WAEC Physics 2020 Paper

Concept:The de Broglie wavelength of a moving particle relates its momentum to Planck's constant using $\lambda = \frac{h}{mv}$.Explanation:The electron has mass $m = 9.1 \times 10^{-31}\,\mathrm{kg}$ and speed $v = 2.0 \times 10^{6}\,\mathrm{m\,s^{-1}}$.Planck's constant is given as $h = 6.6 \times 10^{-34}\,\mathrm{J\,s}$.Substitute these values into the formula:$$\lambda = \frac{h}{mv} = \frac{6.6 \times 10^{-34}}{(9.1 \times 10^{-31})(2.0 \times 10^{6})}$$First multiply the denominator: $mv = 18.2 \times 10^{-25}$.Then calculate the wavelength:$$\lambda = \frac{6.6 \times 10^{-34}}{18.2 \times 10^{-25}} = 3.63 \times 10^{-10}\,\mathrm{m}$$Therefore, the associated wavelength of the electron is $3.63 \times 10^{-10}\,\mathrm{m}$.Answer:$$\lambda = 3.63 \times 10^{-10}\,\mathrm{m}$$This corresponds to option A.