Answer:
de Broglie wavelength of the bullet is 3.093 x 10⁻³⁵ m
Explanation:
Given;
mass of bullet, m = 28 g = 0.028 kg
velocity of the bullet, v = 765 m/s
de Broglie wavelength of the bullet is given by;
λ = h / mv
where;
λ is de Broglie wavelength of the bullet
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
λ = h / mv
λ = (6.626 x 10⁻³⁴ ) / (0.028 x 765)
λ = 3.093 x 10⁻³⁵ m
Therefore, de Broglie wavelength of the bullet is 3.093 x 10⁻³⁵ m
