B. \( 6.1 \times 10^{-31} \, \text{kg} \) Using the de Broglie wavelength: \(\lambda = \frac{h}{p}\) Rearranging gives: \(p = \frac{h}{\lambda}\) Given: \(\lambda = 4.0 \times 10^{-7} \, \text{mm} = 4.0 \times 10^{-10} \, \text{m}\) Now substituting for \( p \): \(p = \frac{6.63 \times 10^{-34}}{4.0 \times 10^{-10}} = 1.6575 \times 10^{-24} \, \text{kg m/s}\) Now, to find mass \( m \): \(m = \frac{p}{v} = \frac{1.6575 \times 10^{-24}}{2.7 \times 10^{8}} \approx 6.1 \times 10^{-31} \, \text{kg}\).