Question

The de Broglie wavelength of an electron traveling at 7.0 x 107m/s is 1.0 x 10-11m. (Remember that me = 9.1 x 10-31kg and h = 6.63 x 10-34J·s)
True
False

Answer 1

The De Broglie's wavelength of a particle is given by:

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

where

$h=6.6 \cdot 10^{-34} Js$ 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:

$p=m_e v=(9.1 \cdot 10^{-31} kg)(7.0 \cdot 10^7 m/s)=6.4 \cdot 10^{-23} kg m/s$

and if we substitute this into the previous equation, we find the De Broglie wavelength of the electron:

$\lambda=\frac{h}{p}=\frac{6.6 \cdot 10^{-34} Js}{6.4 \cdot 10^{-23} kg m/s}=1.0 \cdot 10^{-11} m$

So, the answer is True.