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Chapter 28: Problem 68

A bullet leaves the barrel of a rifle with a speed of $300.0 \mathrm{m} / \mathrm{s} .\( The mass of the bullet is \)10.0 \mathrm{g} .$ (a) What is the de Broglie wavelength of the bullet? (b) Compare \(\lambda\) with the diameter of a proton (about \(1 \mathrm{fm}\) ). (c) Is it possible to observe wave properties of the bullet, such as diffraction? Explain.

Short Answer

Expert verified
Answer: The de Broglie wavelength of the bullet is approximately 2.209 x 10^{-34} m, which is much smaller than the diameter of a proton (1 x 10^{-15} m). It is practically impossible to observe the bullet's wave properties, as the wavelength determines the scale at which wave properties can be significant, and in this case, it is too small compared to typical objects or experimental setups.

Step by step solution

01

Calculate the de Broglie wavelength of the bullet

To calculate the de Broglie wavelength (λ) of the bullet, we will use the formula: λ = h / p, where h is Planck's constant and p is the momentum of the bullet. Planck's constant (h) is approximately 6.626 x 10^{-34} Js. The mass of the bullet is given in grams, so we need to convert it to kilograms to be consistent with SI units: m = 10 g * (1 kg / 1000 g) = 0.01 kg. The momentum of the bullet (p) is the product of its mass and velocity: p = mv = (0.01 kg)(300 m/s) = 3 kg m/s. Now we can calculate the de Broglie wavelength: λ = (6.626 x 10^{-34} Js) / (3 kg m/s) = 2.209 x 10^{-34} m.
02

Compare λ with the diameter of a proton

The diameter of a proton is approximately 1 fm or 1 x 10^{-15} m. We can compare the de Broglie wavelength of the bullet with the diameter of a proton: λ / proton diameter = (2.209 x 10^{-34} m) / (1 x 10^{-15} m) = 2.209 x 10^{-19}. The de Broglie wavelength of the bullet is much smaller than the diameter of a proton.
03

Possibility of observing wave properties of the bullet

Since the de Broglie wavelength of the bullet is much smaller than the diameter of a proton, it is not expected that we can observe diffraction or other wave properties with the bullet. The reason is that the de Broglie wavelength determines the scale at which wave properties can be significant, and in this case, the wavelength is much smaller than typical objects or experimental setups, making it practically impossible to observe the wave properties of the bullet.

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Most popular questions from this chapter

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