During the Winter Olympic biathlon trials, Eric is shooting his rifle at a target. What is the de Broglie wavelength of a 10.0-g bullet fired from the rifle at [tex]500. \, \text{m/s}[/tex]?

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High School

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Answer :

Tom056 Tom056 · Answer 1 · 2025-03-06T08:39:03-05:00

The de Broglie wavelength of a 10.0-g bullet fired from the rifle at 500. m/s is approximately 6.63 x 10^(-34) m.

To calculate the de Broglie wavelength of the bullet, we can use the de Broglie wavelength formula:

λ = h / p,

where λ is the de Broglie wavelength, h is the Planck's constant, and p is the momentum of the bullet.

First, we need to find the momentum of the bullet. The momentum (p) can be calculated using the formula:

p = m * v,

where m is the mass of the bullet and v is its velocity.

Mass of the bullet (m) = 10.0 g = 0.01 kg,

Velocity of the bullet (v) = 500 m/s.

Now, let's calculate the momentum of the bullet:

p = (0.01 kg) * (500 m/s) = 5 kg·m/s.

Next, we can substitute the values of Planck's constant (h) and the momentum (p) into the de Broglie wavelength formula:

λ = (6.63 x 10^(-34) J·s) / (5 kg·m/s).

Calculating this expression gives us:

λ ≈ 1.326 x 10^(-34) m.

Therefore, the de Broglie wavelength of a 10.0-g bullet fired from the rifle at 500. m/s is approximately 6.63 x 10^(-34) m.

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