What is the de Broglie wavelength

What is the de Broglie wavelength, in centimeters, of a 12.4-g hummingbird flying at 1.20 × 102 mph? (1 mile = 1.61 km)

What is the de Broglie wavelength (in nm) associated with a 2.5-g Ping-Pong ball traveling 35 mph?

The Correct Answer and Explanation is :

1. de Broglie Wavelength of a Hummingbird:

The de Broglie wavelength (( \lambda )) is given by the equation:

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

Where:

  • ( h ) is Planck’s constant: ( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} )
  • ( m ) is the mass of the object
  • ( v ) is the velocity of the object

Step 1: Convert mass and velocity to SI units

  • Mass of hummingbird: ( 12.4 \, \text{g} = 12.4 \times 10^{-3} \, \text{kg} )
  • Velocity of hummingbird: ( 1.20 \times 10^2 \, \text{mph} )

First, convert the speed from miles per hour (mph) to meters per second (m/s):

[
1 \, \text{mile} = 1.61 \, \text{km} = 1.61 \times 10^3 \, \text{m}
]
[
1 \, \text{hour} = 3600 \, \text{seconds}
]

Thus:

[
v = 1.20 \times 10^2 \, \text{mph} \times \frac{1.61 \times 10^3 \, \text{m}}{1 \, \text{mile}} \times \frac{1}{3600 \, \text{seconds/hour}}
]
[
v \approx 53.6 \, \text{m/s}
]

Step 2: Calculate the de Broglie wavelength

Now we can calculate the de Broglie wavelength:

[
\lambda = \frac{6.626 \times 10^{-34}}{(12.4 \times 10^{-3})(53.6)}
]

[
\lambda \approx 1.04 \times 10^{-34} \, \text{m}
]

Convert to centimeters:

[
\lambda \approx 1.04 \times 10^{-32} \, \text{cm}
]

2. de Broglie Wavelength of a Ping-Pong Ball:

Step 1: Convert mass and velocity to SI units

  • Mass of Ping-Pong ball: ( 2.5 \, \text{g} = 2.5 \times 10^{-3} \, \text{kg} )
  • Velocity: ( 35 \, \text{mph} )

Convert the velocity to meters per second:

[
v = 35 \, \text{mph} \times \frac{1.61 \times 10^3 \, \text{m}}{1 \, \text{mile}} \times \frac{1}{3600 \, \text{seconds/hour}}
]
[
v \approx 15.6 \, \text{m/s}
]

Step 2: Calculate the de Broglie wavelength

[
\lambda = \frac{6.626 \times 10^{-34}}{(2.5 \times 10^{-3})(15.6)}
]

[
\lambda \approx 1.69 \times 10^{-32} \, \text{m}
]

Convert to nanometers (nm):

[
\lambda \approx 1.69 \times 10^{-23} \, \text{nm}
]

Explanation:

The de Broglie wavelength is a concept from quantum mechanics, which links the wave-like properties of a particle to its momentum. For macroscopic objects like a hummingbird or a Ping-Pong ball, the de Broglie wavelength is extremely small, making their wave-like behavior undetectable in practice. The wavelength is inversely proportional to both the mass and velocity of the object. Since the mass of the objects in these examples is much larger than subatomic particles like electrons, the resulting wavelengths are very tiny.

Even though the hummingbird and Ping-Pong ball are in motion, their wavelengths are so small that they cannot exhibit noticeable wave-like behavior. For comparison, quantum effects are most noticeable in objects of much smaller mass, such as electrons or atoms, where the wavelength becomes significant. This demonstrates how the de Broglie wavelength, though fundamental, has negligible consequences for everyday objects.

Scroll to Top