Conceptual Physics (12th Edition)

$2.4 \times 10^{-11} m$. This is less than an atomic diameter.
The de Broglie wavelength is Planck’s constant/momentum, as stated on page 590. First find the electron’s momentum. $$p = mv = (9.1 \times 10^{-31} kg)(3.0 \times 10^{7} \frac{m}{s})$$ $$= 2.7 \times 10^{-23} \frac{kg m}{s}$$ Now find the de Broglie wavelength. $$\lambda = \frac{h}{p} = (6.63 \times 10^{-34} J s )/(2.7 \times 10^{-23} \frac{kg m}{s} )$$ $$= 2.4 \times 10^{-11} m$$