WAEC Physics 2010 Paper

Concept:The moving electron behaves like a matter wave; its wavelength is given by the de Broglie relation $\lambda = \dfrac{h}{mv}$, where $p = mv$ is the momentum of the electron.Explanation:Given, $m = 9.1 \times 10^{-31}\ \mathrm{kg}$ and $v = 2.0 \times 10^{6}\ \mathrm{m\,s^{-1}}$.First, calculate the momentum of the electron:$$p = mv = (9.1 \times 10^{-31})(2.0 \times 10^{6}) = 1.82 \times 10^{-24}\ \mathrm{kg\,m\,s^{-1}}$$Now apply the de Broglie wavelength formula using $h = 6.6 \times 10^{-34}\ \mathrm{J\,s}$:$$\lambda = \frac{h}{p} = \frac{6.6 \times 10^{-34}}{1.82 \times 10^{-24}} = 3.63 \times 10^{-10}\ \mathrm{m}$$Note: The photon formula $E = \frac{hc}{\lambda}$ is not applicable here because this is a moving electron, not a photon.Answer:$$\lambda = 3.63 \times 10^{-10}\ \mathrm{m}$$