Answer
5.39 nm
Work Step by Step
Recall that de Broglie wavelength is $\lambda=\frac{h}{mv}$ where $h$ is the Planck's constant, $m$ is the mass and $v$ is the velocity.
Substituting values, we have
$ \lambda=\frac{6.626\times10^{-34}\,J\cdot s}{(9.11\times10^{-31}\,kg)(1.35\times10^{5}\,m/s)}$
$=5.39\times10^{-9}\,m=5.39\,nm$
