Question

A 0.22-caliber handgun fires a 28-g bullet at a velocity of 765 m/s. Calculate the de Broglie wavelength of the bullet.

Answer 1

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