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Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 ºC) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.

prasanjeet kumar , 12 Years ago
Grade 8
anser 1 Answers
Saurabh Koranglekar

To determine the typical de Broglie wavelength of a helium (He) atom in helium gas at room temperature, we need to apply the de Broglie wavelength formula and also calculate the mean separation between two atoms under the given conditions. Let's break this down step by step.

Step 1: Formula for de Broglie Wavelength

The de Broglie wavelength is given by the formula:

λ = h / p

Where:

  • λ is the de Broglie wavelength
  • h is Planck's constant = 6.626 × 10⁻³⁴ J·s
  • p is the momentum of the atom = m·v, where m is the mass and v is the velocity of the atom

Step 2: Calculate the Average Velocity of a He Atom

We can use the kinetic theory of gases to estimate the average velocity of a helium atom at room temperature (27°C) and 1 atm pressure. The average kinetic energy of a gas particle is given by:

KE = (3/2) k_B T

Where:

  • k_B is Boltzmann's constant = 1.38 × 10⁻²³ J/K
  • T is the temperature in Kelvin (T = 27°C + 273 = 300 K)
  • KE is the average kinetic energy of one helium atom

The average kinetic energy is also related to the velocity of the atom by:

KE = (1/2) m v²

So, solving for the average velocity:

v = √[(3 k_B T) / m]

The mass of a helium atom is approximately 4.0026 amu, which is equivalent to:

m = 4.0026 × 1.66 × 10⁻²⁷ kg ≈ 6.64 × 10⁻²⁷ kg
Substitute the values for k\_B, T, and m into the equation for v:
v = √[(3 × 1.38 × 10⁻²³ J/K × 300 K) / (6.64 × 10⁻²⁷ kg)] ≈ √[(1.242 × 10⁻²³) / (6.64 × 10⁻²⁷)] ≈ √(1.87 × 10³ m²/s²)

So, the average velocity of the helium atom is approximately:

v ≈ 43.3 m/s

Step 3: Calculate the de Broglie Wavelength

Now that we have the average velocity, we can calculate the de Broglie wavelength:

λ = h / p = h / (m·v)
Substituting the known values for h, m, and v:
λ = (6.626 × 10⁻³⁴ J·s) / (6.64 × 10⁻²⁷ kg × 43.3 m/s) ≈ 2.4 × 10⁻¹⁰ m

So, the typical de Broglie wavelength of a helium atom at room temperature is approximately:

λ ≈ 2.4 × 10⁻¹⁰ m = 0.24 nm

Step 4: Calculate the Mean Separation Between Atoms

To compare the de Broglie wavelength with the mean separation between helium atoms, we can use the ideal gas law, which gives us the number density of atoms:

PV = nRT

Where:

  • P is the pressure (1 atm = 1.013 × 10⁵ Pa)
  • V is the volume
  • n is the number of moles of gas
  • R is the gas constant = 8.314 J/(mol·K)
  • T is the temperature (300 K)

For one mole of helium gas, the volume at 1 atm and 300 K is given by:

V = nRT / P = (1 mol × 8.314 J/(mol·K) × 300 K) / 1.013 × 10⁵ Pa ≈ 0.0245 m³

For a single atom, the number density is:

n_atoms = (1 mol / 6.022 × 10²³ atoms) = 1.66 × 10⁻²⁴ atoms/m³

Finally, the mean separation between atoms is the inverse of the number density:

d = 1 / n_atoms ≈ 1 / (1.66 × 10²⁴ atoms/m³) ≈ 6.0 × 10⁻²⁵ m

Therefore, the mean separation between the helium atoms is approximately:

d ≈ 2.4 × 10⁻¹⁰ m = 0.24 nm

Comparison

The de Broglie wavelength of a helium atom (0.24 nm) is comparable to the mean separation between atoms, also approximately 0.24 nm. This indicates that quantum mechanical effects, such as wave-particle duality, are significant at this scale for helium atoms at room temperature.

Last Activity: 5 Years ago
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