Question

louis de broglie
compare the de broglie wavelength of a baseball moving at 90.0 miles per hour (40.2 m/s) to that of an alpha particle moving at 3.40×10^7 miles per hour (1.52×10^7 m/s) and a proton with a speed of 1.30×10^7 miles per hour (5.81×10^6 m/s).
which region of the electromagnetic spectrum are each of these wavelengths near?
use the references to access important values if needed for this question.

Explanation:

Step1: Recall de Broglie wavelength formula

The de Broglie wavelength formula is $\lambda=\frac{h}{mv}$, where $h = 6.626\times10^{-34}\text{ J}\cdot\text{s}$ (Planck's constant), $m$ is the mass of the particle, and $v$ is the velocity of the particle.

Step2: Calculate wavelength for baseball

For the baseball: $m = 0.140\text{ kg}$, $v=40.2\text{ m/s}$.
$\lambda=\frac{6.626\times 10^{-34}}{0.140\times40.2}=\frac{6.626\times 10^{-34}}{5.628}\approx1.18\times10^{-34}\text{ m}$. This is an extremely small wavelength, much smaller than any atomic - scale or sub - atomic scale lengths.

Step3: Calculate wavelength for proton

For the proton: $m = 1.67\times10^{-27}\text{ kg}$, $v = 5.81\times10^{6}\text{ m/s}$.
$\lambda=\frac{6.626\times 10^{-34}}{1.67\times10^{-27}\times5.81\times10^{6}}=\frac{6.626\times 10^{-34}}{9.6927\times10^{-21}}\approx6.84\times10^{-14}\text{ m}$. This wavelength is on the order of nuclear dimensions (nuclear radii are typically in the range of $10^{-15}-10^{-14}\text{ m}$).

Step4: Calculate wavelength for alpha particle

For the alpha particle: $m = 6.64\times10^{-27}\text{ kg}$, $v = 1.52\times10^{7}\text{ m/s}$.
$\lambda=\frac{6.626\times 10^{-34}}{6.64\times10^{-27}\times1.52\times10^{7}}=\frac{6.626\times 10^{-34}}{1.00928\times10^{-19}}\approx6.57\times10^{-15}\text{ m}$. This wavelength is also on the order of nuclear dimensions.

Answer:

Baseball: $1.18\times 10^{-34}\text{ m}$ (not in the electromagnetic spectrum, non - observable wave - like behavior on the macroscopic scale). Proton: $6.84\times 10^{-14}\text{ m}$ (nuclear region). Alpha particle: $6.57\times 10^{-15}\text{ m}$ (nuclear region).