The De Broglie's wavelength of a particle is given by:
[tex] \lambda=\frac{h}{p} [/tex]
where
[tex] h=6.6 \cdot 10^{-34} Js [/tex] is the Planck constant
p is the momentum of the particle
In this problem, the momentum of the electron is equal to the product between its mass and its speed:
[tex] p=m_e v=(9.1 \cdot 10^{-31} kg)(7.0 \cdot 10^7 m/s)=6.4 \cdot 10^{-23} kg m/s [/tex]
and if we substitute this into the previous equation, we find the De Broglie wavelength of the electron:
[tex] \lambda=\frac{h}{p}=\frac{6.6 \cdot 10^{-34} Js}{6.4 \cdot 10^{-23} kg m/s}=1.0 \cdot 10^{-11} m [/tex]
So, the answer is True.
