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A 0.22 caliber handgun fires a 1.9g bullet at a velocity of 765m/s. Calculate the de Broglie wavelength of the bullet. Is the wave nature of matter significant for the bullets?

Respuesta :

onyebuchinnaji

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.

Cricetus

The wavelength will be "[tex]4.56\times 10^{-34} \ m[/tex]".

Given:

  • Mass, m = 1.9 g or, 0.0019 kg
  • Velocity, v = 765 m
  • Plank's constant, h = 6.626 × 10⁻³⁴ J/s

The De-Broglie wavelength,

→ [tex]\lambda = \frac{h}{mv}[/tex]

By putting the values,

     [tex]= \frac{6.626\times 10^{-34}}{0.0019\times 765}[/tex]

     [tex]= 4.56\times 10^{-34} \ m[/tex]

Thus the response above is right.

Learn more about wavelength here:

https://brainly.com/question/10931065

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