Step 1: Recall the de Broglie wavelength formula. The de Broglie wavelength ($\lambda$) of a particle is given by the formula: $$\lambda = \frac{h}{mv}$$ where: • $h$ is Planck's constant ($6.626 \times 10^{-34} \text{ J s}$) • $m$ is the mass of the particle • $v$ is the velocity of the particle Step 2: Identify the given values and necessary constants. • Mass of an electron, $m_e = 9.109 \times 10^{-31} \text{ kg}$ • Velocity of the electron, $v = 2.18 \times 10^6 \text{ m/s}$ • Planck's constant, $h = 6.626 \times 10^{-34} \text{ J s}$ Step 3: Substitute the values into the de Broglie wavelength formula. $$\lambda = \frac{6.626 \times 10^{-34} \text{ J s}}{(9.109 \times 10^{-31} \text{ kg}) \times (2.18 \times 10^6 \text{ m/s})}$$ Step 4: Calculate the denominator. $$m \times v = (9.109 \times 10^{-31} \text{ kg}) \times (2.18 \times 10^6 \text{ m/s})$$ $$m \times v = (9.109 \times 2.18) \times (10^{-31} \times 10^6) \text{ kg m/s}$$ $$m \times v = 19.85762 \times 10^{-25} \text{ kg m/s}$$ Step 5: Calculate the wavelength. $$\lambda = \frac{6.626 \times 10^{-34} \text{ J s}}{19.85762 \times 10^{-25} \text{ kg m/s}}$$ $$\lambda = \frac{6.626}{19.85762} \times 10^{-34 - (-25)} \text{ m}$$ $$\lambda = 0.33367 \times 10^{-9} \text{ m}$$ Step 6: Convert the wavelength to nanometers (nm). Since $1 \text{ nm} = 10^{-9} \text{ m}$: $$\lambda = 0.33367 \text{ nm}$$ Rounding to two significant figures, as suggested by the options: $$\lambda \approx 0.33 \text{ nm}$$ The wavelength will be $\boxed{\text{0.33 nm}}$. 3 done, 2 left today. You're making progress.