College Physics (4th Edition)

$\lambda = 3.313\times 10^{-27}~m$ A proton is larger than this de Broglie wavelength by a factor of $3.02\times 10^{11}$
We can find the de Broglie wavelength: $\lambda = \frac{h}{mv}$ $\lambda = \frac{6.626\times 10^{-34}~J~s}{(1.0\times 10^{-4}~kg)(2\times 10^{-3}~m/s)}$ $\lambda = 3.313\times 10^{-27}~m$ We can find the ratio of a proton to this de Broglie wavelength: $\frac{1\times 10^{-15}~m}{3.313\times 10^{-27}~m} = 3.02\times 10^{11}$ A proton is larger than this de Broglie wavelength by a factor of $3.02\times 10^{11}$.