Questions: A baseball pitcher's fastballs have been clocked at about 94 mph (1 mile = 1609 m). Part 1 of 2 Calculate the wavelength of a 0.145 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.

Solution Steps

First, we need to convert the speed of the baseball from miles per hour (mph) to meters per second (m/s). Given that \(1 \text{ mile} = 1609 \text{ m}\) and \(1 \text{ hour} = 3600 \text{ seconds}\), we can convert \(94 \text{ mph}\) as follows:

\[ 94 \text{ mph} = 94 \times \frac{1609 \text{ m}}{1 \text{ mile}} \times \frac{1 \text{ hour}}{3600 \text{ s}} = 42.0111 \text{ m/s} \]

Using the de Broglie wavelength formula:

\[ \lambda = \frac{h}{mv} \]

where \(h = 6.626 \times 10^{-34} \text{ m}^2 \text{ kg/s}\) is Planck's constant, \(m = 0.145 \text{ kg}\) is the mass of the baseball, and \(v = 42.0111 \text{ m/s}\) is the velocity.

\[ \lambda = \frac{6.626 \times 10^{-34}}{0.145 \times 42.0111} = 1.086 \times 10^{-34} \text{ m} \]

Assuming the mass of a hydrogen atom is approximately \(1.67 \times 10^{-27} \text{ kg}\), we use the same de Broglie wavelength formula:

\[ \lambda = \frac{h}{mv} \]

where \(m = 1.67 \times 10^{-27} \text{ kg}\) and \(v = 42.0111 \text{ m/s}\).

\[ \lambda = \frac{6.626 \times 10^{-34}}{1.67 \times 10^{-27} \times 42.0111} = 9.438 \times 10^{-12} \text{ m} \]

Final Answer