Through de Broglie's hypothesis, we can express the relationship between a particle's wavelength and momentum through the equation below.
[tex] \lambda = \frac{h}{p} [/tex]
where h is the Plank's constant equivalent to 6.626 x 10⁻³⁴ J.s. Recall that momentum, p, can be computed as
p = mv
where m is the mass and v is the velocity of the particle. Now, we have a rock with a mass of 60.0 g or 0.06 kg and a velocity of 40.0 m/s. Now, its de Broglie wavelength is
[tex] \lambda = \frac{(6.626)(10^{-34})}{(0.06)(40.0)} [/tex]
[tex] \lambda = (2.761) x 10^{-34} [/tex]
Therefore, the rock's de Broglie's wavelength is 2.761 x 10⁻³⁴ m.
Answer: 2.761 x 10⁻³⁴ m
