1. Understand the de Broglie Wavelength
The de Broglie wavelength (λ) of a particle is related to its momentum (p) by the following equation:
λ = h/p
Where:
* λ is the de Broglie wavelength
* h is Planck's constant (6.626 x 10⁻³⁴ Js)
* p is the momentum
2. Calculate the Momentum
The momentum of a particle is given by:
p = mv
Where:
* m is the mass of the particle
* v is the velocity of the particle
To find the velocity, we'll use the concept of the average kinetic energy of a gas molecule at a given temperature.
3. Calculate Average Kinetic Energy
The average kinetic energy (KE) of a gas molecule is related to the temperature (T) by the following equation:
KE = (3/2)kT
Where:
* k is Boltzmann's constant (1.38 x 10⁻²³ J/K)
* T is the temperature in Kelvin
4. Calculate Velocity
Since kinetic energy is also given by KE = (1/2)mv², we can combine this with the average kinetic energy equation to find the velocity:
(1/2)mv² = (3/2)kT
v² = (3kT)/m
v = √((3kT)/m)
5. Plug in the Values
* Mass of an oxygen molecule (O₂): 32 g/mol = 32 x 10⁻³ kg/mol. We need the mass in kg, so divide by Avogadro's number (6.022 x 10²³ molecules/mol): m ≈ 5.31 x 10⁻²⁶ kg
* Room temperature: 25°C = 298 K
Now, calculate the velocity:
v = √((3 * 1.38 x 10⁻²³ J/K * 298 K) / (5.31 x 10⁻²⁶ kg)) ≈ 482 m/s
6. Calculate the de Broglie Wavelength
Finally, calculate the de Broglie wavelength:
λ = h/p = h/(mv) = (6.626 x 10⁻³⁴ Js) / (5.31 x 10⁻²⁶ kg * 482 m/s) ≈ 2.6 x 10⁻¹¹ m
Conclusion
The typical de Broglie wavelength of an oxygen molecule at room temperature is approximately 2.6 x 10⁻¹¹ meters, which is about 0.26 Angstroms. This wavelength is much smaller than the typical size of an atom, which is on the order of 1 Angstrom.
