Question

A baseball pitcher's fastballs have been clocked at about \(100 \mathrm{mph}\). (a) Calculate the wavelength of a \(0.141-\mathrm{kg}\) baseball (in \(\mathrm{nm}\) ) at this speed. (b) What is the wavelength of a hydrogen atom at the same speed \((1 \mathrm{mile}=1609 \mathrm{~m})\) ?

Short answer

(a) The wavelength is approximately \(1.051 \times 10^{-25} \text{ nm}\). (b) The wavelength is approximately \(8.87 \text{ nm}\).

Step-by-step solution

Convert Speed to m/s

First, convert the speed from miles per hour to meters per second. Since 1 mile equals 1609 meters, \[ 100 \text{ mph} = 100 \times 1609 \text{ m/h} = 160900 \text{ m/h} = \frac{160900}{3600} \text{ m/s} = 44.7 \text{ m/s}. \]

Apply De Broglie Wavelength Formula for Baseball

The De Broglie wavelength formula is \( \lambda = \frac{h}{mv} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ m}^2 \text{ kg/s} \), \( m \) is mass, and \( v \) is velocity.For the baseball, \( m = 0.141 \text{ kg} \) and \( v = 44.7 \text{ m/s} \), so \[ \lambda = \frac{6.626 \times 10^{-34}}{0.141 \times 44.7} = \frac{6.626 \times 10^{-34}}{6.3027} \approx 1.051 \times 10^{-34} \text{ m}. \]

Convert Wavelength to Nanometers for Baseball

Convert the wavelength from meters to nanometers by recalling that \( 1 \text{ nm} = 10^{-9} \text{ m} \). Thus, \[ \lambda_{\text{baseball}} = 1.051 \times 10^{-34} \text{ m} \times 10^{9} \text{ nm/m} \approx 1.051 \times 10^{-25} \text{ nm}. \]

Calculate Wavelength for Hydrogen Atom

Assuming the hydrogen atom has the mass approximately equal to \( 1.67 \times 10^{-27} \text{ kg} \), use the De Broglie formula for mass and speed of this atom:\[ v = 44.7 \text{ m/s} \text{ and } m = 1.67 \times 10^{-27} \text{ kg}, \] so \[ \lambda = \frac{6.626 \times 10^{-34}}{1.67 \times 10^{-27} \times 44.7} = \frac{6.626 \times 10^{-34}}{7.4709 \times 10^{-26}} \approx 8.87 \times 10^{-9} \text{ m}. \]

Convert Wavelength to Nanometers for Hydrogen Atom

Convert the wavelength from meters to nanometers for the hydrogen atom:\[ \lambda_{\text{hydrogen}} = 8.87 \times 10^{-9} \text{ m} \times 10^{9} \text{ nm/m} \approx 8.87 \text{ nm}. \]

Key concepts

Wavelength Calculation

When discussing the de Broglie wavelength, one fundamental task is calculating the wavelength of an object in motion. The de Broglie wavelength \( \lambda \) is given by the formula \( \lambda = \frac{h}{mv} \), where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity.
To solve these types of exercises, like finding the wavelengths for a baseball and hydrogen atom, the formula allows us to see how the microscopic (Planck's constant) influences macroscopic properties (mass and speed).

• A crucial part of this calculation involves determining the velocity in the appropriate units. Once the speed and mass are known, you can plug these values into the de Broglie equation to find the wavelength.
• After this step, it's important to remember to perform unit conversions if necessary, such as converting wavelengths from meters to nanometers for more convenient comparisons.

Grasping how these variables interplay is key to understanding how de Broglie wavelengths illuminate the wave-particle duality of matter.

Speed Conversion

Speed conversion is all about making sure the velocity used in calculations is in the right units. In physics problems, speeds are often provided in units like miles per hour (mph), but in scientific calculations, we commonly need them in meters per second (m/s).
To convert from mph to m/s:

• First, convert miles to meters by multiplying by 1609 (since 1 mile = 1609 meters)
• Then, convert hours to seconds by dividing by 3600 (since 1 hour = 3600 seconds)

For example, a speed of 100 mph becomes 44.7 m/s after conversion. This consistent unit in calculations ensures that every measurement aligns correctly, preventing errors in equations like the de Broglie wavelength.

Planck's Constant

Planck's constant is a fundamental constant in quantum mechanics that relates the energy of a photon to its frequency. It is denoted by \( h \) and has a value of \( 6.626 \times 10^{-34} \) m²kg/s.
This tiny constant is key in calculations involving the de Broglie wavelength, emphasizing how quantum effects are typically significant only at the atomic and subatomic levels. Because Planck's constant is so small, it highlights why macroscopic objects like a baseball have imperceptibly tiny wavelengths compared to microscopic particles like a hydrogen atom. This distinction helps students appreciate the quantum behavior of particles, guiding our understanding from macroscopic observations to quantum realities.

Mass of Particles

The mass of particles is a central element in determining their de Broglie wavelength, as it appears in the equation \( \lambda = \frac{h}{mv} \). Different particles, ranging from large objects like a baseball to tiny ones like a hydrogen atom, have vastly different masses.
These differences significantly affect the eventual wavelength outcome.

• For instance, the mass of a baseball is about 0.141 kg, whereas a hydrogen atom's mass is roughly \( 1.67 \times 10^{-27} \) kg.
• This variation in mass is why a baseball's de Broglie wavelength is extremely tiny compared to that of a hydrogen atom at the same speed.

Understanding the importance of mass in these calculations helps to delineate between the classical and quantum mechanical worlds, offering insights into how particle masses impact phenomena like wavelength and, by extension, wave properties.