Question

a baseball pitchers fastballs have been clocked at about 94 mph (1 mile = 1609 m). part 1 of 2 calculate the wavelength of a 0.142 kg baseball at this speed. be sure your answer has the correct number of significant digits. part 2 of 2 what is the wavelength of a hydrogen atom at the same speed? be sure your answer has the correct number of significant digits.

Explanation:

Step1: Convert speed from mph to m/s

First, convert 94 mph to m/s. 1 mile = 1609 m and 1 hour = 3600 s. So, $v = 94\times\frac{1609}{3600}\text{ m/s}\approx42.0\text{ m/s}$.

Step2: Calculate de - Broglie wavelength for baseball

Use the de - Broglie wavelength formula $\lambda=\frac{h}{p}$, where $h = 6.63\times10^{- 34}\text{ J}\cdot\text{s}$ is Planck's constant and $p = mv$ is momentum. For a baseball of mass $m = 0.142\text{ kg}$, $p=0.142\text{ kg}\times42.0\text{ m/s}=5.964\text{ kg}\cdot\text{m/s}$. Then $\lambda=\frac{6.63\times10^{-34}\text{ J}\cdot\text{s}}{5.964\text{ kg}\cdot\text{m/s}}\approx1.11\times10^{-34}\text{ m}$.

Step3: Calculate mass of hydrogen atom

The mass of a hydrogen atom $m_{H}=1.67\times10^{-27}\text{ kg}$.

Step4: Calculate de - Broglie wavelength for hydrogen atom

Since the speed is the same as the baseball's speed $v = 42.0\text{ m/s}$, $p_{H}=m_{H}v=1.67\times10^{-27}\text{ kg}\times42.0\text{ m/s}=7.014\times10^{-26}\text{ kg}\cdot\text{m/s}$. Then $\lambda_{H}=\frac{6.63\times10^{-34}\text{ J}\cdot\text{s}}{7.014\times10^{-26}\text{ kg}\cdot\text{m/s}}\approx9.45\times10^{-9}\text{ m}$.

Answer:

The wavelength of the 0.142 - kg baseball is approximately $1.11\times10^{-34}\text{ m}$ and the wavelength of the hydrogen atom at the same speed is approximately $9.45\times10^{-9}\text{ m}$.