Calculate the de Broglie wavelength of an electron moving at $10^6$ m/s. (Take $h = 6.6 \times 10^{-34}$ J·s, $m_e = 9.1 \times 10^{-31}$ kg)

de Broglie wavelength $\lambda = \dfrac{h}{m v}$. $\lambda = \dfrac{6.6 \times 10^{-34}}{(9.1 \times 10^{-31}) \times 10^6} = \dfrac{6.6 \times 10^{-34}}{9.1 \times 10^{-25}} = 0.725 \times 10^{-9} \approx 7.3 \times 10^{-10}$ m. Unit-wise: J·s / (kg · m/s) = (kg·m²/s²·s) / (kg·m/s) = m ✓. The answer ~ 7 Å is on the atomic scale, which is why wave-like behaviour matters for electrons but not for everyday objects (a 1 kg ball at 1 m/s has $\lambda \sim 10^{-34}$ m, completely undetectable).