Answer:
- de Broglie wavelength of the bullet is 4.56 x 10⁻³⁴ m
- The value of the wavelength shows that wave nature of matter is insignificant for the bullet because it is larger than particles.
Explanation:
Given;
mass of the bullet, m = 1.9 g = 0.0019 kg
velocity of the bullet, v = 765 m/s
de Broglie wavelength of the bullet is given by;
[tex]\lambda = \frac{h}{mv}[/tex]
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
λ is de Broglie wavelength of the bullet
[tex]\lambda = \frac{h}{mv}\\\\ \lambda =\frac{(6.626*10^{-34})}{(0.0019)(765)}\\\\ \lambda =4.56 *10^{-34} \ m[/tex]
Thus, this value of the wavelength shows that wave nature of matter is insignificant for the bullet because it is larger than particles.
