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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

If diffraction were the only limitation on resolution, what would be the smallest structure that could be resolved in an electron microscope using 10-keV electrons?
What is the de Broglie wavelength of a basketball of mass \(0.50 \mathrm{kg}\) when it is moving at \(10 \mathrm{m} / \mathrm{s} ?\) Why don't we see diffraction effects when a basketball passes through the circular aperture of the hoop?
A light-emitting diode (LED) has the property that electrons can be excited into the conduction band by the electrical energy from a battery; a photon is emitted when the electron drops back to the valence band. (a) If the band gap for this diode is \(2.36 \mathrm{eV},\) what is the wavelength of the light emitted by the LED? (b) What color is the light emitted? (c) What is the minimum battery voltage required in the electrical circuit containing the diode?
A beam of neutrons has the same de Broglie wavelength as a beam of photons. Is it possible that the energy of each photon is equal to the kinetic energy of each neutron? If so, at what de Broglie wavelength(s) does this occur? [Hint: For the neutron, use the relativistic energy-momentum relation \(\left.E^{2}=E_{0}^{2}+(p c)^{2} .\right]\)
In the Davisson-Germer experiment (Section \(28.2),\) the electrons were accelerated through a \(54.0-\mathrm{V}\) potential difference before striking the target. (a) Find the de Broglie wavelength of the electrons. (b) Bragg plane spacings for nickel were known at the time; they had been determined through x-ray diffraction studies. The largest plane spacing (which gives the largest intensity diffraction maxima) in nickel is \(0.091 \mathrm{nm} .\) Using Bragg's law [Eq. ( \(25-15\) )], find the Bragg angle for the first-order maximum using the de Broglie wavelength of the electrons. (c) Does this agree with the observed maximum at a scattering angle of \(130^{\circ} ?\) [Hint: The scattering angle and the Bragg angle are not the same. Make a sketch to show the relationship between the two angles.]
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