Question

Compare the de Broglie wavelength of a bullet moving at 700 miles per hour (313 m/s) to that of an alpha particle moving at3.40E7 miles per hour (1.52E7 m/s) and a proton with a speed of 1.30E7 miles per hour (5.81E6 m/s).

Which region of the electromagnetic spectrum are each of these wavelengths near?

A. Ultraviolet 10⁻⁸ to 10⁻⁷ meters
B. X-ray 10⁻¹¹ to 10⁻⁸ meters
C. Gamma ray 10⁻¹⁶ to 10⁻¹¹ meters
D. Smaller than 10⁻²⁰ meters. Cannot detect wave-like properties. Only particle-like behavior will be observable.

Answer 1

Option C: The de Broglie wavelengths of a bullet, alpha particle, and proton differ greatly because of their mass and velocity differences; with the bullet having the shortest and likely undetectable wavelength.

Step-by-step explanation:

The question asks us to compare the de Broglie wavelengths of different particles moving at high speeds and to determine which region of the electromagnetic spectrum these wavelengths are near. The de Broglie wavelength (λ) can be calculated using the formula λ = h/p, where h is Planck's constant and p is the momentum of the particle. Given that the momentum is the product of mass and velocity (p = mv), we see that wavelength is inversely proportional to both mass and velocity.

Without completing the exact calculations, we can qualitatively analyze the situation. A bullet being much more massive compared to subatomic particles such as alpha particles and protons, will have a significantly smaller de Broglie wavelength, likely placing it in option D, as the wave-like properties are not detectable. On the other hand, the alpha particle and proton, with lesser masses and high velocities, will have longer de Broglie wavelengths but still very short in the context of the electromagnetic spectrum. These wavelengths will likely fall into the categories of Gamma rays (C) due to their minuscule wavelengths compared to the bullet's.