Answer:
[tex]2.4\cdot 10^{-11} m[/tex]
Explanation:
The De Broglie wavelength of a particle is given by
[tex]\lambda=\frac{h}{p}[/tex] (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
[tex]p=\gamma m_0 v[/tex] (2)
where:
[tex]\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{\sqrt{1-(\frac{0.10 c}{c})^2}}=1.005[/tex] is the relativistic factor
[tex]m_0 =9.11\cdot 10^{-31} kg[/tex] is the rest mass of the electron
[tex]v=0.10 c=0.10(3\cdot 10^8 m/s)=3\cdot 10^7 m/s[/tex] is the speed of the electron
Substituting into (2), we find
[tex]p=(1.005)(9.11\cdot 10^{-31}kg)(3\cdot 10^7 m/s)=2.75\cdot 10^{-23}kg m/s[/tex]
and substituting into (1), we find the wavelength of the electron
[tex]\lambda=\frac{6.63\cdot 10^{-34}Js}{2.75\cdot 10^{-23} kg m/s}=2.4\cdot 10^{-11} m[/tex]
