Question

a baseball pitchers fastballs have been clocked at about 91 mph (1 mile = 1609 m). part 1 of 2 calculate the wavelength of a 0.149 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 to m/s

Given speed $v = 91\ mph$. Since $1\ mile=1609\ m$ and $1\ hour = 3600\ s$, then $v=91\times\frac{1609}{3600}\ m/s\approx40.7\ m/s$.

Step2: Recall de - Broglie wavelength formula

The de - Broglie wavelength formula is $\lambda=\frac{h}{mv}$, where $h = 6.63\times10^{- 34}\ J\cdot s$ is Planck's constant, $m$ is the mass of the object, and $v$ is the speed.

Step3: Calculate wavelength of baseball

For the baseball with $m = 0.149\ kg$, $\lambda_{baseball}=\frac{h}{mv}=\frac{6.63\times10^{-34}}{0.149\times40.7}\ m\approx1.08\times10^{-34}\ m$.

Step4: Calculate mass of hydrogen atom

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

Step5: Calculate wavelength of hydrogen atom

Using the de - Broglie formula again with the same speed $v\approx40.7\ m/s$, $\lambda_{H}=\frac{h}{m_{H}v}=\frac{6.63\times10^{-34}}{1.67\times10^{-27}\times40.7}\ m\approx9.77\times10^{-9}\ m$.

Answer:

The wavelength of the $0.149\ kg$ baseball is approximately $1.08\times10^{-34}\ m$ and the wavelength of the hydrogen atom is approximately $9.77\times10^{-9}\ m$.