c=0.10(3\cdot 10^8 m/s)=3\cdot 10^7 m/s$ is the speed of the electron

Substituting into (2), we find

$p=(1.005)(9.11\cdot 10^{-31}kg)(3\cdot 10^7 m/s)=2.75\cdot 10^{-23}kg m/s$

and substituting into (1), we find the wavelength of the electron

$\lambda=\frac{6.63\cdot 10^{-34}Js}{2.75\cdot 10^{-23} kg m/s}=2.4\cdot 10^{-11} m$