Answer
$\lambda = 1.84\times 10^{-11}~m$
Work Step by Step
We can find the mass of the electron when it is moving at this relativistic speed:
$m = \gamma~m_0$
$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$
$m = \frac{m_0}{\sqrt{1-\frac{(0.6~c)^2}{c^2}}}$
$m = \frac{m_0}{0.8}$
$m = 1.25~m_0$
We can find the de Broglie wavelength:
$\lambda = \frac{h}{mv}$
$\lambda = \frac{6.626\times 10^{-34}~J~s}{(1.25)(1.6\times 10^{-31}~kg)(0.6)(3.0\times 10^8~m/s)}$
$\lambda = 1.84\times 10^{-11}~m$
