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.

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.

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Nguyệt Ánh 6 months 2021-07-27T16:04:46+00:00 1 Answers 10 views 0

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    2021-07-27T16:06:29+00:00

    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

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